text:
The Normal Distribution

# Using the Normal Distribution

OpenStaxCollege

[latexpage]

The shaded area in the following graph indicates the area to the left of x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x).

The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(Xx) and P(X > x) is the same as P(Xx) for continuous distributions.

# Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.

NOTE

To calculate the probability, use the probability tables provided in [link] without the use of technology. The tables include instructions for how to use them.

If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.

Try It

If the area to the left of x is 0.012, then what is the area to the right?

1 − 0.012 = 0.988

The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

a. Find the probability that a randomly selected student scored more than 65 on the exam.

a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5

Draw a graph.

Then, find P(x > 65).

P(x > 65) = 0.3446

The probability that any student selected at random scores more than 65 is 0.3446.

Go into 2nd DISTR.

After pressing 2nd DISTR, press 2:normalcdf.

The syntax for the instructions are as follows:

normalcdf(lower value, upper value, mean, standard deviation)
For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 1099) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can enter 10^99 instead. The number 1099 is way out in the right tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve.

Historical Note

The TI probability program calculates a z-score and then the probability from the z-score. Before technology, the z-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the z-score was used. You calculate the z-score and look up the area to the left. The probability is the area to the right.

z = $$\frac{65\text{ – 63}}{5}$$ = 0.4

Area to the left is 0.6554.

P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446

Calculate the z-score:

*Press 2nd Distr

*Press 3:invNorm(

*Enter the area to the left of z followed by )

*Press ENTER.

For this Example, the steps are

2nd Distr

3:invNorm(.6554) ENTER

The answer is 0.3999 which rounds to 0.4.

b. Find the probability that a randomly selected student scored less than 85.

b. Draw a graph.

Then find P(x < 85), and shade the graph.

Using a computer or calculator, find P(x < 85) = 1.

normalcdf(0,85,63,5) = 1 (rounds to one)

The probability that one student scores less than 85 is approximately one (or 100%).

c. Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).

c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile.

Let k = the 90th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The variable k is often called a critical value.

k = 69.4

The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:

invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)

For this problem, invNorm(0.90,63,5) = 69.4

d. Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).

d. Find the 70th percentile.

Draw a new graph and label it appropriately. k = 65.6

The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.

invNorm(0.70,63,5) = 65.6

Try It

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.

Find the probability that a randomly selected golfer scored less than 65.

normalcdf(1099,65,68,3) = 0.1587

A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.

a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.

a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5.

Find P(1.8 < x < 2.75).

The probability for which you are looking is the area between x = 1.8 and x = 2.75. P(1.8 < x < 2.75) = 0.5886

normalcdf(1.8,2.75,2,0.5) = 0.5886

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.

b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, k, where P(x < k) = 0.25.

invNorm(0.25,2,0.5) = 1.66

The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

Try It

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.

normalcdf(66,70,68,3) = 0.4950

There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.

a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.

a. normalcdf(23,64.7,36.9,13.9) = 0.8186

b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.

b. normalcdf(–1099,50.8,36.9,13.9) = 0.8413

c. Find the 80th percentile of this distribution, and interpret it in a complete sentence.

c.

invNorm(0.80,36.9,13.9) = 48.6
The 80th percentile is 48.6 years.
80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.
Try It

1. Find the 30th percentile, and interpret it in a complete sentence.
2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old.

Let X = a smart phone user whose age is 13 to 55+. X ~ N(36.9, 13.9)

1. To find the 30th percentile, find k such that P(x < k) = 0.30.

invNorm(0.30, 36.9, 13.9) = 29.6 years

Thirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years.

2. Find P(x < 27)

normalcdf(0,27,36.9,13.9) = 0.2342

(Note that normalcdf(–1099,27,36.9,13.9) = 0.2382. The two answers differ only by 0.0040.)

There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).

a. Calculate the interquartile range (IQR).

a.

IQR = Q3Q1
Calculate Q3 = 75th percentile and Q1 = 25th percentile.
invNorm(0.75,36.9,13.9) = Q3 = 46.2754
invNorm(0.25,36.9,13.9) = Q1 = 27.5246
IQR = Q3Q1 = 18.7508

b. Forty percent of the ages that range from 13 to 55+ are at least what age?

b.

Find k where P(x > k) = 0.40 (“At least” translates to “greater than or equal to.”)
0.40 = the area to the right.
Area to the left = 1 – 0.40 = 0.60.
The area to the left of k = 0.60.
invNorm(0.60,36.9,13.9) = 40.4215.
k = 40.42.
Forty percent of the ages that range from 13 to 55+ are at least 40.42 years.
Try It

Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = 15 points.

1. Calculate the first- and third-quartile scores for this exam.
2. The middle 50% of the exam scores are between what two values?
1. Q1 = 25th percentile = invNorm(0.25,81,15) = 70.9

Q3 = 75th percentile = invNorm(0.75,81,15) = 91.9

2. The middle 50% of the scores are between 70.9 and 91.1.

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.

a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.

a. normalcdf(6,10^99,5.85,0.24) = 0.2660

b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.

b.

1 – 0.20 = 0.80
The tails of the graph of the normal distribution each have an area of 0.40.
Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60).
k1 = invNorm(0.40,5.85,0.24) = 5.79 cm
k2 = invNorm(0.60,5.85,0.24) = 5.91 cm

c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.

c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.

Try It

1. The middle 45% of mandarin oranges from this farm are between ______ and ______.
2. Find the 16th percentile and interpret it in a complete sentence.
1. The middle area = 0.40, so each tail has an area of 0.30.

1 – 0.40 = 0.60

The tails of the graph of the normal distribution each have an area of 0.30.

Find k1, the 30th percentile and k2, the 70th percentile (0.40 + 0.30 = 0.70).

k1 = invNorm(0.30,5.85,0.24) = 5.72 cm

k2 = invNorm(0.70,5.85,0.24) = 5.98 cm

2. normalcdf(5,1099,5.85,0.24) = 0.9998

# References

“Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele’s_rule (accessed May 14, 2013).

“403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013).

“Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013).

“Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).

# Chapter Review

The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ. A special normal distribution, called the standard normal distribution is the distribution of z-scores. Its mean is zero, and its standard deviation is one.

# Formula Review

Normal Distribution: X ~ N(µ, σ) where µ is the mean and σ is the standard deviation.

Standard Normal Distribution: Z ~ N(0, 1).

Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation)

Calculator function for the kth percentile: k = invNorm (area to the left of k, mean, standard deviation)

How would you represent the area to the left of one in a probability statement?

P(x < 1)

What is the area to the right of one?

<!– <solution id=”fs-idp151140256″>
1 – P(x < 1) or P(x < 1)
–>

Is P(x < 1) equal to P(x ≤ 1)? Why?

Yes, because they are the same in a continuous distribution: P(x = 1) = 0

How would you represent the area to the left of three in a probability statement?

<!– <solution id=”fs-idp146985792″>
P(x < 3)
–>

What is the area to the right of three?

1 – P(x < 3) or P(x > 3)

If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x?

<!– <solution id=”fs-idm19166096″>
1 – 0.123 = 0.877 –>

If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x?

1 – 0.543 = 0.457

Use the following information to answer the next four exercises:

X ~ N(54, 8)

Find the probability that x > 56.

<!– <solution id=”fs-idp74510448″>
0.4013
–>

Find the probability that x < 30.

0.0013

Find the 80th percentile.

<!– <solution id=”fs-idp144265408″>
60.73
–>

Find the 60th percentile.

56.03

X ~ N(6, 2)

Find the probability that x is between three and nine.

<!– <solution id=”fs-idp53207040″>
0.8664
–>

X ~ N(–3, 4)

Find the probability that x is between one and four.

0.1186

X ~ N(4, 5)

Find the maximum of x in the bottom quartile.

<!– <solution id=”fs-idp30677232″>
0.6276
–>

Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.

1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P(0 < x < ____________) = ___________ (Use zero for the minimum value of x.)
1. Check student’s solution.
2. 3, 0.1979

Find the probability that a CD player will last between 2.8 and six years.

1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P(__________ < x < __________) = __________

<!– <solution id=”id44443502″>

Check student’s solution
2.8, 6, 0.7694
–>

Find the 70th percentile of the distribution for the time a CD player lasts.

1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70%.
2. P(x < k) = __________ Therefore, k = _________
1. Check student’s solution.
2. 0.70, 4.78 years

# Homework

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

What is the probability of spending more than two days in recovery?

1. 0.0580
2. 0.8447
3. 0.0553
4. 0.9420

<!– <solution id=”id12741946″>
d
–>

The 90th percentile for recovery times is?

1. 8.89
2. 7.07
3. 7.99
4. 4.32

c

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space?

1. Yes
2. No
3. Unable to determine

<!– <solution id=”id15290500″>
a
–>

Find the probability that it takes at least eight minutes to find a parking space.

1. 0.0001
2. 0.9270
3. 0.1862
4. 0.0668

d

Seventy percent of the time, it takes more than how many minutes to find a parking space?

1. 1.24
2. 2.41
3. 3.95
4. 6.05

<!– <solution id=”id16763246″>
c
–>

According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.

1. X ~ _____(_____,_____)
2. Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement.
3. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not,
4. The middle 40% of heights fall between what two values? Sketch the graph, and write the
probability statement.
1. X ~ N(66, 2.5)
2. 0.5404
3. No, the probability that an Asian male is over 72 inches tall is 0.0082

IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual.

1. X ~ _____(_____,_____)
2. Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement.
3. MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement.
4. The middle 50% of IQs fall between what two values? Sketch the graph and write the probability statement.

<!– <solution id=”eip-idp25493760″>
N(100, 15)
The probability that a person has an IQ greater than 120 is 0.0918.
A person has to have an IQ over 130 to qualify for MENSA.
The middle 50% of IQ scores falls between 89.95 and 110.05.
–>

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories.

1. X ~ _____(_____,_____)
2. Find the probability that the percent of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined.
3. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement.
1. X ~ N(36, 10)
2. The probability that a person consumes more than 40% of their calories as fat is 0.3446.
3. Approximately 25% of people consume less than 29.26% of their calories as fat.

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet.

1. If X = distance in feet for a fly ball, then X ~ _____(_____,_____)
2. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability.
3. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.

<!– <solution id=”eip-idp105794048″>
X ~ N(250, 50)
The probability that a fly ball travels less than 220 feet is 0.2743.
Eighty percent of the fly balls will travel less than 292 feet.
–>

In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.

1. In words, define the random variable X.
2. X ~ _____(_____,_____)
3. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.
4. What percent of the children spend over ten hours per day unsupervised?
5. Seventy percent of the children spend at least how long per day unsupervised?
1. X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
2. X ~ N(3, 1.5)
3. The probability that the child spends less than one hour a day unsupervised is 0.0918.
4. The probability that a child spends over ten hours a day unsupervised is less than 0.0001.
5. 2.21 hours

In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district.

1. State the approximate distribution of X.
2. Is 1,956.8 a population mean or a sample mean? How do you know?
3. Find the probability that a randomly selected district had fewer than 1,600 votes for President Clinton. Sketch the graph and write the probability statement.
4. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton.
5. Find the third quartile for votes for President Clinton.

<!– <solution id=”eip-idp28319056″>
X ~ N(1956.8, 572.3)
This is a population mean, because all election districts are included.
The probability that a district had less than 1,600 votes for President Clinton is 0.2676.
0.3798
Seventy-five percent of the districts had fewer than 2,340 votes for President Clinton.
–>

Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days.

1. In words, define the random variable X.
2. X ~ _____(_____,_____)
3. If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement.
4. Sixty percent of all trials of this type are completed within how many days?
1. X = the distribution of the number of days a particular type of criminal trial will take
2. X ~ N(21, 7)
3. The probability that a randomly selected trial will last more than 24 days is 0.3336.
4. 22.77

Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps.

1. In words, define the random variable X.
2. X ~ _____(_____,_____)
3. Find the percent of her laps that are completed in less than 130 seconds.
4. The fastest 3% of her laps are under _____.
5. The middle 80% of her laps are from _______ seconds to _______ seconds.

<!– <solution id=”eip-idp63928720″>
X = the distribution of race times that Terry Vogel produces
X ~ N(129.71, 2.28)
Terri completes 55.17% of her laps in less than 130 seconds.
Terri completes 55.17% of her laps in less than 130 seconds.
124.4 and 135.02
–>

Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. [link] displays the ordered real data (in minutes):

 0.5 4.25 5 6 7.25 1.75 4.25 5.25 6 7.25 2 4.25 5.25 6.25 7.25 2.25 4.25 5.5 6.25 7.75 2.25 4.5 5.5 6.5 8 2.5 4.75 5.5 6.5 8.25 2.75 4.75 5.75 6.5 9.5 3.25 4.75 5.75 6.75 9.5 3.75 5 6 6.75 9.75 3.75 5 6 6.75 10.75
Calculate the sample mean and the sample standard deviation.
Construct a histogram.
Draw a smooth curve through the midpoints of the tops of the bars.
In words, describe the shape of your histogram and smooth curve.
Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____(_____,_____)
Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes.
Determine the cumulative relative frequency for waiting less than 6.1 minutes.
Why aren’t the answers to part f and part g exactly the same?
Why are the answers to part f and part g as close as they are?
If only ten customers has been surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion.
1. mean = 5.51, s = 2.15
2. Check student’s solution.
3. Check student’s solution.
4. Check student’s solution.
5. X ~ N(5.51, 2.15)
6. 0.6029
7. The cumulative frequency for less than 6.1 minutes is 0.64.
8. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.
9. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.
10. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.

Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false.

1. Ricardo’s actual GPA is lower than Anita’s actual GPA.
2. Ricardo is not passing because his z-score is zero.
3. Anita is in the 70th percentile of students at her college.

<!– <solution id=”eip-idp7493680″>
If the average GPA is less at Anita’s school than it is at Ricardo’s, then Ricardo’s actual score could be higher.
Passing can be defined differently at different schools. Also, since Ricardo’s z-score is 0, his GPA is actually
the average for his school, which is typically a passing GPA.
Anita’s percentile is higher than the 70th percentile.
–>

[link] shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums. The table does not include horse-racing or motor-racing stadiums.

 40,000 40,000 45,050 45,500 46,249 48,134 49,133 50,071 50,096 50,466 50,832 51,100 51,500 51,900 52,000 52,132 52,200 52,530 52,692 53,864 54,000 55,000 55,000 55,000 55,000 55,000 55,000 55,082 57,000 58,008 59,680 60,000 60,000 60,492 60,580 62,380 62,872 64,035 65,000 65,050 65,647 66,000 66,161 67,428 68,349 68,976 69,372 70,107 70,585 71,594 72,000 72,922 73,379 74,500 75,025 76,212 78,000 80,000 80,000 82,300
Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums (the data).
Construct a histogram.
Draw a smooth curve through the midpoints of the tops of the bars of the histogram.
In words, describe the shape of your histogram and smooth curve.
Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____(_____,_____).
Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators.
Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint: Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample.
Why aren’t the answers to part f and part g exactly the same?
1. mean = 60,136

s = 10,468

5. X ~ N(60136, 10468)
6. 0.7440
7. The cumulative relative frequency is 43/60 = 0.717.
8. The answers for part f and part g are not the same, because the normal distribution is only an
approximation.

An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability.

<!– <solution id=”eip-102″>
For x = 240,

X−μ

σ

=

240−280

13

=−3.0769

For x = 306,

306−280

13

=2

P(240 < x < 306) = P(–3.0769 < z < 2) = normalcdf(–3.0769,2,0,1) = 0.9762.
According to the scenario given, this means that there is a 97.62% chance that he is not the father.
To answer the second part of the question, there is a 1 – 0.9762 = 0.0238 = 2.38% chance that he is the father. –>

A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.

• n = 100; p = 0.1; q = 0.9
• μ = np = (100)(0.10) = 10
• σ = $$\sqrt{npq}$$ = $$\sqrt{\text{(100)(0}\text{.1)(0}\text{.9)}}$$ = 3
1. z = ±1: x1 = µ + = 10 + 1(3) = 13 and x2 = µ = 10 – 1(3) = 7. 68% of the defective cars will fall between seven and 13.
2. z = ±2: x1 = µ + = 10 + 2(3) = 16 and x2 = µ = 10 – 2(3) = 4. 95 % of the defective cars will fall between four and 16
3. z = ±3: x1 = µ + = 10 + 3(3) = 19 and x2 = µ = 10 – 3(3) = 1. 99.7% of the defective cars will fall between one and 19.

We flip a coin 100 times (n = 100) and note that it only comes up heads 20% (p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following:

1. There is about a 68% chance that the number of heads will be somewhere between ___ and ___.
2. There is about a ____chance that the number of heads will be somewhere between 12 and 28.
3. There is about a ____ chance that the number of heads will be somewhere between eight and 32.

<!– <solution id=”eip-171″>
There is about a 68% chance that the number of heads will be somewhere between 16 and 24. z = ±1: x1 = µ + zσ = 20 + 1(4) = 24 and x2 = µ-zσ = 20 – 1(4) = 16.
There is about a 95% chance that the number of heads will be somewhere between 12 and 28. For this problem: normalcdf(12,28,20,4) = 0.9545 = 95.45%
There is about a 99.73% chance that the number of heads will be somewhere between eight and 32. For this problem: normalcdf(8,32,20,4) = 0.9973 = 99.73%.
–>

A 💲1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are

1. somewhere between 34 and 54 prizes.
2. somewhere between 54 and 64 prizes.
3. more than 64 prizes.
• n = 190; p = $$\frac{1}{5}$$
= 0.2; q = 0.8
• μ = np = (190)(0.2) = 38
• σ = $$\sqrt{npq}$$ = $$\sqrt{\text{(190)(0}\text{.2)(0}\text{.8)}}$$ = 5.5136
1. For this problem: P(34 < x < 54) = normalcdf(34,54,48,5.5136) = 0.7641
2. For this problem: P(54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018
3. For this problem: P(x > 64) = normalcdf(64,1099,48,5.5136) = 0.0000012 (approximately 0)

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.

On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.

1. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
2. Find the 95th percentile, and express it in a sentence.

<!– <solution id=”eip-11″>
X = the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning.
X ~ N(28, 5)
P(x ≥ 30) = 0.3446; normalcdf(30,1EE99,28,5) = 0.3446

invNorm(0.95,0.28,0.05) = 0.3622.95% of the percent of 18 to 34 year olds who check Facebook before getting out of bed in the morning is at most 36.22%.
P(25 < x < 55).
P(25 < x < 55) = normalcdf(25,55,28,5) = 0.7257(0.7257)(400) = 290.28
–>


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# HW#3 Create a spreadsheet to solve the following problems. I want to see the formulas (FV, PV, PMT,.

HW#3 Create a spreadsheet to solve the following problems. I want to see the formulas (FV, PV, PMT, NPV, IRR, NPER, and EXP where appropriate). Each student must submit his or her own unique file. Please number your answers clearly in bold or highlighted. Future Value (10 points) 1. If I won a lump-sum lottery today which gave me $45,000 and I decided to put it into a 5 year investment earning an annual rate of 4%, how much would it be worth at the end of 5 years? 2. What if I decided to put my winnings in the same investment for 10 years? Present Value (10 points) 3. If I won a lump-sum lottery which promises to give me Document Preview: HW#3 Create a spreadsheet to solve the following problems. I want to see the formulas (FV, PV, PMT, NPV, IRR, NPER, and EXP where appropriate). Each student must submit his or her own unique file. Please number your answers clearly in bold or highlighted. Future Value (10 points) 1. If I won a lump-sum lottery today which gave me$45,000 and I decided to put it into a 5 year investment earning an annual rate of 4%, how much would it be worth at the end of 5 years? 2. What if I decided to put my winnings in the same investment for 10 years? Present Value (10 points) 3. If I won a lump-sum lottery which promises to give me $45,000 5 years from now and the annual interest rate is 2%, how much would it be worth today (present value)? 4. What if the rate is 4%? Annuities (10 points) 5. What is the present value of a car payment of$299 per month for 5 years with a 3.50% annual rate? (In other words, how much cash would it take today to buy the car outright instead of making monthly payments?) 6. What is the future value of a payment of $299 per month for 5 years with a 3.50% annual rate? (In other words, if you had put the monthly payments in the bank earning 3.50%, how much would it be worth at the end of 5 years?) Annuity Payment and Number of Periods (20 points) 7. If I bought a house for$350,000 and mortgaged 80% of the cost using a 30-year loan at a fixed annual rate of 4.0%, what would my annual mortgage payment be? 8. What would my monthly mortgage payment be? 9. If you double your monthly mortgage payments from above, how long will it take to pay off your mortgage in months? 10. Why does doubling your monthly mortgage payment reduce the length of the loan by more than half? Deferred Annuities (25 points) Suppose your 7 year old daughter has announced that she would like to attend college. You believe it will cost \$22,000 per year, for 4 years, for her to attend state college 11 years from now. You are giving yourself 11...

Attachments:

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# Normalization of integral bounds

Say we have a difficult integral of the form ##\displaystyle \int_a^{b}f(x) ~dx##. Let ##t = \frac{x-a}{b-x}##. Then ##\displaystyle \int_0^{\infty}f \left( \frac{bt+a}{t+1} \right)\frac{1-a}{(t+1)^2} ~dt##. My idea is that making this change of variables transforms the integral into a form where we could potentially use the gamma function, Laplace transform, etc, to evaluate the integral. In practice is this not the case, and the integral just ends up getting messier?


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# Math Help - questions about interest rates

1. ## questions about interest rates

How much more interest will be earned if $2,000 is invested for 4 years at 8% compounded continuously, instead? How much more interest will be earned if$2,000 is invested for 4 years at 8% compounded continuously, instead of at 8% compounded quarterly?

A parent wants to have $400,000 available for her child's education. Find the amount that would have to? a parent wants to have$400,000 available for her child's education. Find the amount that would have to

Suppose an individual wants to have \$200,000 available for her child's education. Find the amount that would have to be invested at 12%, compounded continuously, if the number of years until college is:
1 ) 6 years
2 ) 15 years
3) Does leaving the money invested twice as long mean that only half as much is needed initially? Explain why or why not.

thank you.

2. The first is an excellent exploration.

Try this: $2000*\left(1\;+\;\frac{0.08}{m}\right)^{4*m}$

What is it for m = 1? Annually?
What is it for m = 2? Semi-Annually?
What is it for m = 4? Quarterly?
What is it for m = 12? Monthly?
What is it for m = 52? Weekly?
What is it for m = 365? Daily?
What is it for m = 365*24? Hourly?

Is there a limit to the values you are getting? Let's remember a little calculus and track it down.


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Cell Transport and Homeostasis | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Biology Workbook Go to the latest version.

# 3.3: Cell Transport and Homeostasis

Created by: CK-12

## Lesson 3.3: True or False

Name___________________ Class______________ Date________

Write true if the statement is true or false if the statement is false.

_____ 1. Passive transport needs energy.

_____ 2. Active transport needs energy.

_____ 3. Carrier proteins change shape when they transport substances.

_____ 4. Diffusion does not require any help from other molecules.

_____ 5. Facilitated diffusion does not require any help from other molecules.

_____ 6. Endocytosis removes large molecules from the cell.

_____ 7. In diffusion, substances move from an area of lower concentration to an area of higher concentration.

_____ 8. The sodium-potassium pump is a type of channel protein.

_____ 9. Ions can easily flow through a carrier protein.

_____ 10. Diffusion is the osmosis of water.

_____ 11. Endocytosis and exocytosis are types of vesicle transport.

_____ 12. Channel proteins form small “holes” in the plasma membrane.

_____ 13. Transport of substances across the cell membrane helps maintain homeostasis by keeping the cell's conditions within normal ranges.

_____ 14. Channel proteins and carrier proteins are both transport proteins.

_____ 15. The plasma membrane controls what enters and leaves the cell.

Name___________________ Class______________ Date________

Read these passages from the text and answer the questions that follow.

Passive Transport

Passive transport occurs when substances cross the plasma membrane without any input of energy from the cell. No energy is needed because the substances are moving from an area where they have a higher concentration to an area where they have a lower concentration. Concentration refers to the number of particles of a substance per unit of volume. The more particles of a substance in a given volume, the higher the concentration. A substance always moves from an area where it is more concentrated to an area where it is less concentrated. It’s a little like a ball rolling down a hill. It goes by itself without any input of extra energy.

Simple Diffusion

Diffusion is the movement of a substance across a membrane, due to a difference in concentration, without any help from other molecules. The substance simply moves from the side of the membrane where it is more concentrated to the side where it is less concentrated. Substances that can squeeze between the lipid molecules in the plasma membrane by simple diffusion are generally very small, hydrophobic molecules, such as molecules of oxygen and carbon dioxide.

Osmosis

Osmosis is a special type of diffusion — the diffusion of water molecules across a membrane. Like other molecules, water moves from an area of higher concentration to an area of lower concentration. Water moves in or out of a cell until its concentration is the same on both sides of the plasma membrane.

Facilitated Diffusion

Water and many other substances cannot simply diffuse across a membrane. Hydrophilic molecules, charged ions, and relatively large molecules, such as glucose, all need help with diffusion. The help comes from special proteins in the membrane known as transport proteins. Diffusion with the help of transport proteins is called facilitated diffusion. There are several types of transport proteins, including channel proteins and carrier proteins.

• Channel proteins form pores, or tiny holes, in the membrane. This allows water molecules and small ions to pass through the membrane without coming into contact with the hydrophobic tails of the lipid molecules in the interior of the membrane.
• Carrier proteins bind with specific ions or molecules, and in doing so, they change shape. As carrier proteins change shape, they carry the ions or molecules across the membrane.

Questions

1. Explain why passive transport does not require energy.

$\\\\\\\\\\$

2. What is a main difference between diffusion and facilitated diffusion?

$\\\\\\\\\\$

3. Describe how simple diffusion proceeds. What kind of molecules can move across the membrane by simple diffusion?

$\\\\\\\\\\$

4. How is water transported across the membrane?

$\\\\\\\\\\$

5. What are the two types of transport proteins? Describe how they function.

$\\\\\\\\\\$

## Lesson 3.3: Multiple Choice

Name___________________ Class______________ Date________

Circle the letter of the correct choice.

1. Controlling what enters and leaves the cell in an important function of the
1. nucleus.
2. vesicle.
3. plasma membrane.
4. Golgi apparatus.
2. During diffusion, substances move from an area of ____________ concentration to an area of ____________ concentration.
1. higher, lower
2. lower, higher
3. higher, equal
4. lower, equal
3. A channel protein does which of the following?
1. Carries ions or molecules across the membrane.
2. Forms tiny holes in the membrane.
3. Changes shape as it transports molecules.
4. all of the above
4. The sodium-potassium pump
1. uses energy to move sodium ions out of the cell and potassium ions into the cell.
2. uses energy to move potassium ions out of the cell and sodium ions into the cell.
3. moves sodium ions out of the cell and potassium ions into the cell without using energy.
4. moves potassium ions out of the cell and sodium ions into the cell without using energy.
5. Osmosis
1. is the diffusion of water.
2. is the diffusion of water and other small molecules.
3. is the diffusion of water and small ions.
4. is the diffusion of small molecules and ions.
6. Types of passive transport include which of the following? (1) simple diffusion, (2) osmosis, (3) facilitated diffusion, (4) active transport, and (5) vesicle transport.
1. 1 and 2
2. 1, 2, and 3
3. 4 and 5
4. 1, 2, 3, 4, and 5
7. Endocytosis and exocytosis
1. are both a type of vesicle transport.
2. move very large molecules either in or out of the cell.
3. are both a form of active transport.
4. all of the above
8. Which of the following needs energy? (1) passive transport, (2) active transport, (3) exocytosis, and (4) osmosis.
1. 1 only
2. 2 only
3. 2 and 3
4. 2, 3, and 4

## Lesson 3.3: Vocabulary I

Name___________________ Class______________ Date________

Match the vocabulary word with the proper definition.

Definitions

_____ 1. transport across a membrane without any additional energy requirement

_____ 2. the diffusion of water

_____ 3. type of vesicle transport that moves a substance into the cell

_____ 4. type of vesicle transport that moves a substance out of the cell

_____ 5. special proteins in the membrane that aid diffusion

_____ 6. membrane protein that forms a small hole that allows ions to pass through

_____ 7. an active transport protein

_____ 8. diffusion with the help of transport proteins

_____ 9. the movement of a substance across a membrane without any help from other molecules

_____ 10. the transport of very large molecules, such as proteins

_____ 11. transport across a membrane in which energy is required

Terms

a. active transport

b. channel protein

c. diffusion

d. endocytosis

e. exocytosis

f. facilitated diffusion

g. osmosis

h. passive transport

i. sodium-potassium pump

j. transport protein

k. vesicle transport

## Lesson 3.3: Vocabulary II

Name___________________ Class______________ Date________

Fill in the blank with the appropriate term.

1. By moving substances into and out of cells, ____________ , the process of keeping stable conditions inside a cell, is maintained.

2. A ____________ protein changes shape as it carries ions or molecules across the membrane.

3. Exocytosis is the type of ____________ transport that moves a substance out of the cell.

4. ____________ transport is movement across the plasma membrane that does not require an input of energy.

5. The sodium-potassium ____________ is involved in the active-transport of ions.

6. Facilitated diffusion needs the help of ____________ proteins

7. ____________ refers to the number of particles of a substance per unit of volume.

8. ____________ is the type of vesicle transport that moves a substance into the cell.

9. Energy for active transport is supplied by molecules of ____________.

10. ____________ is the diffusion of water.

11. During active transport, a substance is moving from an area of ____________ concentration to an area of ____________ concentration.

12. Moving molecules in and out of the cell is an important role of the ____________.

## Lesson 3.3: Critical Writing

Name___________________ Class______________ Date________

Thoroughly answer the question below. Use appropriate academic vocabulary and clear and complete sentences.

Discuss passive and active transport. Describe the main differences between these two types of transport, and provide examples of each type.

Feb 23, 2012

May 30, 2014


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Probabilistic bisimulation

In theoretical computer science, probabilistic bisimulation is an extension of the concept of bisimulation for fully probabilistic transition systems first described by K.G. Larsen and A. Skou.[1]

A discrete probabilistic transition system is a triple

${\displaystyle S=(\operatorname {St} ,\operatorname {Act} ,\tau :\operatorname {St} \times \operatorname {Act} \times \operatorname {St} \rightarrow [0,1])}$

where ${\displaystyle \tau (s,a,t)}$ gives the probability of starting in the state s, performing the action a and ending up in the state t. The set of states is assumed to be countable. There is no attempt to assign probabilities to actions. It is assumed that the actions are chosen nondeterministically by an adversary or by the environment. This type of system is fully probabilistic, there is no other indeterminacy.

The definition of a probabilistic bisimulation on a system S is an equivalence relation R on the state space St, such that for every pair s,t in St with sRt and for every action a in Act and for every equivalence class C of R ${\displaystyle \tau (s,a,C)=\tau (t,a,C).}$ Two states are said to be probabilistically bisimilar if there is some such R relating them.

When applied to Markov chains, probabilistic bisimulation is the same concept as lumpability.[2][3] Probabilistic bisimulation extends naturally to weighted bisimulation.[4]

References

1. ^ K. G. Larsen and A. Skou and appeared in the article Bisimulation through Probabilistic Testing, published in Information and Computation, vol 94, pages 1-28, 1991
2. ^ Probabilistic Noninterference through Weak Probabilistic Bisimulation by Geoffrey Smith Proceedings of the 16th IEEE Computer Security Foundations Workshop (CSFW’03) 1063-6900/03
3. ^ Kemeny, John G.; Snell, J. Laurie (July 1976) [1960]. Gehring, F. W.; Halmos, P. R., eds. Finite Markov Chains (Second ed.). New York Berlin Heidelberg Tokyo: Springer-Verlag. p. 224. ISBN 978-0-387-90192-3.
4. ^ Oliveira, J.N. (2013). "Weighted Automata as Coalgebras in Categories of Matrices". Int. J. Found. Comput. Sci. 24: 709–728. doi:10.1142/S0129054113400145.


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# The variety generated by all the ordinal sums of perfect MV-chains

We present the logic $\displaystyle BL_{Chang}$ , an axiomatic extension of BL (see [23]) whose corresponding algebras form the smallest variety containing all the ordinal sums of perfect MV-chains. We will analyze this logic and the corresponding algebraic semantics in the propositional and in the first-order case. As we will see, moreover, the variety of $\displaystyle BL_{Chang}$ -algebras will be strictly connected to the one generated by Chang’s MV-algebra (that is, the variety generated by all the perfect MV-algebras): we will also give some new results concerning these last structures and their logic.


text:
# Maximum likelihood transition matrix¶

Frank Noefrank.noe@fu-berlin.de
FU Berlin, Arnimallee 6, 14195 Berlin, Germany

 Summary: We have observed one or several trajectories amongst $n$ discrete states. At lag time $\tau$, there have been $c_{ij}=c_{ij}(\tau)$ transitions between these states, as encoded in a count matrix $\mathbf{C}\in\mathbb{R}^{n\times n}$. We aim at estimating a Markov transition matrix, $\hat{\mathbf{P}}\in\mathbb{R}^{n\times n}$. Mathematically, \begin{aligned} \hat{\mathbf{P}} & = & \arg\max_{\mathbf{P}}p(\mathbf{C}\mid\mathbf{P}).\end{aligned} It turns out that the maximum likelihood estimator is identical to the intuitive fraction of counts, $\hat{p}_{ij}=c_{ij}/\sum_{k=1}^{n}c_{ik}$.

Consider that we have observed a trajectory $[s_{1},...,s_{T}]$ amongst $n$ discrete states labeled $1$ through $n$. The number of transitions between any pair of states $i$ and $j$ are denoted $\mathbf{C}=[c_{ij}^{\mathrm{obs}}]$, i.e.: $$c_{ij}^{\mathrm{obs}}=\left|\{s_{t}=i\wedge s_{t+1}=j\mid t=1,...,T-1\}\right| \:\:\:\:(1)$$ and the total number of transitions out of state $i$ is $c_{i}^{\mathrm{obs}}$: $$c_{i}^{\mathrm{obs}}=\sum_{j=1}^{n}c_{ij}^{\mathrm{obs}}. \:\:\:\:(2)$$ The likelihood of a Markov chain with transition matrix $\mathbf{P}=[p_{ij}]\in\mathbb{R}^{n\times n}$ is given by: $$p(\mathbf{C}^{\mathrm{obs}}|\mathbf{T})=\prod_{i,j=1}^{n}p_{ij}^{c_{ij}^{\mathrm{obs}}} \:\:\:\:(3)$$ Our aim is to estimate the transition matrix that maximizes (3). In order to generalize our approach we also consider the maximization of a posterior probability which uses a conjugate prior, i.e. a prior with the same functional form as the likelihood: $$p(\mathbf{T}|\mathbf{C}^{\mathrm{obs}})\propto\prod_{i,j=1}^{n}p_{ij}^{c_{ij}^{\mathrm{prior}}+c_{ij}^{\mathrm{obs}}}=\prod_{i,j=1}^{n}p_{ij}^{c_{ij}}, \:\:\:\:(4)$$ Where $c_{ij}^{\mathrm{prior}}$ are prior counts. We now seek the matrix that maximizes (4). For the choice $c_{ij}^{\mathrm{prior}}\equiv0$ (uniform prior), this is identical to the maximum likelihood estimator. The probability (4) is difficult to work with due to the product. For optimization purposes it is therefore a common “trick” to instead work with the logarithm of the likelihood (log-likelihood): $$Q=\log p(\mathbf{P}\mid\mathbf{C})=\sum_{i,j}c_{ij}\log p_{ij}. \:\:\:\:(5)$$ This is useful since the logarithm is a monotonic function: as a result, the maximum of $\log f$ is also the maximum of $f$. However, this function is not bounded from above, since for $p_{ij}\rightarrow\infty$, $Q\rightarrow\infty$. Of course, we somehow need to restrict ourselves to sets of variables which actually form transition matrices, i.e., they satisfy the constraint: $$\sum_{j}p_{ij}=1. \:\:\:\:(6)$$ When optimizing with equality constraints, one uses Langrangian multipliers. The Lagrangian for $Q$ is given by: $$F=Q+\lambda_{1}(\sum_{j}p_{1j}-1)+...+\lambda_{m}(\sum_{j}p_{mj}-1). \:\:\:\:(7)$$ This function is maximized by the maximum likelihood transition matrix. It turns out that $F$ only has a single stationary point, which can be easily found by setting the partial derivatives, $$\frac{\partial\log F}{\partial p_{ij}}=\frac{c_{ij}}{p_{ij}}+\lambda_{i}, \:\:\:\:(8)$$ to zero: \begin{aligned} \frac{c_{ij}}{\hat{p}_{ij}}+\lambda_{i} & =0\\ \lambda_{i} & \hat{p}_{ij}=-c_{ij}. \:\:\:\:(9)\end{aligned} We now sum over $j$ on both sides and make use of the transition matrix property: $$\lambda_{i}\sum_{j=1}^{m}\hat{T}_{ij}=\lambda_{i}=-\sum_{j=1}^{m}c_{ij}=-c_{i}. \:\:\:\:(10)$$ Now we inserte the Langrange multipliers, $\lambda_{i}=-c_{i}$, in Eq. (9) and thus: \begin{aligned} \frac{c_{ij}}{\hat{p}_{ij}}-c_{i} & =0\\ \hat{p}_{ij} & =\frac{c_{ij}}{c_{i}}. \:\:\:\:(11)\end{aligned} It turns out that $\hat{\mathbf{P}}(\tau)$, as provided by Eq. (11), is the maximum of $p(\mathbf{P}|\mathbf{C}^{\mathrm{obs}})$ and thus also of $p(\mathbf{C}^{\mathrm{obs}}|\mathbf{P})$ when transition matrices are assumed to be uniformly distributed a priori.

In the limit of infinite sampling, i.e., trajectory length $T\rightarrow\infty$, $p(\mathbf{P}|\mathbf{C}^{\mathrm{obs}})$ converges towards a Dirac delta distribution with its peak at $\hat{\mathbf{P}}(\tau)$. Given that $\pi_{i}$ is the equilibrium probability of state $i$, the expected number of transitions out of state $i$ is $$\mathbb{E}[c_{ij}]=N\pi_{i}p_{ij} \:\:\:\:(12)$$ and thus $$\mathbb{E}[c_{i}]=N\pi_{i}. \:\:\:\:(13)$$ In this case the prior contribution vanishes: $$\lim_{N\rightarrow\infty}\hat{p}_{ij}=\lim_{N\rightarrow\infty}\frac{c_{ij}}{c_{i}}=\lim_{N\rightarrow\infty}\frac{c_{ij}^{\mathrm{prior}}+N\pi_{i}p_{ij}}{c_{ij}^{\mathrm{prior}}+\pi_{i}p_{ij}}=p_{ij}, \:\:\:\:(14)$$

i.e., the estimator is “asymptotically unbiased”.


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# Meaning of hypothesis test for "μ = 25"; isn't it impossible?

I'm working through a stats textbook and have a question of the form:

You will perform a significance test of $H_0: μ=25$ based on an SRS of $n=25$. Assume $σ=5$.

I'm stuck on the 'equals' part of $μ=25$. Given that we're working in the set of reals, isn't the probability of $μ$ equaling 25 zero? (Assume a non-textbook dataset; e.g. the odds of $μ=180$ for the height of adult humans is zero).

I think what the hypothesis is trying to say is "$μ$ is within bounds dictated by your confidence level, but mean$(μ)=25$".

The alternate hypothesis is something like $H_a: μ \neq 25$, which I know is true for any non-hypothetical dataset.

So, what does $H_0: μ=25$ actually mean?

I don't see any reason why having $\mu=25$ is impossible. After all, $\mu$ has to equal some number. It is common in statistical circles to hear people say that the null hypothesis, $\text H_0: \mu_{\text{treatment}}=\mu_{\text{control}}$, must be false (which is a statement that I don't necessarily agree with, but that's another discussion). Your situation, however, is (superficially similar, but) different from this.

Another issue that might be confusing you is that you refer to "the probability of μ equaling 25". This is a very common misconception of classical (Frequentist) hypothesis testing. Properly speaking, there is no 'probability' of $\mu$ equaling 25, or any other number. You could say that the probability is either $1$ or $0$, but you don't know which. Nonetheless, $\mu$ isn't a random variable; it's a parameter, that is, an unknown constant. The only probability distribution that can be assigned to this is the degenerate distribution.

So, what are you doing? When you perform a hypothesis test like what I suspect you are performing, you are checking to see what is the probability of getting a value (specifically a sample mean) as extreme or more extreme than yours, if the null hypothesis is true. The answer to that question is your p-value. After having done this, there are at least two valid ways of interpreting that p-value.

First, Fisher thought you should interpret it as a continuous measure of evidence against the null hypothesis. (Note that there is no alternative hypothesis or 'bright line' here, .04 is about the same as .06.)

Another possibility is to use the p-value as part of a decision making process, in keeping with the ideas of Neyman and Pearson. From this perspective, you are trying to differentiate between two hypotheses, and you designed your study explicitly to do so. For example, you conducted a power analysis, and set $\alpha$ and $\beta$ to levels of type I and type II error rates that you are willing to live with in the long run. After having analyzed your data and made a decision, you will either have made one of those errors or a correct decision, but you will never know which; you will only know that over all the times you engaged in this activity, you will have made a certain percentage of type I (type II) errors for those cases when the null hypothesis was actually true (false).

• This helps a lot. I think where I was going wrong was confusing 'check if H0 is correct' and 'assume H0 is correct, then see if the sample fits'. At least, that's how I think I'm supposed to interpret it... Feb 12 '13 at 6:51
• You're welcome, Ian. And yes, that's exactly how the logic works / how you're supposed to interpret it. Feb 12 '13 at 14:18

There are some details you've left out, and I am not sure what sort of answer you are looking for, but here goes...

Assuming you are performing something along the lines of a t-test, you are not looking for the probability of "mu = 25," but instead answering the question "is it reasonable to conclude that mu is some number other than (!=) 25?" So the idea that mu equals exactly 25 is not problematic in the sense that you were referring to in your question. If you conduct the test, you will either conclude that you can reject the null assumption as unreasonable or not, or you will calculate a p-value which answers a different, conditional, question ("what is the probability of seeing data as extreme as mine, in terms of its sample mean, if the data actually came from a distribution centered at 25?") which will allow you to draw a similar conclusion regarding the reasonableness of the assumed mu.

If you are asking the different question: "what is the true mean of this (given) data?" then you could build a confidence interval. There your concerns would be a consideration and handled in the interval's construction.

In classical/frequentist statistics, parameters like mu are treated as fixed, exact numbers; so asking "what is the probability of mu being 25?" is a sort of "nonsense" question in that philosophical context. A Bayesian approach would allow you to formulate the question in that way, but given the context of the question I don't think that is what you are working on. In that approach, there are ways to fix up a tiny interval around mu to get a probability.

You sound as though you are coming to an introductory stats book with a little extra background in probability already under your belt. Sometimes that makes the overly simple books harder to follow.

• +1 I see you have been acquainted with our site for a long time, but I would like nevertheless to welcome you on the occasion of posting your first answers. This is a nice clear explanation that gets to the heart of the matter.
– whuber
Feb 12 '13 at 0:31


text:
# Heisenberg imcertainty principle (get it!)

by mrfeathers
Tags: heisenberg, imcertainty, principle
 P: 6 I dont understand why it is so hard to find the exact position and velocity of orbiting electron. And also, why would we want to know it, if it is always moving? im not trying to disprove it or anything, so dont make fun of me, i am an uneducated peon
 P: 42 whats your idea of finding both the exact position and velocity of the electron? how do you find the location of your keyboard? shine light on it (lamp or diffuse sunlight through the window) what happens if you try do the same for something tiny like an electron ?
 P: 6 you should do that then, just prove the heisenberg uncertainty therom but just looking at the orbiting electrons with your own eyes
P: 42

## Heisenberg imcertainty principle (get it!)

when you look at your keyboard, the photons of light that 'interact' with it (and ten hit our eyes and give us what we percieve as colour) are much much smaller than the keyboard.
but if we are trying to see an electron using a photon..
the wavelength ("size") of an photon is alot larger than that of an electron, what implications does this have?
 P: 543 For complete understanding of Uncertainity Principle , check out : http://www.doxlab.co.nr/ TOPIC#2 on the above site is Heisenberg's..
Mentor
P: 28,457
 Quote by Dr.Brain For complete understanding of Uncertainity Principle , check out : http://www.doxlab.co.nr/ TOPIC#2 on the above site is Heisenberg's..
I would not recommend the site listed above. It perpetuates the misconception of the HUP that I have written about[1], that it is the uncertainty in a single measurement. I have seen this mistake repeated several times within the past week on here.

The HUP is NOT about the uncertainty in INSTRUMENTATION or measurement. One can easily verify this by looking at HOW we measure certain quantities. It is silly for Heisenberg to know about technological advances in the future and how much more accurate we can measure things. This is NOT what the HUP is describing. The HUP is NOT describing how well we know about the quantities in a single measurement. I can make as precise of a measurement of the position and momentum of an electron as arbitrary as I want simultaneously, limited to the technology I have on hand. I can make improvements in my accuracy of one without affecting the accuracy of measurement of the other.

What the HUP is telling you is the difference between a classical system and a quantum system. In a classical system, if you have a set of identical initial condition, and you measure ONE observable, and then you measure another observable, you will continue to get the SAME value of that 2nd observable everytime you measure the same value of that 1st observable. The more accurate you measure the 1st observable, the more accuract you can predict the value of the 2nd observable the next time you want to do such a measurement. The only limitation to how accurate you can determine these observable is the limitation to your measuring instruments. But these limitations do NOT scale like the HUP. You don't make one worse as the other one becomes better, because these are technical issues and are not related to one another.

On the other hand, in a quantum system, under the IDENTICAL initial conditions, even if you measure a series of identical values for the 1st observable, the 2nd observable may NOT yield the identical result each time. In fact, as you narrow down the uncertainty of the 1st observable, the 2nd observable may start showing wildly different values as you do this REPEATEDLY. Therefore, unlike the classical system, your ability to know and predict what is going to be the outcome of the 2nd observable goes progressively WORSE as you improve your knowledge about the 1st observable!

Again, it has NOTHING to do with the uncertainty in a SINGLE measurement! It doesn't mean that if you measure with utmost accuracy the position of an electron, that that electron momentum is "spread out" all over the place. This is wrong! I can STILL make an accurate determination of that electron's momentum - only my instrument will limit my accuracy of determining that. However, my ability to know what its momentum is going to be the NEXT time I measure it under the idential situation is what is dictated by the HUP!

Zz

[1] [11-15-2004 09:26 AM] - Misconception of the Heisenberg Uncertainty Principle
 P: 90 time is also an observable so why do we always take Δt= 0 while Δx*Δp >= hbar/2 ?
Mentor
P: 28,457
 Quote by matness time is also an observable so why do we always take Δt= 0 while Δx*Δp >= hbar/2 ?
Who is this "we"?

You will note that in typical school problems, you often work with a solution to the time-INDEPENDENT solution. It doesn't mean that the uncertainty in the time period is zero. It means that for that case, it is irrelevant since the description does not contain any time dynamics.

Zz.
 P: 188 I've a simple answer, hope u won't consider it naive... Most of the working theories right now weren't based on solid proofs but because we needed them, and some observations needed explanations...Theorists make theories, without confitmations, but we assume they r correct untill they r proven wrong, technology gives us the chance to make sure that we r on the right track, we can't yet find the electron and the uncertainty principle is what really works right now... See u when they find out something strong that would give us the opportunity to observe electrons..
Mentor
P: 28,457
 Quote by Nomy-the wanderer I've a simple answer, hope u won't consider it naive... Most of the working theories right now weren't based on solid proofs but because we needed them, and some observations needed explanations...Theorists make theories, without confitmations, but we assume they r correct untill they r proven wrong, technology gives us the chance to make sure that we r on the right track, we can't yet find the electron and the uncertainty principle is what really works right now... See u when they find out something strong that would give us the opportunity to observe electrons..
What are "solid proofs"? Would the statment that says "IT WORKS" be considered as "solid proof"? How about if I point to you your modern electronics? Would that be considered as "solid proof"?

Most people forget that the HUP is a CONSEQUENCE, not the origin, of quantum mechanics! To find a problem in the HUP is to find a problem in QM. And unless people also forgot about the centenial year of QM in 1999, let me remind you that there was an almost universal acclaimed by physicists that QM is THE most successful theory SO FAR in the history of human civilization!

This means that if you think QM does not have "solid proofs", then other parts of physics suffer from even a worse level of lacking of solid proofs!

Please keep in mind that NO part of physics is considered to be accepted and valid until there are sufficient experimental/empirical agreement! In fact, many theories and ideas originally came out of unexpected experimental observation in the first place!

There is no lacking of "solid proofs" for QM, and the HUP. Why this is even brought up here, I have no idea.

Zz.
 P: 90 "i" was thinking these as you said , until i see the word 'simultaneous' for measurements in the defn of HUP. if we say nearly simultaneous then i think there will be no problem(i hope so...) Also there are different explanations for HUP, and maybe this the problem about understanding it. at first i was thinking i get it , but it didnt take a long time for me to confuse (because i am a beginner only) Zz 's article is very helpful but if anyone can send a sketch of proof for HUP it will be more clear thanks n
Mentor
P: 28,457
 Quote by matness "i" was thinking these as you said , until i see the word 'simultaneous' for measurements in the defn of HUP. if we say nearly simultaneous then i think there will be no problem(i hope so...) Also there are different explanations for HUP, and maybe this the problem about understanding it. at first i was thinking i get it , but it didnt take a long time for me to confuse (because i am a beginner only) Zz 's article is very helpful but if anyone can send a sketch of proof for HUP it will be more clear
I don't understand. You want a "sketch of proof" for the HUP? What is this?

I think every student has either done, or seen the diffraction from a single slit. To me, this is a VERY clear example of the HUP! It just happens that we typically use wave description of light to account for such effects. But with the photon picture, the identical diffraction pattern can be directly obtained and the spreading is a direct consequence of the HUP!

More? The deBoer effect that is very pronounced in noble gasses is a direct consequence of the HUP. This leads to a correction to the internal energy (and thus, the specific heat capacity) of the gasses at very low temperatures. Only via taking into account such corrections can one obtain the experimental values!

But I think people pay waaaaay too much attention at disecting the consequences and forgetting the principles that CAUSE such consequences. The fact that this came out of "First Quantization" principle of QM that is based on [A,B] operations of two non-commuting observables is less understood by many who do not understand the formalism of QM. This is a crucial part of elementary QM with which a whole slew of consequences are built upon!

Zz.
 P: 90 what i wonder is the mathematical part : [A,B]= C --> ΔA * ΔB >= ||/2 is it just a thm about standart deviation?
P: 318
 Quote by mrfeathers I dont understand why it is so hard to find the exact position and velocity of orbiting electron. And also, why would we want to know it, if it is always moving? im not trying to disprove it or anything, so dont make fun of me, i am an uneducated peon

In QM, we may define the position of a particle. We may also define the momentum of a particle. However momentum is not the velocity of the particle. Many people in this forum always mix the momentum with the velocity (due to their equality for average values: i.e. Erhenfest Theorem) and tend to make incorrect deductions.
QM tells one thing: we cannot associate a classical path to a particle, hence we cannot define a couple (position, velocity) to the particle.
The HUP property applied to the (position, momentum) observables just highligh this fact: we cannot find a particle where both position and momentum have "defined" values equal to their mean values (i.e. through the Erhenfest Theorem, they have a classical path if is the case).

Seratend.
P: 318
 Quote by matness what i wonder is the mathematical part : [A,B]= C --> ΔA * ΔB >= ||/2 is it just a thm about standart deviation?
Yes.

Seratend.
P: 264
 Quote by ZapperZ I can make as precise of a measurement of the position and momentum of an electron as arbitrary as I want simultaneously, limited to the technology I have on hand.
What sort of technology do we have to achieve this?
P: 701
 Quote by seratend The second part answers to the first part of your post. In QM, we may define the position of a particle. We may also define the momentum of a particle. However momentum is not the velocity of the particle. Many people in this forum always mix the momentum with the velocity (due to their equality for average values: i.e. Erhenfest Theorem) and tend to make incorrect deductions. QM tells one thing: we cannot associate a classical path to a particle, hence we cannot define a couple (position, velocity) to the particle. The HUP property applied to the (position, momentum) observables just highligh this fact: we cannot find a particle where both position and momentum have "defined" values equal to their mean values (i.e. through the Erhenfest Theorem, they have a classical path if is the case). Seratend.
my (rather naive) understanding of it is that in a classical system:

$$xp - px = 0$$

but accord to HUP:

$$xp - px \neq 0$$

so that measuring the position and then measuring the momentum, is not the same as measuring the momentum and then measuring the position. infact, they will always differ by $$\frac{ih}{2\pi}$$

position and momentum are not seen as real values, but as non-commutative operators. you can derive the schrodinger wave equation in a straightforward manner from this.

(this is one interpretation).
P: 219
 so that measuring the position and then measuring the momentum, is not the same as measuring the momentum and then measuring the position. infact, they will always differ by
No, that's not correct. The HUP doesn't says something about only one simultaneoulsy measurment. It says something about a serie of simultaneously measurements (always the same conditions). You see?
If you make 1000 simultaneoulsy measurments (position and momentum) and you measure the position at each experiment exactly then you will get at each measurement of momentum a completely different value.

 Related Discussions Quantum Physics 41 Quantum Physics 3 General Physics 44 Introductory Physics Homework 0


text:
# Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 04 (a)/Solution 1

First, we find ${\displaystyle x'(0)}$.
Since we have the differential equation ${\displaystyle x'(t)=rx(t)(1-(x(t))^{2})}$, plugging ${\displaystyle t=0}$ and using ${\displaystyle x(0)=3}$, we get
${\displaystyle x'(0)=rx(0)(1-(x(0))^{2})=3r(1-3^{2})=-24r}$
Then, by its definition in the Hint, we get the speed of the particle at ${\displaystyle t=0}$ as ${\displaystyle \color {blue}|x'(0)|=24r}$.


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# If the average (arithmetic mean) of seven consecutive integers is k +

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If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12
[Reveal] Spoiler: OA

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Re: If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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13 Sep 2017, 03:45
Let first integer of series = a
Difference between integers = d= 1 (consecutive integers given)
Here no. of terms in series = n=7
Average = K+2

Average = Arithmetic mean =$$\frac{sum of all 7 numbers}{7}$$

by formula Sum of series =$$\frac{n(2a+(n-1)d)}{2} = \frac{7(2a+(7-1)*1)}{2} = \frac{7(2a+6)}{2} = 7(a+3)$$

As Average = k+2 = $$\frac{sum of all 7 numbers}{7}$$ = $$\frac{7(a+3)}{7}$$
=> k+2 =a+3
=> a= k-1

Now as nth term of series = a+(n-1)d => last term of given series = a+(7-1)*1 = (k-1)+6 = k+5

Product of first and last term = (k-1)*(k+5) = k^2+5k-k-5 = k^2+4k-5

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Re: If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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13 Sep 2017, 03:52
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Bunuel wrote:
If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

Let the first integer be x;
Sum of the 7 integers x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)= 7x+21

Average = Sum/n= 7(x+3)/7=x+3;

x+3= k+2 ==>x= k-1;
Greatest integer = k+5;

Product x*x+6 =k^2+4k-5

Ans:C

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If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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13 Sep 2017, 08:41
Bunuel wrote:
If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

Since there are 7 consecutive integers, K+2 is the fourth number in the series. So, the series should be k-1, k, k+1, k+2, k+3, k+4, k+5.

then the product of the greatest and least integer =(k+5)* (k-1)= k^2+4k-5

Ans is C
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If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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13 Sep 2017, 08:50
Given data: A set with 7 consecutive numbers has mean k+2

Lets assume the 7 numbers to be 1,2,3,4,5,6,7
The mean of consecutive integers in a set with odd number of elements is the middle number.
In this set, the 4th element which is 4 is the mean.
k+2 = 4 => k = 2
The product of the least and the greatest number is 1*7(7)

Evaluating the answer options(s.t k = 2)
A. k^2 - 9 = 4 -9 = -5
B. k^2 - 2k + 1 = 4 - 4 + 1 = 1
C. k^2 + 4k - 5 = 4 + 8 - 5 = 7

Since Option C gives us the same value for the product of the smallest and greatest number,
we need not evaluate the other 2 answer options. (Option C) is our answer!
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If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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15 Sep 2017, 00:40
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Total 7 consecutive nos and K+2 is the average which means 3 nos each are on the right and left of this no.
K-1, k,k+1,k+2,k+3, k+4, k+5
Smallest no=k-1
largest no=k+5

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If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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15 Sep 2017, 11:33
Bunuel wrote:
If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

I used algebra, erred, and took a long time to correct. So I timed myself using what I thought was a time-consuming method. Wrong. I was well under a minute.

I used only one sketch, and no fancy lines or asterisks. I used them here for clarity.

1. List the integers

x|x+1|x+2|x+3|x+4|x+5|x+6|

2. Place (k+2) under the median

"The arithmetic mean is k + 2." Median = mean for evenly spaced set. So

x|x+1|x+2|x+3|x+4|x+5|x+6|
*|***|***|k+2|***|***|***|

3. Find k. From above:
k + 2 = x + 3
k = x + 1, so

x|x+1|x+2|x+3|x+4|x+5|x+6|
*|_k_|***|k+2|***|***|***|

4. Find answer. The product of the greatest and least integer is? We need x, and (x+6), in terms of k, so

_x_|x+1|x+2|x+3|x+4|x+5|x+6|
k-1|_k_ |***|k+2|***|***|k+5|

(k - 1)(k + 5) = k$$^2$$ + 4k - 5

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Re: If the average (arithmetic mean) of seven consecutive integers is k + [#permalink]

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13 Oct 2017, 13:20
Bunuel wrote:
If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

Is this a 500 lvl question? really doesn't feel like it

anyway

Sum of the numbers = 7k + 14
let x = smallest number
21 + x = 7k + 14
x= k-1

product of smallest and biggest = x(x+6) = x^2 + 6x = k^2 - 2k + 1 + 6k - 6 = k^2 +4k - 5

Not a hard problem, but I don't think it is level 500.

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# The figure above shows a path around a triangular piece of land. Mary

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The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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14 Oct 2015, 21:14
Bunuel wrote:

The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

Kudos for a correct solution.

Attachment:
2015-10-15_0825.png

PR = $$\sqrt{8^2+6^2} = 10$$

DIstance walked by Mary = 8+6 = 14

% Exceeded distance of Mary = (14-10)*100/10 = 40%

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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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15 Oct 2015, 00:52
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1
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

The total distance Mary walked is 8+6=14 miles, while the distance Ted walked is 10 miles( by the Pythagorean theorem √(8^2 +6^2)=√(100)=10).
So Mary walked 4 miles more than Ted. That means Mary walked 40%((4/10)*100) more walked than Ted. So the answer is (B).
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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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17 Nov 2016, 08:36
1
Bunuel wrote:

The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

Kudos for a correct solution.

Attachment:
2015-10-15_0825.png

The standard measurement of 90 degree scale is 3-4-5
So, if we multiply the measurement by 2 then I’ll get it as 6-8-10.
That means, PQ=10,
So, Ted’s walk distance is 10.
Distance walked by Mary = 8+6 = 14.
So Mary walked 4 miles more than Ted. If I want to make 100% of Mary’s walk, I have to multiply it by 10. Simultaneously, Ted’s walk will be 4*10=40%. So the answer is (B)
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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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30 May 2017, 15:13
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Attached is a visual that should help.
Attachments

Screen Shot 2017-05-30 at 4.07.21 PM.png [ 133.22 KiB | Viewed 16438 times ]

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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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10 Dec 2017, 13:29
Area = 1/2 bh
Area = 1/2 8*6
Area = 1/2 48
Area = 24

24 - 8 - 6 = distance Ted walked PR =10

Mary = 14
Ted = 10

Difference 4
Percentage difference 4/10
40%

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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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27 Jan 2018, 22:15

GMATinsight Bunuel niks18 chetan2u
The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

Quote:
PR = $$\sqrt{8^2+6^2} = 10$$

DIstance walked by Mary = 8+6 = 14

% Exceeded distance of Mary = (14-10)*100/10 = 40%

Can experts suggest why we took 10 in denominator and not 14?
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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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27 Jan 2018, 22:27
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GMATinsight Bunuel niks18 chetan2u
The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

Quote:
PR = $$\sqrt{8^2+6^2} = 10$$

DIstance walked by Mary = 8+6 = 14

% Exceeded distance of Mary = (14-10)*100/10 = 40%

Can experts suggest why we took 10 in denominator and not 14?

% will be (extra distance walked by Mary)/(Distance walked by Ted)

We need a % over distance walked by Ted,
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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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04 Jun 2019, 07:46
We're told that the figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R and Ted walked directly from P to R. We're asked by what PERCENT the distance that Mary walked exceeded the distance that Ted walked. This question is based around Geometry patterns and the Percent Change Formula.
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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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04 Jun 2019, 23:01
Formula to be used:

Pythagoras theorem: $$Hypotenuse^2$$ = $$Base^2$$ + $$Perpendicular^2$$

Total distance walked by Mary = $$PQ + QR = 8 + 6 = 14 miles$$

Total distance walked by Ted = PR

Applying Pythagoras theorem in the Triangle PQR:

$$PR^2$$ = $$PQ^2$$ + $$QR^2$$

$$PR^2$$ = $$8^2$$ + $$6^2$$ $$= 64 + 36 = 100$$

PR = +10 OR -10

Since PR is a distance and cannot be negative, PR = 10 miles

The difference in the distance walked by Mary and Ted = 14 - 10 = 4 miles

Percent of the distance that Mary walked more than Ted walked = $$\frac{4}{10}$$ * 100 = 40%

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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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09 Jun 2019, 17:53
Bunuel wrote:

The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

Kudos for a correct solution.

Attachment:
2015-10-15_0825.png

Since we have a right triangle, we see that we have a 6-8-10 right triangle.

Mary’s distance was 14, and Ted’s distance was 10, so Mary traveled 4 miles more than Ted.

4/10 = 40%

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Re: The figure above shows a path around a triangular piece of land. Mary  [#permalink]

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23 Jun 2019, 05:17
Bunuel wrote:

The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%

Kudos for a correct solution.

Attachment:
2015-10-15_0825.png

First, the real amount of time that this question should take us. 8 and 6,

that means that this is a 10, and this is a 14. 14 exceeds 10 by 4,

out of the original number which is a 10, that is 40%, that's it! The correct

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Re: The figure above shows a path around a triangular piece of land. Mary   [#permalink] 23 Jun 2019, 05:17