Is It Unusual? A population is normally distributed , with a mean of 100 and a standard deviation of 15. Determine whether cither event is unusual. Explain your reasoning. a. The mean of a sample of 3 is 112 or more. b. The mean of a sample of 75 is 105 or more.

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Answer 1

Textbook Question

Chapter 5, Problem 1UA

Is It Unusual? A population is normally distributed, with a mean of 100 and a standard deviation of 15. Determine whether cither event is unusual. Explain your reasoning.

a. The mean of a sample of 3 is 112 or more.
b. The mean of a sample of 75 is 105 or more.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Whether the event mean of sample of 3 is 112 or more is unusual or not.

Answer to Problem 1UA

No, the event mean of sample of 3 is 112 or more is not unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 3.

Calculation:

The variable x represents random variable.

The notation x¯ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by μx¯ and σx¯ , respectively.

The distribution of sample means is normal with the parameters μx¯ and σx¯ by using central limit theorem.

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is, σx¯=σn .

Substitute 15 for σ and 3 for n

σx¯=153=151.732=8.6605

Thus, the standard deviation of the sampling distribution of sample means is σx¯=8.6605

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 112 for x¯ , 100 for μx¯ and 8.6605 for σx¯

That is,

z=112−1008.6605=128.6605=1.39

The probability that the mean of sample of 3 is 112 or more is obtained by finding the area to the right of 1.39. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 1.39.

Procedure:

Locate 1.3 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<1.39)=0.9177

The area to the right of 1.39 is,

P(z>1.39)=1−P(z<1.39)=1−0.9177=0.0823

Thus, the probability that the mean of sample of 3 is 112 or more is 0.0823.

Interpretation:

Here, the probability that the mean of sample of 3 is 112 or more is greater than 0.05. That is 0.0823>0.05 . Thus, the event mean of sample of 3 is 112 or more is not unusual.

b.

Expert Solution

To determine

Whether the event mean of sample of 75 is 105 or more is unusual or not.

Answer to Problem 1UA

Yes, the event mean of sample of 3 is 112 or more is unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 75.

Calculation:

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is,

σx¯=σn

Substitute 15 for σ and 75 for n

σx¯=1575=158.6602=1.7321

Thus, the standard deviation of the sampling distribution of sample means is σx¯=1.7321

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 105 for x¯ , 100 for μx¯ and 1.7321 for σx¯

That is,

z=105−1001.7321=51.7321=2.89

The probability that the mean of sample of 75 is 105 or more is obtained by finding the area to the right of 2.89. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 2.89.

Procedure:

Locate 2.8 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<2.89)=0.9981

The area to the right of 2.89 is,

P(z>2.89)=1−P(z<2.89)=1−0.9981=0.0019

Thus, the probability that the mean of sample of 75 is 105 or more is 0.0019.

Interpretation:

Here, the probability that the mean of sample of 75 is 105 or more is less than 0.05. That is 0.0019<0.05 . Thus, the event mean of sample of 75 is 105 or more is unusual.

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Answer 2

Textbook Question

Chapter 5, Problem 1UA

Is It Unusual? A population is normally distributed, with a mean of 100 and a standard deviation of 15. Determine whether cither event is unusual. Explain your reasoning.

a. The mean of a sample of 3 is 112 or more.
b. The mean of a sample of 75 is 105 or more.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Whether the event mean of sample of 3 is 112 or more is unusual or not.

Answer to Problem 1UA

No, the event mean of sample of 3 is 112 or more is not unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 3.

Calculation:

The variable x represents random variable.

The notation x¯ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by μx¯ and σx¯ , respectively.

The distribution of sample means is normal with the parameters μx¯ and σx¯ by using central limit theorem.

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is, σx¯=σn .

Substitute 15 for σ and 3 for n

σx¯=153=151.732=8.6605

Thus, the standard deviation of the sampling distribution of sample means is σx¯=8.6605

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 112 for x¯ , 100 for μx¯ and 8.6605 for σx¯

That is,

z=112−1008.6605=128.6605=1.39

The probability that the mean of sample of 3 is 112 or more is obtained by finding the area to the right of 1.39. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 1.39.

Procedure:

Locate 1.3 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<1.39)=0.9177

The area to the right of 1.39 is,

P(z>1.39)=1−P(z<1.39)=1−0.9177=0.0823

Thus, the probability that the mean of sample of 3 is 112 or more is 0.0823.

Interpretation:

Here, the probability that the mean of sample of 3 is 112 or more is greater than 0.05. That is 0.0823>0.05 . Thus, the event mean of sample of 3 is 112 or more is not unusual.

b.

Expert Solution

To determine

Whether the event mean of sample of 75 is 105 or more is unusual or not.

Answer to Problem 1UA

Yes, the event mean of sample of 3 is 112 or more is unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 75.

Calculation:

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is,

σx¯=σn

Substitute 15 for σ and 75 for n

σx¯=1575=158.6602=1.7321

Thus, the standard deviation of the sampling distribution of sample means is σx¯=1.7321

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 105 for x¯ , 100 for μx¯ and 1.7321 for σx¯

That is,

z=105−1001.7321=51.7321=2.89

The probability that the mean of sample of 75 is 105 or more is obtained by finding the area to the right of 2.89. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 2.89.

Procedure:

Locate 2.8 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<2.89)=0.9981

The area to the right of 2.89 is,

P(z>2.89)=1−P(z<2.89)=1−0.9981=0.0019

Thus, the probability that the mean of sample of 75 is 105 or more is 0.0019.

Interpretation:

Here, the probability that the mean of sample of 75 is 105 or more is less than 0.05. That is 0.0019<0.05 . Thus, the event mean of sample of 75 is 105 or more is unusual.

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Answer 3

Textbook Question

Chapter 5, Problem 1UA

Is It Unusual? A population is normally distributed, with a mean of 100 and a standard deviation of 15. Determine whether cither event is unusual. Explain your reasoning.

a. The mean of a sample of 3 is 112 or more.
b. The mean of a sample of 75 is 105 or more.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Whether the event mean of sample of 3 is 112 or more is unusual or not.

Answer to Problem 1UA

No, the event mean of sample of 3 is 112 or more is not unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 3.

Calculation:

The variable x represents random variable.

The notation x¯ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by μx¯ and σx¯ , respectively.

The distribution of sample means is normal with the parameters μx¯ and σx¯ by using central limit theorem.

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is, σx¯=σn .

Substitute 15 for σ and 3 for n

σx¯=153=151.732=8.6605

Thus, the standard deviation of the sampling distribution of sample means is σx¯=8.6605

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 112 for x¯ , 100 for μx¯ and 8.6605 for σx¯

That is,

z=112−1008.6605=128.6605=1.39

The probability that the mean of sample of 3 is 112 or more is obtained by finding the area to the right of 1.39. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 1.39.

Procedure:

Locate 1.3 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<1.39)=0.9177

The area to the right of 1.39 is,

P(z>1.39)=1−P(z<1.39)=1−0.9177=0.0823

Thus, the probability that the mean of sample of 3 is 112 or more is 0.0823.

Interpretation:

Here, the probability that the mean of sample of 3 is 112 or more is greater than 0.05. That is 0.0823>0.05 . Thus, the event mean of sample of 3 is 112 or more is not unusual.

b.

Expert Solution

To determine

Whether the event mean of sample of 75 is 105 or more is unusual or not.

Answer to Problem 1UA

Yes, the event mean of sample of 3 is 112 or more is unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 75.

Calculation:

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is,

σx¯=σn

Substitute 15 for σ and 75 for n

σx¯=1575=158.6602=1.7321

Thus, the standard deviation of the sampling distribution of sample means is σx¯=1.7321

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 105 for x¯ , 100 for μx¯ and 1.7321 for σx¯

That is,

z=105−1001.7321=51.7321=2.89

The probability that the mean of sample of 75 is 105 or more is obtained by finding the area to the right of 2.89. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 2.89.

Procedure:

Locate 2.8 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<2.89)=0.9981

The area to the right of 2.89 is,

P(z>2.89)=1−P(z<2.89)=1−0.9981=0.0019

Thus, the probability that the mean of sample of 75 is 105 or more is 0.0019.

Interpretation:

Here, the probability that the mean of sample of 75 is 105 or more is less than 0.05. That is 0.0019<0.05 . Thus, the event mean of sample of 75 is 105 or more is unusual.

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Answer 4

Textbook Question

Chapter 5, Problem 1UA

Is It Unusual? A population is normally distributed, with a mean of 100 and a standard deviation of 15. Determine whether cither event is unusual. Explain your reasoning.

a. The mean of a sample of 3 is 112 or more.
b. The mean of a sample of 75 is 105 or more.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Whether the event mean of sample of 3 is 112 or more is unusual or not.

Answer to Problem 1UA

No, the event mean of sample of 3 is 112 or more is not unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 3.

Calculation:

The variable x represents random variable.

The notation x¯ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by μx¯ and σx¯ , respectively.

The distribution of sample means is normal with the parameters μx¯ and σx¯ by using central limit theorem.

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is, σx¯=σn .

Substitute 15 for σ and 3 for n

σx¯=153=151.732=8.6605

Thus, the standard deviation of the sampling distribution of sample means is σx¯=8.6605

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 112 for x¯ , 100 for μx¯ and 8.6605 for σx¯

That is,

z=112−1008.6605=128.6605=1.39

The probability that the mean of sample of 3 is 112 or more is obtained by finding the area to the right of 1.39. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 1.39.

Procedure:

Locate 1.3 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<1.39)=0.9177

The area to the right of 1.39 is,

P(z>1.39)=1−P(z<1.39)=1−0.9177=0.0823

Thus, the probability that the mean of sample of 3 is 112 or more is 0.0823.

Interpretation:

Here, the probability that the mean of sample of 3 is 112 or more is greater than 0.05. That is 0.0823>0.05 . Thus, the event mean of sample of 3 is 112 or more is not unusual.

b.

Expert Solution

To determine

Whether the event mean of sample of 75 is 105 or more is unusual or not.

Answer to Problem 1UA

Yes, the event mean of sample of 3 is 112 or more is unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 75.

Calculation:

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is,

σx¯=σn

Substitute 15 for σ and 75 for n

σx¯=1575=158.6602=1.7321

Thus, the standard deviation of the sampling distribution of sample means is σx¯=1.7321

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 105 for x¯ , 100 for μx¯ and 1.7321 for σx¯

That is,

z=105−1001.7321=51.7321=2.89

The probability that the mean of sample of 75 is 105 or more is obtained by finding the area to the right of 2.89. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 2.89.

Procedure:

Locate 2.8 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<2.89)=0.9981

The area to the right of 2.89 is,

P(z>2.89)=1−P(z<2.89)=1−0.9981=0.0019

Thus, the probability that the mean of sample of 75 is 105 or more is 0.0019.

Interpretation:

Here, the probability that the mean of sample of 75 is 105 or more is less than 0.05. That is 0.0019<0.05 . Thus, the event mean of sample of 75 is 105 or more is unusual.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Whether the event mean of sample of 3 is 112 or more is unusual or not.

Answer to Problem 1UA

No, the event mean of sample of 3 is 112 or more is not unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 3.

Calculation:

The variable x represents random variable.

The notation x¯ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by μx¯ and σx¯ , respectively.

The distribution of sample means is normal with the parameters μx¯ and σx¯ by using central limit theorem.

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is, σx¯=σn .

Substitute 15 for σ and 3 for n

σx¯=153=151.732=8.6605

Thus, the standard deviation of the sampling distribution of sample means is σx¯=8.6605

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 112 for x¯ , 100 for μx¯ and 8.6605 for σx¯

That is,

z=112−1008.6605=128.6605=1.39

The probability that the mean of sample of 3 is 112 or more is obtained by finding the area to the right of 1.39. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 1.39.

Procedure:

Locate 1.3 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<1.39)=0.9177

The area to the right of 1.39 is,

P(z>1.39)=1−P(z<1.39)=1−0.9177=0.0823

Thus, the probability that the mean of sample of 3 is 112 or more is 0.0823.

Interpretation:

Here, the probability that the mean of sample of 3 is 112 or more is greater than 0.05. That is 0.0823>0.05 . Thus, the event mean of sample of 3 is 112 or more is not unusual.

b.

Expert Solution

To determine

Whether the event mean of sample of 75 is 105 or more is unusual or not.

Answer to Problem 1UA

Yes, the event mean of sample of 3 is 112 or more is unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 75.

Calculation:

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is,

σx¯=σn

Substitute 15 for σ and 75 for n

σx¯=1575=158.6602=1.7321

Thus, the standard deviation of the sampling distribution of sample means is σx¯=1.7321

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 105 for x¯ , 100 for μx¯ and 1.7321 for σx¯

That is,

z=105−1001.7321=51.7321=2.89

The probability that the mean of sample of 75 is 105 or more is obtained by finding the area to the right of 2.89. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 2.89.

Procedure:

Locate 2.8 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<2.89)=0.9981

The area to the right of 2.89 is,

P(z>2.89)=1−P(z<2.89)=1−0.9981=0.0019

Thus, the probability that the mean of sample of 75 is 105 or more is 0.0019.

Interpretation:

Here, the probability that the mean of sample of 75 is 105 or more is less than 0.05. That is 0.0019<0.05 . Thus, the event mean of sample of 75 is 105 or more is unusual.

Answer 8

a.

Expert Solution

To determine

Whether the event mean of sample of 3 is 112 or more is unusual or not.

Answer to Problem 1UA

No, the event mean of sample of 3 is 112 or more is not unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 3.

Calculation:

The variable x represents random variable.

The notation x¯ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by μx¯ and σx¯ , respectively.

The distribution of sample means is normal with the parameters μx¯ and σx¯ by using central limit theorem.

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is, σx¯=σn .

Substitute 15 for σ and 3 for n

σx¯=153=151.732=8.6605

Thus, the standard deviation of the sampling distribution of sample means is σx¯=8.6605

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 112 for x¯ , 100 for μx¯ and 8.6605 for σx¯

That is,

z=112−1008.6605=128.6605=1.39

The probability that the mean of sample of 3 is 112 or more is obtained by finding the area to the right of 1.39. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 1.39.

Procedure:

Locate 1.3 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<1.39)=0.9177

The area to the right of 1.39 is,

P(z>1.39)=1−P(z<1.39)=1−0.9177=0.0823

Thus, the probability that the mean of sample of 3 is 112 or more is 0.0823.

Interpretation:

Here, the probability that the mean of sample of 3 is 112 or more is greater than 0.05. That is 0.0823>0.05 . Thus, the event mean of sample of 3 is 112 or more is not unusual.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Whether the event mean of sample of 3 is 112 or more is unusual or not.

Answer 13

Answer to Problem 1UA

No, the event mean of sample of 3 is 112 or more is not unusual.

Answer 14

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 3.

Calculation:

The variable x represents random variable.

The notation x¯ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by μx¯ and σx¯ , respectively.

The distribution of sample means is normal with the parameters μx¯ and σx¯ by using central limit theorem.

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is, σx¯=σn .

Substitute 15 for σ and 3 for n

σx¯=153=151.732=8.6605

Thus, the standard deviation of the sampling distribution of sample means is σx¯=8.6605

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 112 for x¯ , 100 for μx¯ and 8.6605 for σx¯

That is,

z=112−1008.6605=128.6605=1.39

The probability that the mean of sample of 3 is 112 or more is obtained by finding the area to the right of 1.39. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 1.39.

Procedure:

Locate 1.3 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<1.39)=0.9177

The area to the right of 1.39 is,

P(z>1.39)=1−P(z<1.39)=1−0.9177=0.0823

Thus, the probability that the mean of sample of 3 is 112 or more is 0.0823.

Interpretation:

Here, the probability that the mean of sample of 3 is 112 or more is greater than 0.05. That is 0.0823>0.05 . Thus, the event mean of sample of 3 is 112 or more is not unusual.

Answer 15

b.

Expert Solution

To determine

Whether the event mean of sample of 75 is 105 or more is unusual or not.

Answer to Problem 1UA

Yes, the event mean of sample of 3 is 112 or more is unusual.

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 75.

Calculation:

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is,

σx¯=σn

Substitute 15 for σ and 75 for n

σx¯=1575=158.6602=1.7321

Thus, the standard deviation of the sampling distribution of sample means is σx¯=1.7321

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 105 for x¯ , 100 for μx¯ and 1.7321 for σx¯

That is,

z=105−1001.7321=51.7321=2.89

The probability that the mean of sample of 75 is 105 or more is obtained by finding the area to the right of 2.89. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 2.89.

Procedure:

Locate 2.8 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<2.89)=0.9981

The area to the right of 2.89 is,

P(z>2.89)=1−P(z<2.89)=1−0.9981=0.0019

Thus, the probability that the mean of sample of 75 is 105 or more is 0.0019.

Interpretation:

Here, the probability that the mean of sample of 75 is 105 or more is less than 0.05. That is 0.0019<0.05 . Thus, the event mean of sample of 75 is 105 or more is unusual.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Whether the event mean of sample of 75 is 105 or more is unusual or not.

Answer 20

Answer to Problem 1UA

Yes, the event mean of sample of 3 is 112 or more is unusual.

Answer 21

Explanation of Solution

Given info:

The population follows normal distribution with mean 100 and standard deviation 15. The sample of size is 75.

Calculation:

The mean of the sampling distribution of sample means is, μx¯=μ .

Here, the population mean is, μ=100 . Therefore, the sampling distribution of the sample mean is, μx¯=100 .

The formula to find the standard deviation of the sample means (σx¯) is,

σx¯=σn

Substitute 15 for σ and 75 for n

σx¯=1575=158.6602=1.7321

Thus, the standard deviation of the sampling distribution of sample means is σx¯=1.7321

The formula to convert the x¯ value into z score is,

z=x¯−μx¯σx¯

Substitute 105 for x¯ , 100 for μx¯ and 1.7321 for σx¯

That is,

z=105−1001.7321=51.7321=2.89

The probability that the mean of sample of 75 is 105 or more is obtained by finding the area to the right of 2.89. But, the Table 4: Standard normal distribution applies only for cumulative areas from the left.

Use Table 4: Standard normal distribution to find the area to the left of 2.89.

Procedure:

Locate 2.8 in the left column of the Table 4.
Obtain the value in the corresponding row below 0.09.

That is, P(z<2.89)=0.9981

The area to the right of 2.89 is,

P(z>2.89)=1−P(z<2.89)=1−0.9981=0.0019

Thus, the probability that the mean of sample of 75 is 105 or more is 0.0019.

Interpretation:

Here, the probability that the mean of sample of 75 is 105 or more is less than 0.05. That is 0.0019<0.05 . Thus, the event mean of sample of 75 is 105 or more is unusual.

Answer 22

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Is It Unusual? A population is normally distributed , with a mean of 100 and a standard deviation of 15. Determine whether cither event is unusual. Explain your reasoning. a. The mean of a sample of 3 is 112 or more. b. The mean of a sample of 75 is 105 or more.

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