Using the Normal Distribution, The Normal Distribution | Statistics

Learning objectives

The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X < x). Normal tables, computers, and calculators are used to provide or calculate the probability P(X < x).

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows a normal distribution bell curve in which the area represented by the probability P(X < x) is shaded.

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(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.

Suppose the graph above were to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61.

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 m79718 without the use of technology. The tables include instructions for how to use them.

The probability is represented by the area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-table under the z-column. Most z-tables show the area under the normal curve to the left of z. Others show the mean to z area. The method used will be indicated on the table.

We will discuss the z-table that represents the area under the normal curve to the left of z. Once you have located the z-score, locate the corresponding area. This will be the area under the normal curve, to the left of the z-score. This area can be used to find the area to the right of the z-score, or by subtracting from 1 or the total area under the normal curve. These areas can also be used to determine the area between two z-scores.

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

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.

Calculate the z-score:

z = \frac{x - μ}{σ} = \frac{65 - 63}{5} = \frac{2}{5} = .40

The z-table shows that the area to the left of z is 0.6554. Subtracting this area from 1 gives 0.3446.

Then, find P(x > 65).

P (x > 65) = 0 .3446

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows a normal distribution bell curve graph in which the area representing the probability P(x > 65) = 0.3446 is shaded.

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 is 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 keya 2nd keyand 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. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values under the curve to fall below it. This is an arbitrary value and one that works well, for our purpose.

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 percent.

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

c. Find the 90^th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90^th percentile. This time, we are looking for a score that corresponds to a given area under the curve.

Let k = the 90^th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90^th 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 10 percent are the same or higher. The variable k is often called a critical value.

We know the mean, standard deviation, and area under the normal curve. We need to find the z-score that corresponds to the area of 0.9 and then substitute it with the mean and standard deviation, into our z-score formula. The z-table shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.9. Thus, we can write the following:

$1.28 = \frac{x - 63}{5}$

Multiplying each side of the equation by 5 gives

$6.4 = x - 63$

Adding 63 to both sides of the equation gives

$69.4 = x.$

Thus, our score, k, is 69.4.

k = 69 .4

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. A point, k, is labeled to the right of 63. A vertical line extends from k to the curve. The area under the curve to the left of k is shaded. This represents the probability that x is less than k: P(x < k) = 0.90

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

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

d. Find the 70^th percentile.

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

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

invNorm(0.70,63,5) = 65.6

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).

First, calculate the z-scores for each x-value.

$\begin{matrix} z = \frac{1.8 - 2}{0.5} = \frac{- 0.2}{0.5} = - 0.40 \\ z = \frac{2.75 - 2}{0.5} = \frac{0.75}{0.5} = 1.5 \end{matrix}$

Now, use the Z-table to locate the area under the normal curve to the left of each of these z-scores.

The area to the left of the z-score of −0.40 is 0.3446. The area to the left of the z-score of 1.5 is 0.9332. The area between these scores will be the difference in the two areas, or $0.9332 - 0.3446$ , which equals 0.5886.

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.

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 25^th percentile, k, where P(x < k) = 0.25.

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The area under the left tail of the curve is shaded. The shaded area shows that the probability that x is less than k is 0.25. It follows that k = 1.67.

invNorm(0.25,2,0.5) = 1.66

We use invNorm because we are looking for the k-value.

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

In the United States smartphone users between the ages of 13 and 55+ 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

The z-scores are calculated as

$\begin{matrix} z = \frac{23 - 36.9}{13.9} = \frac{- 13.9}{13.9} = - 1 \\ z = \frac{64.7 - 36.9}{13.9} = \frac{27.8}{13.9} = 2 \end{matrix}$

The Z-table shows the area to the left of a z-score with an absolute value of 1 to be 0.1587. It shows the area to the left of a z-score of 2 to be 0.9772. The difference in the two areas is 0.8185.

This is slightly different than the area given by the calculator, due to rounding.

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 80^th percentile of this distribution, and interpret it in a complete sentence.

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

Try It

Use the information in eip-562 to answer the following questions:

Find the 30^th percentile, and interpret it in a complete sentence.
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)

To find the 30^th 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.
Find P(x < 27)

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 36.9 on the horizontal axis. The point 27 is also labeled. A vertical line extends from 27 to the curve. The area under the curve to the left of 27 is shaded. The shaded area shows that P(x < 27) = 0.2342.

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.)

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).

IQR = Q₃ – Q₁
Calculate Q₃ = 75^th percentile and Q₁ = 25^th percentile.
Recall that we can use invNorm to find the k-value. We can use this to find the quartile values.
invNorm(0.75,36.9,13.9) = Q₃ = 46.2754
invNorm(0.25,36.9,13.9) = Q₁ = 27.5246
IQR = Q₃ – Q₁ = 18.8

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

Find k where P(x ≥ k) = 0.40. At least translates to greater than or equal to.
0.40 = the area to the right
The 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 = 40.4.
Forty percent of the ages that range from 13 to 55+ are at least 40.4 years.

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

View original figure (external; image not reproduced in this text edition)

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

1 – 0.20 = 0.80. Outside of the middle 20 percent will be 80 percent of the values.
The tails of the graph of the normal distribution each have an area of 0.40.
Find k₁, the 40^th percentile, and k₂, the 60^th percentile (0.40 + 0.20 = 0.60). This leaves the middle 20 percent, in the middle of the distribution.
k₁ = invNorm(0.40,5.85,0.24) = 5.79 cm
k₂ = invNorm(0.60,5.85,0.24) = 5.91 cm

So, the middle 20 percent of mandarin oranges have diameters between 5.79 cm and 5.91 cm.

c. Find the 90^th 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 6.16 cm.

References

Chicago Public Media & Ira Glass. (2013). 403: NUMMI. Retrieved from http://www.thisamericanlife.org/radioarchives/episode/403/nummi

lauramitchell347. (2012, Dec. 28). Smart phone users, by the numbers. Visually. Retrieved from http://visual.ly/smart-phone-users-numbers

Statistics Brain Research Institute. (2013). Facebook company statistics statistic brain. Retrieved from http://www.statisticbrain.com/facebook-statistics/

Wikipedia (2013). Naegele's rule. Retrieved from http://en.wikipedia.org/wiki/Naegele's_rule

Win at the Lottery. (2013). Scratch-off lottery ticket playing tips. Retrieved from www.winatthelottery.com/public/department40.cfm

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 k^th 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?

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph showing a bell shaped curve of normal distribution with a vertical line to the right of center labeled with a 3. The axes are unlabeled.

P(x < 1)

What is the area to the right of one?

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph showing a bell shaped curve of normal distribution with a vertical line to the right of center labeled with a 3. The axes are unlabeled.

Is P(x < 1) equal to P(x ≤ 1)? Why or why not?

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?

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph showing a bell shaped curve of normal distribution with a vertical line to the right of center labeled with a 3. The axes are unlabeled.

What is the area to the right of three?

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph showing a bell shaped curve of normal distribution with a vertical line to the right of center labeled with a 3. The axes are unlabeled.

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?

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.

Find the probability that x < 30.

0.0013

Find the 80^th percentile.

Find the 60^th percentile.

56.03

X ~ N(6, 2)

Find the probability that x is between three and nine.

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.

Use the following information to answer the next three exercises: 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.

Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.

View original figure (external; image not reproduced in this text edition)
Original figure description: Empty normal distribution curve.
P(0 < x < ____________) = ___________. Use zero for the minimum value of x.

Check student’s solution
3, 0.1979

Find the probability that a CD player will last between 2.8 and 6 years.

Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
View original figure (external; image not reproduced in this text edition)
Original figure description: Empty normal distribution curve.
P(__________ < x < __________) = __________

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

Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70 percent.
View original figure (external; image not reproduced in this text edition)
Original figure description: Empty normal distribution curve.
P(x < k) = __________. Therefore, k = _________.

Check student’s solution
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?

0.0580
0.8447
0.0553
0.9420

The 90^th percentile for recovery times is –

8.89
7.07
7.99
4.32

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

Yes
No
Unable to determine

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

0.0001
0.9270
0.1862
0.0668

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

1.24
2.41
3.95
6.05

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.

X ~ _____ (_____ ,_____)
Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement.
Would you expect to meet many Asian adult males taller than 72 inches? Explain why or why not, and numerically justify your answer.
The middle 40 percent of heights fall between what two values? Sketch the graph, and write the probability statement.

X ~ N(66, 2.5)
0.5404
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.

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

The percent of fat calories that a person in the United States 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 = percentage of fat calories.

X ~ _____ (_____, _____)
Find the probability that the percentage of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined.
Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement.

X ~ N(36, 10)
The probability that a person consumes more than 40 percent of their calories as fat is 0.3446.
Approximately 25 percent of people consume less than 29.26 percent 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.

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

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 that child spends alone per day.

In words, define the random variable X.
X ~ _____ (_____, _____)
Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.
What percentage of the children spend more than 10 hours per day unsupervised?
Seventy percent of the children spend at least how long per day unsupervised?

X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
X ~ N(3, 1.5)
The probability that the child spends less than one hour a day unsupervised is 0.0918.
The probability that a child spends over 10 hours a day unsupervised is less than 0.0001.
2.21 hours

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

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

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.

In words, define the random variable X.
X ~ _____ (_____, _____)
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.
Sixty percent of all trials of this type are completed within how many days?

X = the distribution of the number of days a particular type of criminal trial will take
X ~ N(21, 7)
The probability that a randomly selected trial will last more than 24 days is 0.3336.
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.

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

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. element-694 displays the ordered real data, in minutes.

0.50	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 10 customers were 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.

mean = 5.51, s = 2.15
Check student's solution.
Check student's solution.
Check student's solution.
X ~ N(5.51, 2.15)
0.6029
The cumulative frequency for less than 6.1 minutes is 0.64.
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.
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.
The approximation would have been less accurate, because the smaller sample size means that the data does not fit a 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:

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

element-666 shows a sample of the maximum capacitymaximum number of spectatorsof 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.
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. HintOrder 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?

mean = 60,136
s = 10,468
Answers will vary
Answers will vary
Answers will vary
X ~ N(60136, 10468)
0.7440
The cumulative relative frequency is 43/60 = 0.717.
The answers for part f and part g are not the same because the normal distribution is only an approximation.

The length of a pregnancy of a certain female animal is normally distributed with a mean of 280 days and a standard deviation of 13 days. The father was not present from 240 to 306 days before the birth of the offspring, so the pregnancy would have been less than 240 days or more than 306 days long, if he was the father. 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.

A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10 percent 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 ruleone standard deviation, two standard deviations, and three standard deviations from the mean 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 = $\sqrt{n p q}$ = $\sqrt{(100)(0 .1)(0 .9)}$ = 3

z = ±1: x₁ = µ + zσ = 10 + 1(3) = 13 and x2 = µ – zσ = 10 – 1(3) = 7. 68 percent of the defective cars will fall between seven and 13
z = ±2: x₁ = µ + zσ = 10 + 2(3) = 16 and x2 = µ – zσ = 10 – 2(3) = 4. 95 percent of the defective cars will fall between four and 16
z = ±3: x₁ = µ + zσ = 10 + 3(3) = 19 and x2 = µ – zσ = 10 – 3(3) = 1. 99.7 percent 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 percent (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 = 4verify the mean and standard deviation. Solve the following:

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

A child playing a carnival game will be a winner one out of five times. If 190 games are played, what is the probability that there are

somewhere between 34 and 54 wins
somewhere between 54 and 64 wins
more than 64 wins

n = 190; p = $\frac{1}{5}$ = 0.2; q = 0.8
μ = np = (190)(0.2) = 38
σ = $\sqrt{n p q}$ = $\sqrt{(190)(0 .2)(0 .8)}$ = 5.5136

For this problem: P(34 < x < 54) = normalcdf(34,54,48,5.5136) = 0.7641
For this problem: P(54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018
For this problem: P(x > 64) = normalcdf(64,10⁹⁹,48,5.5136) = 0.0000012 (approximately 0)

A social media site provides a variety of statistics on its website that detail the growth and popularity of the site.

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

Find the probability that the percentage of 18- to 34-year-olds who check the social media website before getting out of bed in the morning is at least 30.
Find the 95^th percentile, and express it in a sentence.

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