Chapter 6.5, Problem 15P

(a)

To determine

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

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 $\mu$ is measured using the z-score or z value. The formula for z score is,

$z=\frac{x-\mu }{\sigma }$

In the formula, x is the raw score, $\mu$ is the mean and $\sigma$ is the standard deviation.

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

Substitute x as 40, $\mu$ as 85 and $\sigma$ as 25 in the formula of z score.

$\begin{array}{c}z=\frac{40-85}{25}\\ =\frac{-45}{25}\\ =-1.8\end{array}$

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,

$\begin{array}{c}P\left(x<-1.8\right)=P\left(z<-1.8\right)\\ =0.0359\end{array}$

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

(b)

To determine

Comment on the probability distribution of $\overline{x}$.

Find the probability that $\overline{x}<40$ when two tests taken about a week apart.

Answer to Problem 15P

The probability of $\overline{x}<40$ when two tests taken about a week apart is 0.0054.

Explanation of Solution

Calculation:

Sampling distribution:

For x distribution the sampling distribution of $\overline{x}$ is,

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\end{array}$

In the formula $\mu$ denotes the population mean, $\sigma$ denotes the population standard deviation, and n denotes the sample size.

The $\overline{x}$ can be converted into standard normal distribution as,

$z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

In the formula, $\overline{x}$ is the mean raw score, ${\mu }_{\overline{x}}$ is the mean of the sampling distribution and ${\sigma }_{\overline{x}}$ is the standard deviation of the sampling distribution.

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

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

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =85\end{array}$

The value of ${\mu }_{\overline{x}}$ is 85.

The standard deviation of the distribution is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{25}{\sqrt{2}}\\ =\frac{25}{1.4142}\\ =17.68\end{array}$

The value of ${\sigma }_{\overline{x}}$ is 17.68.

Substitute $\overline{x}$ as 40, ${\mu }_{\overline{x}}$ as 85 and ${\sigma }_{\overline{x}}$ as 17.68 in the formula of z score.

$\begin{array}{c}z=\frac{40-85}{17.68}\\ =\frac{-45}{17.68}\\ =-2.55\end{array}$

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,

$\begin{array}{c}P\left(\overline{x}<40\right)=P\left(z<-2.55\right)\\ =0.0054\end{array}$

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

(c)

To determine

Comment on the probability distribution of $\overline{x}$.

Find the probability of $\overline{x}<40$ when three tests taken about a week apart.

Answer to Problem 15P

The probability of $\overline{x}<40$ 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 $\overline{x}$ is normal regard less of the sample size considered. Hence, the $\overline{x}$ distribution of sample size 3 is normally distributed.

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

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =85\end{array}$

The value of ${\mu }_{\overline{x}}$ is 85.

The standard deviation of the distribution is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{25}{\sqrt{3}}\\ =\frac{25}{1.7321}\\ =14.43\end{array}$

The value of ${\sigma }_{\overline{x}}$ is 14.43.

Substitute $\overline{x}$ as 40, ${\mu }_{\overline{x}}$ as 85 and ${\sigma }_{\overline{x}}$ as 14.43 in the formula of z score.

$\begin{array}{c}z=\frac{40-85}{14.43}\\ =\frac{-45}{14.43}\\ =-3.12\end{array}$

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,

$\begin{array}{c}P\left(\overline{x}<40\right)=P\left(z<-3.12\right)\\ =0.0009\end{array}$

Hence, the probability of $\overline{x}<40$ when three tests taken about a week is 0.0009.

(d)

To determine

Comment on the probability distribution of $\overline{x}$.

Find the probability of $\overline{x}<40$ when five tests taken about a week apart.

Answer to Problem 15P

The probability of $\overline{x}<40$ 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 $\overline{x}$ is normal regard less of the sample size considered. Hence, the $\overline{x}$ distribution of sample size 5 is normally distributed.

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

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =85\end{array}$

The value of ${\mu }_{\overline{x}}$ is 85.

The standard deviation of the distribution is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{25}{\sqrt{5}}\\ =\frac{25}{2.2361}\\ =11.18\end{array}$

The value of ${\sigma }_{\overline{x}}$ is 11.18.

Substitute $\overline{x}$ as 40, ${\mu }_{\overline{x}}$ as 85 and ${\sigma }_{\overline{x}}$ as 11.18 in the formula of z score.

$\begin{array}{c}z=\frac{40-85}{11.18}\\ =\frac{-45}{11.18}\\ =-4.03\end{array}$

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,

$\begin{array}{c}P\left(\overline{x}<40\right)=P\left(z<-4.03\right)\\ =0.0000\end{array}$

Hence, the probability of $\overline{x}<40$ 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 $\overline{x}<40$ based on five tests is extremely rare event or more likely event.

Explanation of Solution

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

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

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

From part (d), the probability of $\overline{x}<40$ 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 $\overline{x}<40$ 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 $\overline{x}<40$ 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 $\overline{x}<40$ based on five tests is more likely event.