Find the probability that on a single test x < 40 .

Question

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Answer 1

Question

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.

Chapter 6.5, Problem 15P

(a)

To determine

Find the probability that on a single test $x<40$ .

(a)

Expert Solution

Answer to Problem 15P

The probability that a single test, $x<40$ is 0.0359.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $μ$ is measured using the z-score or z value. The formula for z score is,

$z=x−μσ$

In the formula, x is the raw score, $μ$ is the mean and $σ$ is the standard deviation.

The variable x denotes the level of glucose in the blood after a 12-hour fast.

Substitute x as 40, $μ$ as 85 and $σ$ as 25 in the formula of z score.

$z=40−8525=−4525=−1.8$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.8.

Locate the value –1.8 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0359.

The probability is,

$P(x<−1.8)=P(z<−1.8)=0.0359$

Hence, the probability that a single test, $x<40$ is 0.0359.

(b)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

(b)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $x¯$ is,

$μx¯=μσx¯=σn$

In the formula $μ$ denotes the population mean, $σ$ denotes the population standard deviation, and n denotes the sample size.

The $x¯$ can be converted into standard normal distribution as,

$z=x¯−μx¯σx¯$

In the formula, $x¯$ is the mean raw score, $μx¯$ is the mean of the sampling distribution and $σx¯$ is the standard deviation of the sampling distribution.

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 2 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=252=251.4142=17.68$

The value of $σx¯$ is 17.68.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 17.68 in the formula of z score.

$z=40−8517.68=−4517.68=−2.55$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.55.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0054.

The probability is,

$P(x¯<40)=P(z<−2.55)=0.0054$

Hence, the probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

(c)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

(c)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when three tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 3 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=253=251.7321=14.43$

The value of $σx¯$ is 14.43.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 14.43 in the formula of z score.

$z=40−8514.43=−4514.43=−3.12$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –3.12.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0009.

The probability is,

$P(x¯<40)=P(z<−3.12)=0.0009$

Hence, the probability of $x¯<40$ when three tests taken about a week is 0.0009.

(d)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

(d)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when five tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 5 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=255=252.2361=11.18$

The value of $σx¯$ is 11.18.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 11.18 in the formula of z score.

$z=40−8511.18=−4511.18=−4.03$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –4.03.

For values of z less than –3.49 the probability is approximately 0.000.

The probability is,

$P(x¯<40)=P(z<−4.03)=0.0000$

Hence, the probability of $x¯<40$ when five tests taken about a week is approximately 0.0000.

(e)

To determine

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

(e)

Expert Solution

Explanation of Solution

From part (a), the probability that a single test, $x<40$ is 0.0359.

From part (b), the probability of $x¯<40$ for $n=2$ tests taken about a week apart is 0.0054.

From part (c), the probability of $x¯<40$ for $n=3$ tests taken about a week apart is 0.0009.

From part (d), the probability of $x¯<40$ for $n=5$ tests taken about a week apart is 0.0000.

It can be observed that, as the number of tests taken about a week apart is increasing the probability of $x¯<40$ is decreasing.

The variable x denotes the level of glucose in the blood after a 12-hour fast. If the value of x is less than 40 then there would be severe excess insulin. If 5 tests were taken in a week then the probability value for $x¯<40$ is 0.000. That is, about 0% of times x would be less than 40 if 5 tests were taken indicating that it is certain to have excess insulin.

Hence, the test result of $x¯<40$ based on five tests is more likely event.

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Answer 2

Question

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.

Chapter 6.5, Problem 15P

(a)

To determine

Find the probability that on a single test $x<40$ .

(a)

Expert Solution

Answer to Problem 15P

The probability that a single test, $x<40$ is 0.0359.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $μ$ is measured using the z-score or z value. The formula for z score is,

$z=x−μσ$

In the formula, x is the raw score, $μ$ is the mean and $σ$ is the standard deviation.

The variable x denotes the level of glucose in the blood after a 12-hour fast.

Substitute x as 40, $μ$ as 85 and $σ$ as 25 in the formula of z score.

$z=40−8525=−4525=−1.8$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.8.

Locate the value –1.8 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0359.

The probability is,

$P(x<−1.8)=P(z<−1.8)=0.0359$

Hence, the probability that a single test, $x<40$ is 0.0359.

(b)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

(b)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $x¯$ is,

$μx¯=μσx¯=σn$

In the formula $μ$ denotes the population mean, $σ$ denotes the population standard deviation, and n denotes the sample size.

The $x¯$ can be converted into standard normal distribution as,

$z=x¯−μx¯σx¯$

In the formula, $x¯$ is the mean raw score, $μx¯$ is the mean of the sampling distribution and $σx¯$ is the standard deviation of the sampling distribution.

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 2 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=252=251.4142=17.68$

The value of $σx¯$ is 17.68.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 17.68 in the formula of z score.

$z=40−8517.68=−4517.68=−2.55$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.55.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0054.

The probability is,

$P(x¯<40)=P(z<−2.55)=0.0054$

Hence, the probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

(c)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

(c)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when three tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 3 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=253=251.7321=14.43$

The value of $σx¯$ is 14.43.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 14.43 in the formula of z score.

$z=40−8514.43=−4514.43=−3.12$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –3.12.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0009.

The probability is,

$P(x¯<40)=P(z<−3.12)=0.0009$

Hence, the probability of $x¯<40$ when three tests taken about a week is 0.0009.

(d)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

(d)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when five tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 5 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=255=252.2361=11.18$

The value of $σx¯$ is 11.18.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 11.18 in the formula of z score.

$z=40−8511.18=−4511.18=−4.03$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –4.03.

For values of z less than –3.49 the probability is approximately 0.000.

The probability is,

$P(x¯<40)=P(z<−4.03)=0.0000$

Hence, the probability of $x¯<40$ when five tests taken about a week is approximately 0.0000.

(e)

To determine

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

(e)

Expert Solution

Explanation of Solution

From part (a), the probability that a single test, $x<40$ is 0.0359.

From part (b), the probability of $x¯<40$ for $n=2$ tests taken about a week apart is 0.0054.

From part (c), the probability of $x¯<40$ for $n=3$ tests taken about a week apart is 0.0009.

From part (d), the probability of $x¯<40$ for $n=5$ tests taken about a week apart is 0.0000.

It can be observed that, as the number of tests taken about a week apart is increasing the probability of $x¯<40$ is decreasing.

The variable x denotes the level of glucose in the blood after a 12-hour fast. If the value of x is less than 40 then there would be severe excess insulin. If 5 tests were taken in a week then the probability value for $x¯<40$ is 0.000. That is, about 0% of times x would be less than 40 if 5 tests were taken indicating that it is certain to have excess insulin.

Hence, the test result of $x¯<40$ based on five tests is more likely event.

Want to see more full solutions like this?

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

Answer 3

Question

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.

Chapter 6.5, Problem 15P

(a)

To determine

Find the probability that on a single test $x<40$ .

(a)

Expert Solution

Answer to Problem 15P

The probability that a single test, $x<40$ is 0.0359.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $μ$ is measured using the z-score or z value. The formula for z score is,

$z=x−μσ$

In the formula, x is the raw score, $μ$ is the mean and $σ$ is the standard deviation.

The variable x denotes the level of glucose in the blood after a 12-hour fast.

Substitute x as 40, $μ$ as 85 and $σ$ as 25 in the formula of z score.

$z=40−8525=−4525=−1.8$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.8.

Locate the value –1.8 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0359.

The probability is,

$P(x<−1.8)=P(z<−1.8)=0.0359$

Hence, the probability that a single test, $x<40$ is 0.0359.

(b)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

(b)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $x¯$ is,

$μx¯=μσx¯=σn$

In the formula $μ$ denotes the population mean, $σ$ denotes the population standard deviation, and n denotes the sample size.

The $x¯$ can be converted into standard normal distribution as,

$z=x¯−μx¯σx¯$

In the formula, $x¯$ is the mean raw score, $μx¯$ is the mean of the sampling distribution and $σx¯$ is the standard deviation of the sampling distribution.

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 2 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=252=251.4142=17.68$

The value of $σx¯$ is 17.68.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 17.68 in the formula of z score.

$z=40−8517.68=−4517.68=−2.55$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.55.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0054.

The probability is,

$P(x¯<40)=P(z<−2.55)=0.0054$

Hence, the probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

(c)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

(c)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when three tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 3 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=253=251.7321=14.43$

The value of $σx¯$ is 14.43.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 14.43 in the formula of z score.

$z=40−8514.43=−4514.43=−3.12$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –3.12.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0009.

The probability is,

$P(x¯<40)=P(z<−3.12)=0.0009$

Hence, the probability of $x¯<40$ when three tests taken about a week is 0.0009.

(d)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

(d)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when five tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 5 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=255=252.2361=11.18$

The value of $σx¯$ is 11.18.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 11.18 in the formula of z score.

$z=40−8511.18=−4511.18=−4.03$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –4.03.

For values of z less than –3.49 the probability is approximately 0.000.

The probability is,

$P(x¯<40)=P(z<−4.03)=0.0000$

Hence, the probability of $x¯<40$ when five tests taken about a week is approximately 0.0000.

(e)

To determine

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

(e)

Expert Solution

Explanation of Solution

From part (a), the probability that a single test, $x<40$ is 0.0359.

From part (b), the probability of $x¯<40$ for $n=2$ tests taken about a week apart is 0.0054.

From part (c), the probability of $x¯<40$ for $n=3$ tests taken about a week apart is 0.0009.

From part (d), the probability of $x¯<40$ for $n=5$ tests taken about a week apart is 0.0000.

It can be observed that, as the number of tests taken about a week apart is increasing the probability of $x¯<40$ is decreasing.

The variable x denotes the level of glucose in the blood after a 12-hour fast. If the value of x is less than 40 then there would be severe excess insulin. If 5 tests were taken in a week then the probability value for $x¯<40$ is 0.000. That is, about 0% of times x would be less than 40 if 5 tests were taken indicating that it is certain to have excess insulin.

Hence, the test result of $x¯<40$ based on five tests is more likely event.

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

Answer 4

Question

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.

Chapter 6.5, Problem 15P

(a)

To determine

Find the probability that on a single test $x<40$ .

(a)

Expert Solution

Answer to Problem 15P

The probability that a single test, $x<40$ is 0.0359.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $μ$ is measured using the z-score or z value. The formula for z score is,

$z=x−μσ$

In the formula, x is the raw score, $μ$ is the mean and $σ$ is the standard deviation.

The variable x denotes the level of glucose in the blood after a 12-hour fast.

Substitute x as 40, $μ$ as 85 and $σ$ as 25 in the formula of z score.

$z=40−8525=−4525=−1.8$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.8.

Locate the value –1.8 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0359.

The probability is,

$P(x<−1.8)=P(z<−1.8)=0.0359$

Hence, the probability that a single test, $x<40$ is 0.0359.

(b)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

(b)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $x¯$ is,

$μx¯=μσx¯=σn$

In the formula $μ$ denotes the population mean, $σ$ denotes the population standard deviation, and n denotes the sample size.

The $x¯$ can be converted into standard normal distribution as,

$z=x¯−μx¯σx¯$

In the formula, $x¯$ is the mean raw score, $μx¯$ is the mean of the sampling distribution and $σx¯$ is the standard deviation of the sampling distribution.

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 2 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=252=251.4142=17.68$

The value of $σx¯$ is 17.68.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 17.68 in the formula of z score.

$z=40−8517.68=−4517.68=−2.55$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.55.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0054.

The probability is,

$P(x¯<40)=P(z<−2.55)=0.0054$

Hence, the probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

(c)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

(c)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when three tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 3 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=253=251.7321=14.43$

The value of $σx¯$ is 14.43.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 14.43 in the formula of z score.

$z=40−8514.43=−4514.43=−3.12$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –3.12.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0009.

The probability is,

$P(x¯<40)=P(z<−3.12)=0.0009$

Hence, the probability of $x¯<40$ when three tests taken about a week is 0.0009.

(d)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

(d)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when five tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 5 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=255=252.2361=11.18$

The value of $σx¯$ is 11.18.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 11.18 in the formula of z score.

$z=40−8511.18=−4511.18=−4.03$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –4.03.

For values of z less than –3.49 the probability is approximately 0.000.

The probability is,

$P(x¯<40)=P(z<−4.03)=0.0000$

Hence, the probability of $x¯<40$ when five tests taken about a week is approximately 0.0000.

(e)

To determine

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

(e)

Expert Solution

Explanation of Solution

From part (a), the probability that a single test, $x<40$ is 0.0359.

From part (b), the probability of $x¯<40$ for $n=2$ tests taken about a week apart is 0.0054.

From part (c), the probability of $x¯<40$ for $n=3$ tests taken about a week apart is 0.0009.

From part (d), the probability of $x¯<40$ for $n=5$ tests taken about a week apart is 0.0000.

It can be observed that, as the number of tests taken about a week apart is increasing the probability of $x¯<40$ is decreasing.

The variable x denotes the level of glucose in the blood after a 12-hour fast. If the value of x is less than 40 then there would be severe excess insulin. If 5 tests were taken in a week then the probability value for $x¯<40$ is 0.000. That is, about 0% of times x would be less than 40 if 5 tests were taken indicating that it is certain to have excess insulin.

Hence, the test result of $x¯<40$ based on five tests is more likely event.

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Answer 5

Chapter 6.5, Problem 15P

(a)

To determine

Find the probability that on a single test $x<40$ .

(a)

Expert Solution

Answer to Problem 15P

The probability that a single test, $x<40$ is 0.0359.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $μ$ is measured using the z-score or z value. The formula for z score is,

$z=x−μσ$

In the formula, x is the raw score, $μ$ is the mean and $σ$ is the standard deviation.

The variable x denotes the level of glucose in the blood after a 12-hour fast.

Substitute x as 40, $μ$ as 85 and $σ$ as 25 in the formula of z score.

$z=40−8525=−4525=−1.8$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.8.

Locate the value –1.8 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0359.

The probability is,

$P(x<−1.8)=P(z<−1.8)=0.0359$

Hence, the probability that a single test, $x<40$ is 0.0359.

(b)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

(b)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $x¯$ is,

$μx¯=μσx¯=σn$

In the formula $μ$ denotes the population mean, $σ$ denotes the population standard deviation, and n denotes the sample size.

The $x¯$ can be converted into standard normal distribution as,

$z=x¯−μx¯σx¯$

In the formula, $x¯$ is the mean raw score, $μx¯$ is the mean of the sampling distribution and $σx¯$ is the standard deviation of the sampling distribution.

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 2 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=252=251.4142=17.68$

The value of $σx¯$ is 17.68.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 17.68 in the formula of z score.

$z=40−8517.68=−4517.68=−2.55$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.55.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0054.

The probability is,

$P(x¯<40)=P(z<−2.55)=0.0054$

Hence, the probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

(c)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

(c)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when three tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 3 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=253=251.7321=14.43$

The value of $σx¯$ is 14.43.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 14.43 in the formula of z score.

$z=40−8514.43=−4514.43=−3.12$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –3.12.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0009.

The probability is,

$P(x¯<40)=P(z<−3.12)=0.0009$

Hence, the probability of $x¯<40$ when three tests taken about a week is 0.0009.

(d)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

(d)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when five tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 5 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=255=252.2361=11.18$

The value of $σx¯$ is 11.18.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 11.18 in the formula of z score.

$z=40−8511.18=−4511.18=−4.03$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –4.03.

For values of z less than –3.49 the probability is approximately 0.000.

The probability is,

$P(x¯<40)=P(z<−4.03)=0.0000$

Hence, the probability of $x¯<40$ when five tests taken about a week is approximately 0.0000.

(e)

To determine

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

(e)

Expert Solution

Explanation of Solution

From part (a), the probability that a single test, $x<40$ is 0.0359.

From part (b), the probability of $x¯<40$ for $n=2$ tests taken about a week apart is 0.0054.

From part (c), the probability of $x¯<40$ for $n=3$ tests taken about a week apart is 0.0009.

From part (d), the probability of $x¯<40$ for $n=5$ tests taken about a week apart is 0.0000.

It can be observed that, as the number of tests taken about a week apart is increasing the probability of $x¯<40$ is decreasing.

The variable x denotes the level of glucose in the blood after a 12-hour fast. If the value of x is less than 40 then there would be severe excess insulin. If 5 tests were taken in a week then the probability value for $x¯<40$ is 0.000. That is, about 0% of times x would be less than 40 if 5 tests were taken indicating that it is certain to have excess insulin.

Hence, the test result of $x¯<40$ based on five tests is more likely event.

Answer 6

Chapter 6.5, Problem 15P

(a)

To determine

Find the probability that on a single test $x<40$ .

(a)

Expert Solution

Answer to Problem 15P

The probability that a single test, $x<40$ is 0.0359.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $μ$ is measured using the z-score or z value. The formula for z score is,

$z=x−μσ$

In the formula, x is the raw score, $μ$ is the mean and $σ$ is the standard deviation.

The variable x denotes the level of glucose in the blood after a 12-hour fast.

Substitute x as 40, $μ$ as 85 and $σ$ as 25 in the formula of z score.

$z=40−8525=−4525=−1.8$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.8.

Locate the value –1.8 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0359.

The probability is,

$P(x<−1.8)=P(z<−1.8)=0.0359$

Hence, the probability that a single test, $x<40$ is 0.0359.

Answer 7

Chapter 6.5, Problem 15P

(a)

To determine

Find the probability that on a single test $x<40$ .

Answer 8

(a)

Expert Solution

Answer to Problem 15P

The probability that a single test, $x<40$ is 0.0359.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $μ$ is measured using the z-score or z value. The formula for z score is,

$z=x−μσ$

In the formula, x is the raw score, $μ$ is the mean and $σ$ is the standard deviation.

The variable x denotes the level of glucose in the blood after a 12-hour fast.

Substitute x as 40, $μ$ as 85 and $σ$ as 25 in the formula of z score.

$z=40−8525=−4525=−1.8$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.8.

Locate the value –1.8 in column z.
Locate the value 0.00 in top row.
The intersecting value of row and column is 0.0359.

The probability is,

$P(x<−1.8)=P(z<−1.8)=0.0359$

Hence, the probability that a single test, $x<40$ is 0.0359.

Answer 13

(b)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

(b)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $x¯$ is,

$μx¯=μσx¯=σn$

In the formula $μ$ denotes the population mean, $σ$ denotes the population standard deviation, and n denotes the sample size.

The $x¯$ can be converted into standard normal distribution as,

$z=x¯−μx¯σx¯$

In the formula, $x¯$ is the mean raw score, $μx¯$ is the mean of the sampling distribution and $σx¯$ is the standard deviation of the sampling distribution.

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 2 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=252=251.4142=17.68$

The value of $σx¯$ is 17.68.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 17.68 in the formula of z score.

$z=40−8517.68=−4517.68=−2.55$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.55.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0054.

The probability is,

$P(x¯<40)=P(z<−2.55)=0.0054$

Hence, the probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Answer 14

(b)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

Answer 15

(b)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $x¯$ is,

$μx¯=μσx¯=σn$

In the formula $μ$ denotes the population mean, $σ$ denotes the population standard deviation, and n denotes the sample size.

The $x¯$ can be converted into standard normal distribution as,

$z=x¯−μx¯σx¯$

In the formula, $x¯$ is the mean raw score, $μx¯$ is the mean of the sampling distribution and $σx¯$ is the standard deviation of the sampling distribution.

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 2 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=252=251.4142=17.68$

The value of $σx¯$ is 17.68.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 17.68 in the formula of z score.

$z=40−8517.68=−4517.68=−2.55$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.55.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0054.

The probability is,

$P(x¯<40)=P(z<−2.55)=0.0054$

Hence, the probability of $x¯<40$ when two tests taken about a week apart is 0.0054.

Answer 20

(c)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

(c)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when three tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 3 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=253=251.7321=14.43$

The value of $σx¯$ is 14.43.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 14.43 in the formula of z score.

$z=40−8514.43=−4514.43=−3.12$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –3.12.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0009.

The probability is,

$P(x¯<40)=P(z<−3.12)=0.0009$

Hence, the probability of $x¯<40$ when three tests taken about a week is 0.0009.

Answer 21

(c)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

Answer 22

(c)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when three tests taken about a week is 0.0009.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 3 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=253=251.7321=14.43$

The value of $σx¯$ is 14.43.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 14.43 in the formula of z score.

$z=40−8514.43=−4514.43=−3.12$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –3.12.

Locate the value –2.5 in column z.
Locate the value 0.05 in top row.
The intersecting value of row and column is 0.0009.

The probability is,

$P(x¯<40)=P(z<−3.12)=0.0009$

Hence, the probability of $x¯<40$ when three tests taken about a week is 0.0009.

Answer 27

(d)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

(d)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when five tests taken about a week is 0.0009.

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 5 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=255=252.2361=11.18$

The value of $σx¯$ is 11.18.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 11.18 in the formula of z score.

$z=40−8511.18=−4511.18=−4.03$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –4.03.

For values of z less than –3.49 the probability is approximately 0.000.

The probability is,

$P(x¯<40)=P(z<−4.03)=0.0000$

Hence, the probability of $x¯<40$ when five tests taken about a week is approximately 0.0000.

Answer 28

(d)

To determine

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

Answer 29

(d)

Expert Solution

Answer to Problem 15P

The probability of $x¯<40$ when five tests taken about a week is 0.0009.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

When the x distribution follows normal distribution the sampling distribution of $x¯$ is normal regard less of the sample size considered. Hence, the $x¯$ distribution of sample size 5 is normally distributed.

The mean of the x distribution is $μ=85$ . The sampling distribution has the mean same as the population x distribution. That is,

$μx¯=μ=85$

The value of $μx¯$ is 85.

The standard deviation of the distribution is,

$σx¯=255=252.2361=11.18$

The value of $σx¯$ is 11.18.

Substitute $x¯$ as 40, $μx¯$ as 85 and $σx¯$ as 11.18 in the formula of z score.

$z=40−8511.18=−4511.18=−4.03$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –4.03.

For values of z less than –3.49 the probability is approximately 0.000.

The probability is,

$P(x¯<40)=P(z<−4.03)=0.0000$

Hence, the probability of $x¯<40$ when five tests taken about a week is approximately 0.0000.

Answer 34

(e)

To determine

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

(e)

Expert Solution

Explanation of Solution

From part (a), the probability that a single test, $x<40$ is 0.0359.

From part (b), the probability of $x¯<40$ for $n=2$ tests taken about a week apart is 0.0054.

From part (c), the probability of $x¯<40$ for $n=3$ tests taken about a week apart is 0.0009.

From part (d), the probability of $x¯<40$ for $n=5$ tests taken about a week apart is 0.0000.

It can be observed that, as the number of tests taken about a week apart is increasing the probability of $x¯<40$ is decreasing.

The variable x denotes the level of glucose in the blood after a 12-hour fast. If the value of x is less than 40 then there would be severe excess insulin. If 5 tests were taken in a week then the probability value for $x¯<40$ is 0.000. That is, about 0% of times x would be less than 40 if 5 tests were taken indicating that it is certain to have excess insulin.

Hence, the test result of $x¯<40$ based on five tests is more likely event.

Answer 35

(e)

To determine

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

Answer 36

(e)

Expert Solution

Explanation of Solution

From part (a), the probability that a single test, $x<40$ is 0.0359.

From part (b), the probability of $x¯<40$ for $n=2$ tests taken about a week apart is 0.0054.

From part (c), the probability of $x¯<40$ for $n=3$ tests taken about a week apart is 0.0009.

From part (d), the probability of $x¯<40$ for $n=5$ tests taken about a week apart is 0.0000.

It can be observed that, as the number of tests taken about a week apart is increasing the probability of $x¯<40$ is decreasing.

The variable x denotes the level of glucose in the blood after a 12-hour fast. If the value of x is less than 40 then there would be severe excess insulin. If 5 tests were taken in a week then the probability value for $x¯<40$ is 0.000. That is, about 0% of times x would be less than 40 if 5 tests were taken indicating that it is certain to have excess insulin.

Hence, the test result of $x¯<40$ based on five tests is more likely event.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

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Find the probability that on a single test x < 40 .

Concept explainers

Videos

Find the probability that on a single test $x<40$ .

Answer to Problem 15P

Explanation of Solution

Comment on the probability distribution of $x¯$ .

Find the probability that $x¯<40$ when two tests taken about a week apart.

Answer to Problem 15P

Explanation of Solution

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when three tests taken about a week apart.

Answer to Problem 15P

Explanation of Solution

Comment on the probability distribution of $x¯$ .

Find the probability of $x¯<40$ when five tests taken about a week apart.

Answer to Problem 15P

Explanation of Solution

Identify whether the probabilities decreased as n increased or not.

Explain whether the test result of $x¯<40$ based on five tests is extremely rare event or more likely event.

Explanation of Solution

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Chapter 6 Solutions

Find the probability that on a single test x &lt; 40 .

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Find the probability that on a single test x<40<m:math><m:mrow><m:mi>x</m:mi><m:mo>&lt;</m:mo><m:mn>40</m:mn></m:mrow></m:math>.

Answer to Problem 15P

Explanation of Solution

Answer to Problem 15P

Explanation of Solution

Answer to Problem 15P

Explanation of Solution

Answer to Problem 15P

Explanation of Solution

Explanation of Solution

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Chapter 6 Solutions

Find the probability that on a single test x < 40 .

Find the probability that on a single test $x<40$ .