HW #13 – Normal Distribution II

  1. The mean amount of time people spend working out each day is 45 minutes with a standard deviation of 25 minutes. If 40 people are randomly selected, what is the probability that the mean amount of time they spend working out is more than 40 minutes?

If there are two answers, list both answers separated by a comma.

  • The lengths of songs on Pandora are normally distributed with mean 161 seconds and standard deviation 38 seconds. If 15 songs are randomly selected, there is a 76% probability that the selected songs have mean length between what two values?

If there are two answers, list both answers separated by a comma.

  • The mean battery life of the iPhone is 14 hours with a standard deviation of 1.31 hours. What is the probability that a randomly selected iPhone has a battery life less than 11.8 hours?

If there are two answers, list both answers separated by a comma.

  • Scores on the SAT college entrance exam are normally distributed with a mean of 1045 and a standard deviation of 213. There is a 34% probability that the mean of 38 randomly chosen scores is greater than what value?

If there are two answers, list both answers separated by a comma.

  • The reading speed of adults is approximately normal, with a mean speed of 229 words per minute and a standard deviation of 24 words per minute. What is the reading speed of an adult who is at the 94th percentile?

If there are two answers, list both answers separated by a comma.

  • Scores on the ACT college entrance exam are normally distributed with a mean of 19.9 and a standard deviation of 5.3. What percent of scores are between 11 and 23?

If there are two answers, list both answers separated by a comma.

  • The heights of adult women are normally distributed with mean 59.2 inches and standard deviation 10.3 inches. Find the height that separates the tallest 17% of adult women.

If there are two answers, list both answers separated by a comma.

  • According to a recent national survey, the mean number of hours that adult Americans watch Netflix each week is 16.62 hours with a standard deviation of 7.67 hours. If a random sample of 37 adult Americans is selected, what is the probability that their mean number of hours watching Netflix each week is less than 18 hours.

If there are two answers, list both answers separated by a comma.

  • Human reaction speeds are approximately normal with mean 220.1 milliseconds and standard deviation 40.2 milliseconds. The middle 93% of human reaction speeds are between what two values?

If there are two answers, list both answers separated by a comma.

  1. According to a study done by the FAA, checked luggage loaded onto airplanes have a mean weight of 34.2 pounds and a standard deviation of 10.1 pounds. What is the probability that the mean weight of 31 randomly selected bags is between 31 pounds and 37 pounds?

If there are two answers, list both answers separated by a comma.

  1. Wait times at the DMV during peak hours are approximately normal with mean 42.4 minutes and standard deviation of 12.76 minutes. What percent of people going to the DMV during peak hours wait more than 38 minutes?

If there are two answers, list both answers separated by a comma.

  1. The weights of Atlantic bluefin tuna are normally distributed with a mean of 596 pounds and a standard deviation of 61 pounds. Researchers randomly select 17 fishes. There is a 23% probability that the mean weight of the selected fish is less than what value?

If there are two answers, list both answers separated by a comma.

HW #13 – Normal Distribution II

  1. The mean amount of time people spend working out each day is 45 minutes, with a standard deviation of 25 minutes. If 40 people are randomly selected, what is the probability that the mean amount of time they spend working out is more than 40 minutes? If there are two answers, list both answers separated by a comma.

Solution

Z = ( – μ)/(σ/√n), where;

X̄ = 40

μ = 45

σ = 25

n = 40

Thus, Z = (40 – 45)/(25/√40),

Z = -1.26

From the Z table, P(Z < -1.26) = 0.1038

Therefore, P(Z > -1.26) = 1 – 0.1038 = 0.8962.

Thus, the probability that the mean amount of time they spend working out is more than 40 minutes is 0.8962.

If there are two answers, list both answers separated by a comma.

Solution

The z scores corresponding to the middle 76% from the z table are -1.17 and 1.17

Z = (X – μ)/(σ/√n), where;

μ = 161

σ = 38

n = 15

For Z = -1.17, we compute the X value as follows;

-1.17 = (X – 161)/(38/√15),

-1.17*9.811558 = (X – 161)

-11.4795 = X – 161

X = 161 – 11.4795 = 149.5205

X = 149.5205 = 149.52

For Z = 1.17, we compute the X value as follows;

1.17 = (X – 161)/(38/√15),

1.17*9.811558 = (X – 161)

11.4795 = X – 161

X = 161 + 11.4795 = 172.4795

X = 172.4795 = 172.48

Thus, there is a 76% probability that the selected songs have a mean length between 149.52 and 172.48 seconds.

  • The mean battery life of the iPhone is 14 hours, with a standard deviation of 1.31 hours. What is the probability that a randomly selected iPhone has a battery life of less than 11.8 hours? If there are two answers, list both answers separated by a comma.

Solution

Z = (X – μ)/σ, where;

X = 11.8

μ = 14

σ = 1.31

Thus, Z = (11.8 – 14)/1.31

Z = -1.68

From the Z table, P(Z < -1.68) = 0.0465

Therefore, the probability that a randomly selected iPhone has a battery life of less than 11.8 hours is 0.0465.

  • Scores on the SAT college entrance exam are normally distributed with a mean of 1045 and a standard deviation of 213. There is a 34% probability that the mean of 38 randomly chosen scores is greater than what value? If there are two answers, list both answers separated by a comma.

Solution

Z = (X – μ)/(σ/√n), where;

X = unknown

μ = 1045

σ = 213

n = 38

Probability = 34% = 0.34

1 – 0.34 = 0.66

Therefore, we look for the value of the Z score corresponding to the probability of 0.66

From the Z table, Z score = 0.41

We now compute the value of X as follows;

0.41 = (X – 1045)/(213/√38),

0.41*34.55316 = (X – 1045)

14.1668 = (X – 1045)

X = 14.1668 + 1045

X = 1059

Answer = 1059

  • The reading speed of adults is approximately normal, with a mean speed of 229 words per minute and a standard deviation of 24 words per minute. What is the reading speed of an adult who is at the 94th percentile? If there are two answers, list both answers separated by a comma.

Solution

z = (x – μ)/σ

Here, we are interested in computing the value of X.

From the Z table, the Z score corresponding to the 94th percentile is equal to 1.555

Therefore, Z = 1.555

μ = 229

σ = 24

Thus, 1.555 = (X – 229)/24

1.555*24 = X – 229

37.32 = X – 229

X = 37.32 + 229 = 266.32

X = 266 words per minute.

  • Scores on the ACT college entrance exam are normally distributed with a mean of 19.9 and a standard deviation of 5.3. What percent of scores are between 11 and 23? If there are two answers, list both answers separated by a comma.

Solution

z = (x – μ)/σ, where;

x = 11 and 23

μ = 19.9

σ = 5.3

Therefore, for X = 11

Z = (11 – 19.9)/5.3

Z = -1.68

For X = 23

Z = (23 – 19.9)/5.3

Z = 0.58

We are interested in P(11 < X < 23) = P(-1.68 < Z < 0.58) = P(Z < 0.58) – P(Z < -1.68)

From the Z-table, P(Z < 0.58) = 0.7190

P(Z < -1.68) = 0.0465

Thus, P(11 < X < 23) = P(Z < 0.58) – P(Z < -1.68)= 0.7190 – 0.0465 = 0.6725

in terms of percentages, P(11 < X < 23) = 0.6725*100 = 67.25%

Therefore, the percentage of scores between 11 and 23 is 67.25%.

  • The heights of adult women are normally distributed with a mean of 59.2 inches and a standard deviation of 10.3 inches. Find the height that separates the tallest 17% of adult women.

If there are two answers, list both answers separated by a comma.

Solution

From the Z table, the Z score associated with the highest 17% is Z = 0.954

In this case, we are interested in finding the height (X)

z = (x – μ)/σ, where;

Z = 0.954

μ = 59.2

σ = 10.3

Thus, 0.954 = (X – 59.2)/10.3

0.954*10.3 = X – 59.2

9.8262 = X – 59.2

X = 9.8262 + 59.2

X = 69.0262

Thus, the height that separates the tallest 17% of adult women equals 69.0262 inches.

  • According to a recent national survey, the mean number of hours that adult Americans watch Netflix each week is 16.62 hours, with a standard deviation of 7.67 hours. If a random sample of 37 adult Americans is selected, what is the probability that their mean number of hours watching Netflix each week is less than 18 hours? If there are two answers, list both answers separated by a comma.

Solution

Z = ( – μ)/(σ/√n), where;

X̄ = 18

μ = 16.62

σ = 7.67

n = 37

Thus, Z = (18 – 16.62)/(7.67/√37),

Z = 1.09

From the Z table, P(Z < 1.09) = 0.8621

Thus, the probability that their mean number of hours watching Netflix each week is less than 18 hours equals 0.8621.

  • Human reaction speeds are approximately normal, with a mean of 220.1 milliseconds and a standard deviation of 40.2 milliseconds. The middle 93% of human reaction speeds are between what two values? If there are two answers, list both answers separated by a comma.

Solution

The z scores corresponding to the middle 93% from the z table are -1.81 and 1.81

z = (x – μ)/σ, where;

μ = 220.1

σ = 40.2

For Z = -1.81, we compute the X value as follows;

-1.81 = (X – 220.1)/40.2

-1.81*40.2 = X – 220.1

-72.762 = X – 220.1

X = 220.1 – 72.762 = 147.3199

X = 147.3199

For Z = 1.81, we compute the X value as follows;

1.81 = (X – 220.1)/40.2

1.81*40.2 = X – 220.1

72.762 = X – 220.1

X = 220.1 + 72.762 = 292.862

X = 292.862

Thus, the middle 93% of human reaction speeds are between 147.3 and 292.9 milliseconds.

  1. According to a study done by the FAA, checked luggage loaded onto airplanes has a mean weight of 34.2 pounds and a standard deviation of 10.1 pounds. What is the probability that the mean weight of 31 randomly selected bags is between 31 pounds and 37 pounds? If there are two answers, list both answers separated by a comma.

Solution

Z = (X – μ)/(σ/√n), where;

X = 31 and 37

μ = 34.2

σ = 10.1

n = 31

Therefore, for X = 31

Z = (31 – 34.2)/(10.1/√31)

Z = -1.76

Therefore, for X = 37

Z = (37 – 34.2)/(10.1/√31)

Z = 1.54

We are interested in P(31 < X < 37) = P(-1.76 < Z < 1.54) = P(Z < 1.54) – P(Z < -1.76)

From the Z-table, P(Z < 1.54) = 0.9382

P(Z < -1.76) =0.0392

Thus, P(31 < X < 37) = P(Z < 1.54) – P(Z < -1.76) = 0.9382 – 0.0392 = 0.8990

Therefore, the probability that the mean weight of 31 randomly selected bags is between 31 pounds and 37 pounds equals 0.8990.

  1. Wait times at the DMV during peak hours are approximately normal, with a mean of 42.4 minutes and a standard deviation of 12.76 minutes. What percent of people going to the DMV during peak hours wait more than 38 minutes? If there are two answers, list both answers separated by a comma.

Solution

Z = (X – μ)/σ, where;

X = 38

μ = 42.4

σ = 12.76

n = 40

Thus, Z = (38 – 42.4)/12.76

Z = -0.34

From the Z table, P(Z < -0.34) = 0.3669

Therefore, P(Z > -0.34) = 1 – 0.3669 = 0.6331

In terms of percentage, P(Z > -0.34) = 0.6331*100 = 63.31%

Thus, the percentage of people going to the DMV during peak hours who wait for more than 38 minutes equals 63.31%.

  1. The weights of Atlantic bluefin tuna are normally distributed with a mean of 596 pounds and a standard deviation of 61 pounds. Researchers randomly select 17 fishes. There is a 23% probability that the mean weight of the selected fish is less than what value?

If there are two answers, list both answers separated by a comma.

Solution

Z = (X – μ)/(σ/√n), where;

X = unknown

μ = 596

σ = 61

n = 17

Probability = 23% = 0.23

Therefore, we look for the value of the Z score corresponding to the probability of 0.23

From the Z table, Z score = -0.74

We now compute the value of X as follows;

-0.74 = (X – 596)/(61/√17),

-0.74*14.79467 = (X – 596)

-10.9481 = (X – 596)

X = 596 – 10.9481

X = 585.0519 = 585.05

Therefore, there is a 23% probability that the mean weight of the selected fish is less than 585.05 pounds.

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