Chapter 8.2, Problem 17E

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm.

1. a. If the distribution of interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x̄ for these 25 will be between 64 and 67 mm? at least 68 mm? (Hint: See Examples 8.5 and 8.6.)
2. b. Suppose that a random sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 mm? at least 68 mm?

To determine

Find the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males will be between 64 mm and 67 mm.

Find the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males at least 68 mm.

Answer to Problem 17E

The probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males will be between 64 mm and 67 mm is 0.8185.

The probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males at least 68 mm is 0.0013.

Explanation of Solution

Calculation:

It is given that, a random sample of size of 25 adult males is selected from a population with mean value of interpupillary distance $\mu =65\text{mm}$ and standard deviation of $\sigma =5\text{mm}$.

Assume that a random variable X is drawn from a population with mean $\mu$ and standard deviation $\sigma$. Now, the sample mean $\overline{x}$ for sample of size n has mean ${\mu }_{\overline{x}}\left(=\mu \right)$ and standard deviation ${\sigma }_{\overline{x}}\left(=\frac{\sigma }{\sqrt{n}}\right)$.

It is also given that the distribution of interpupillary distance is normal.

Consider, that the random variable x denotes the interpupillary distance of adult male.

Thus, for $n=25$ the mean of $\overline{x}$ is,

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =65\end{array}$.

Thus, for $n=25$ the standard deviation of $\overline{x}$ is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{5}{\sqrt{25}}\\ =\frac{5}{5}\\ =1\end{array}$

Therefore, the mean and standard deviation of sampling distribution of $\overline{x}$ for sample of size 25 are 65 and 1, respectively.

The z-score of sampling mean $\overline{x}$ is defined as $z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$.

Now, the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males will be between 64 mm and 67 mm, implies that, $P\left(64<\overline{x}<67\right)$.

Thus,

$\begin{array}{c}P\left(64<\overline{x}<67\right)=P\left(\frac{64-65}{\frac{5}{\sqrt{25}}}<\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}<\frac{67-65}{\frac{5}{\sqrt{25}}}\right)\\ =P\left(\frac{-1}{1}<z<\frac{2}{1}\right)\\ =P\left(-1<z<2\right)\\ =P\left(z<2\right)-P\left(z<-1\right)\end{array}$

The step by step procedure to obtain $P\left(z<2\right)$ from the table “Standard Normal Probabilities (Cumulative z Curve Areas)” is given below:

• Consider the 1^st column of .00 from the table.
• Consider the row of 2.0 from the table.
• Choose the intersection value of 1^st column and row of 2.0.

Hence, $P\left(z<2\right)=0.9772$.

The step by step procedure to obtain $P\left(z<-1\right)$ from the table “Standard Normal Probabilities (Cumulative z Curve Areas)” is given below:

• Consider the 1^st column of .00 from the table.
• Consider the row of –1.0, from the table.
• Choose the intersection value of 1^st column and row of –1.0.

Hence, $P\left(z<-1\right)=0.1587$.

Thus,

$\begin{array}{c}P\left(64<\overline{x}<67\right)=P\left(z<2\right)-P\left(z<-1\right)\\ =0.9772-0.1587\\ =0.8185\end{array}$

Therefore, the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males will be between 64 mm and 67 mm is 0.8185.

Now, the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males at least 68 mm, implies that, $P\left(\overline{x}\ge 68\right)$.

Thus,

$\begin{array}{c}P\left(\overline{x}\ge 68\right)=P\left(\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\ge \frac{68-65}{\frac{5}{\sqrt{25}}}\right)\\ =P\left(z\ge \frac{3}{1}\right)\\ =1-P\left(z<3\right)\end{array}$

The step by step procedure to obtain $P\left(z<3\right)$ from the table “Standard Normal Probabilities (Cumulative z Curve Areas)” is given below:

• Consider the 1^st column of .00 from the table.
• Consider the row of 3.0 from the table.
• Choose the intersection value of 1^st column and row of 2.0.

Hence, $P\left(z<3\right)=0.9987$.

Thus,

$\begin{array}{c}P\left(\overline{x}\ge 68\right)=1-P\left(z<3\right)\\ =1-0.9987\\ =0.0013\end{array}$

Therefore, the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males at least 68 mm is 0.0013.

To determine

Find the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males will be between 64 mm and 67 mm.

Find the probability that the sample mean distance $\overline{x}$ for the randomly selected 25 adult males at least 68 mm.

Answer to Problem 17E

The probability that the sample mean distance $\overline{x}$ for the randomly selected 100 adult males will be between 64 mm and 67 mm is 0.9772.

The probability that the sample mean distance $\overline{x}$ for the randomly selected 100 adult males at least 68 mm is 0.

Explanation of Solution

Calculation:

It is given that a random sample of 100 adult males is selected and the distribution of interpupillary distance does not follow normal distribution.

It is known that, for a sample size equal to or more than 30, the sampling distribution would follows normal distribution approximately.

Thus, as the sample size is 100$\left(>30\right)$, then the distribution of $\overline{x}$ will follow approximately normal distribution.

Thus, for $n=100$ the mean of $\overline{x}$ is,

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =65\end{array}$.

Thus, for $n=100$ the standard deviation of $\overline{x}$ is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{5}{\sqrt{100}}\\ =\frac{5}{10}\\ =0.5\end{array}$

Therefore, the mean and standard deviation of sampling distribution of $\overline{x}$ for sample of size 100 are 65 and 0.5, respectively.

Now, the probability that the sample mean distance $\overline{x}$ for the randomly selected 100 adult males will be between 64 mm and 67 mm, implies that, $P\left(64<\overline{x}<67\right)$.

Thus,

$\begin{array}{c}P\left(64<\overline{x}<67\right)=P\left(\frac{64-65}{0.5}<\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}<\frac{67-65}{0.5}\right)\\ =P\left(\frac{-1}{0.5}<z<\frac{2}{0.5}\right)\\ =P\left(-2<z<4\right)\\ =P\left(z<4\right)-P\left(z<-2\right)\end{array}$

From the table “Standard Normal Probabilities (Cumulative z Curve Areas)” it is found that $P\left(z<3.89\right)=1$, which is maximum according to this table.

However, $P\left(z<3.89\right)<P\left(z<4\right)$.

Thus, $P\left(z<4\right)\approx 1$.

The step by step procedure to obtain $P\left(z<-2\right)$ from the table “Standard Normal Probabilities (Cumulative z Curve Areas)” is given below:

• Consider the 1^st column of .00 from the table.
• Consider the row of –2.0, from the table.
• Choose the intersection value of 1^st column and row of –2.0.

Hence, $P\left(z<-2\right)=0.0228$.

Thus,

$\begin{array}{c}P\left(64<\overline{x}<67\right)=P\left(z<4\right)-P\left(z<-2\right)\\ =1-0.0228\\ =0.9772\end{array}$

Therefore, the probability that the sample mean distance $\overline{x}$ for the randomly selected 100 adult males will be between 64 mm and 67 mm is 0.9772.

Now, the probability that the sample mean distance $\overline{x}$ for the randomly selected 100 adult males at least 68 mm, implies that, $P\left(\overline{x}\ge 68\right)$.

Thus,

$\begin{array}{c}P\left(\overline{x}\ge 68\right)=P\left(\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\ge \frac{68-65}{0.5}\right)\\ =P\left(z\ge \frac{3}{0.5}\right)\\ =1-P\left(z<6\right)\end{array}$

From the table “Standard Normal Probabilities (Cumulative z Curve Areas)” it is found that $P\left(z<3.89\right)=1$, which is maximum according to this table.

However, $P\left(z<3.89\right)<P\left(z<6\right)$.

Thus, $P\left(z<6\right)\approx 1$.

Thus,

$\begin{array}{c}P\left(\overline{x}\ge 68\right)=1-P\left(z<6\right)\\ =1-1\\ =0\end{array}$

Therefore, the probability that the sample mean distance $\overline{x}$ for the randomly selected 100 adult males at least 68 mm is 0.