Scores from a questionnaire measuring social anxiety form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 10. What is the probability of obtaining a sample mean greater than M = 53. a. for a random sample of n = 4 people? b. for a random sample of n = 16 people? c. for a random sample of n = 25 people?

Question

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Answer 1

Textbook Question

Chapter 7, Problem 11P

Scores from a questionnaire measuring social anxiety form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 10. What is the probability of obtaining a sample mean greater than M = 53.

a. for a random sample of n = 4 people?
b. for a random sample of n = 16 people?
c. for a random sample of n = 25 people?

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

The probability of obtaining a sample mean greater than

$M=53$ for given sample.

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.2743.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=4$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=4M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=104=5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 1

From the SPSS output, $P(X¯<53)$ is 0.7257.

Using $(1)$ p is calculated as:

$p=1−P(X¯<53)=1−0.7257=0.2743$

Thus, the probability of obtaining a mean greater than $M=53$ for given sample is 0.2743.

b.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample of

$n=16$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.1151.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=16$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=16M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1016=2.5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2.5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 2

From the SPSS output, $P(X¯<53)$ is 0.8849.

Thus,

$p=1−P(X¯<53)=1−0.8849=0.1151$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.1151.

c.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample

$n=25$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.0668.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=25$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=25M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1025=2$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 3

From the SPSS output, $P(X¯<53)$ is 0.9332.

Thus,

$p=1−P(X¯<53)=1−0.9332=0.0668$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.0668.

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Answer 2

Textbook Question

Chapter 7, Problem 11P

Scores from a questionnaire measuring social anxiety form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 10. What is the probability of obtaining a sample mean greater than M = 53.

a. for a random sample of n = 4 people?
b. for a random sample of n = 16 people?
c. for a random sample of n = 25 people?

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

The probability of obtaining a sample mean greater than

$M=53$ for given sample.

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.2743.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=4$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=4M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=104=5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 1

From the SPSS output, $P(X¯<53)$ is 0.7257.

Using $(1)$ p is calculated as:

$p=1−P(X¯<53)=1−0.7257=0.2743$

Thus, the probability of obtaining a mean greater than $M=53$ for given sample is 0.2743.

b.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample of

$n=16$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.1151.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=16$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=16M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1016=2.5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2.5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 2

From the SPSS output, $P(X¯<53)$ is 0.8849.

Thus,

$p=1−P(X¯<53)=1−0.8849=0.1151$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.1151.

c.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample

$n=25$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.0668.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=25$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=25M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1025=2$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 3

From the SPSS output, $P(X¯<53)$ is 0.9332.

Thus,

$p=1−P(X¯<53)=1−0.9332=0.0668$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.0668.

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Answer 3

Textbook Question

Chapter 7, Problem 11P

Scores from a questionnaire measuring social anxiety form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 10. What is the probability of obtaining a sample mean greater than M = 53.

a. for a random sample of n = 4 people?
b. for a random sample of n = 16 people?
c. for a random sample of n = 25 people?

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

The probability of obtaining a sample mean greater than

$M=53$ for given sample.

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.2743.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=4$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=4M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=104=5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 1

From the SPSS output, $P(X¯<53)$ is 0.7257.

Using $(1)$ p is calculated as:

$p=1−P(X¯<53)=1−0.7257=0.2743$

Thus, the probability of obtaining a mean greater than $M=53$ for given sample is 0.2743.

b.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample of

$n=16$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.1151.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=16$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=16M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1016=2.5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2.5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 2

From the SPSS output, $P(X¯<53)$ is 0.8849.

Thus,

$p=1−P(X¯<53)=1−0.8849=0.1151$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.1151.

c.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample

$n=25$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.0668.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=25$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=25M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1025=2$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 3

From the SPSS output, $P(X¯<53)$ is 0.9332.

Thus,

$p=1−P(X¯<53)=1−0.9332=0.0668$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.0668.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 7, Problem 11P

Scores from a questionnaire measuring social anxiety form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 10. What is the probability of obtaining a sample mean greater than M = 53.

a. for a random sample of n = 4 people?
b. for a random sample of n = 16 people?
c. for a random sample of n = 25 people?

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

The probability of obtaining a sample mean greater than

$M=53$ for given sample.

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.2743.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=4$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=4M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=104=5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 1

From the SPSS output, $P(X¯<53)$ is 0.7257.

Using $(1)$ p is calculated as:

$p=1−P(X¯<53)=1−0.7257=0.2743$

Thus, the probability of obtaining a mean greater than $M=53$ for given sample is 0.2743.

b.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample of

$n=16$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.1151.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=16$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=16M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1016=2.5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2.5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 2

From the SPSS output, $P(X¯<53)$ is 0.8849.

Thus,

$p=1−P(X¯<53)=1−0.8849=0.1151$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.1151.

c.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample

$n=25$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.0668.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=25$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=25M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1025=2$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 3

From the SPSS output, $P(X¯<53)$ is 0.9332.

Thus,

$p=1−P(X¯<53)=1−0.9332=0.0668$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.0668.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample.

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.2743.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=4$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=4M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=104=5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 1

From the SPSS output, $P(X¯<53)$ is 0.7257.

Using $(1)$ p is calculated as:

$p=1−P(X¯<53)=1−0.7257=0.2743$

Thus, the probability of obtaining a mean greater than $M=53$ for given sample is 0.2743.

b.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample of

$n=16$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.1151.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=16$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=16M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1016=2.5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2.5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 2

From the SPSS output, $P(X¯<53)$ is 0.8849.

Thus,

$p=1−P(X¯<53)=1−0.8849=0.1151$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.1151.

c.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample

$n=25$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.0668.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=25$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=25M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1025=2$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 3

From the SPSS output, $P(X¯<53)$ is 0.9332.

Thus,

$p=1−P(X¯<53)=1−0.9332=0.0668$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.0668.

Answer 8

a.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample.

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.2743.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=4$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=4M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=104=5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 1

From the SPSS output, $P(X¯<53)$ is 0.7257.

Using $(1)$ p is calculated as:

$p=1−P(X¯<53)=1−0.7257=0.2743$

Thus, the probability of obtaining a mean greater than $M=53$ for given sample is 0.2743.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample.

Answer 13

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.2743.

Answer 14

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=4$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=4M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=104=5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 1

From the SPSS output, $P(X¯<53)$ is 0.7257.

Using $(1)$ p is calculated as:

$p=1−P(X¯<53)=1−0.7257=0.2743$

Thus, the probability of obtaining a mean greater than $M=53$ for given sample is 0.2743.

Answer 15

b.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample of

$n=16$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.1151.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=16$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=16M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1016=2.5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2.5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 2

From the SPSS output, $P(X¯<53)$ is 0.8849.

Thus,

$p=1−P(X¯<53)=1−0.8849=0.1151$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.1151.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample of

$n=16$ .

Answer 20

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.1151.

Answer 21

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=16$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=16M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1016=2.5$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2.5 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 2

From the SPSS output, $P(X¯<53)$ is 0.8849.

Thus,

$p=1−P(X¯<53)=1−0.8849=0.1151$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.1151.

Answer 22

c.

Expert Solution

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample

$n=25$ .

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.0668.

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=25$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=25M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1025=2$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2 in the braces of CDF.NORMAL
Choose OK.

Output using the SPSS software is given below:

Essentials of Statistics for The Behavioral Sciences (MindTap Course List), Chapter 7, Problem 11P , additional homework tip 3

From the SPSS output, $P(X¯<53)$ is 0.9332.

Thus,

$p=1−P(X¯<53)=1−0.9332=0.0668$

Hence, the probability of obtaining a mean greater than $M=53$ for given sample is 0.0668.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The probability of obtaining a sample mean greater than

$M=53$ for given sample

$n=25$ .

Answer 27

Answer to Problem 11P

The probability of obtaining a sample mean greater than $M=53$ for given sample is 0.0668.

Answer 28

Explanation of Solution

Given info:

Numbers of people in a random sample are $n=25$ .

Population mean is $μ=50$ .

Population standard deviation is $σ=10$ .

Sample mean is $M=53$ .

Calculation:

If $μ$ and $σ$ represents the population mean and standard deviation respectively. Let n represents numbers of scores in each sample. Let M represents given sample mean. Then,

$μ=50σ=10n=25M=53$

Let $X¯$ represents the random sample mean from given distribution. Let $μM$ and $σM$ represents mean and standard error of $X¯$ Then, expected mean of $X¯$ will be same as population mean and standard error is calculated as:

$μM=50σM=σn=1025=2$

Let p represents the probability that random mean is greater than $M=53$ .

$p=P(X¯>53)=1−P(X¯<53)$

Software procedure:

Step-by-step procedure to obtain the $P(X¯<53)$ using the SPSS software:

Click on first empty block of data view.
Go to Transform>Choose Compute variable>Enter Probability in Target Variable.
Choose CDF& Noncentral CDF under Function group> choose CDF.Normal> drag it to the numeric expression.
Enter 53,50,2 in the braces of CDF.NORMAL
Choose OK.