Chapter 5, Problem 1UA

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

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

To determine

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

Answer to Problem 1UA

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

Explanation of Solution

Given info:

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

Calculation:

The variable x represents random variable.

The notation $\overline{x}$ represents mean of the random variable.

The mean and standard deviation of the sampling distribution of sample means is represented by ${\mu }_{\overline{x}}$ and ${\sigma }_{\overline{x}}$, respectively.

The distribution of sample means is normal with the parameters ${\mu }_{\overline{x}}$ and ${\sigma }_{\overline{x}}$ by using central limit theorem.

The mean of the sampling distribution of sample means is, ${\mu }_{\overline{x}}=\mu$.

Here, the population mean is, $\mu =100$. Therefore, the sampling distribution of the sample mean is, ${\mu }_{\overline{x}}=100$.

The formula to find the standard deviation of the sample means $\left({\sigma }_{\overline{x}}\right)$ is, ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Substitute 15 for $\sigma$ and 3 for n

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{15}{\sqrt{3}}\\ =\frac{15}{1.732}\\ =8.6605\end{array}$

Thus, the standard deviation of the sampling distribution of sample means is ${\sigma }_{\overline{x}}=8.6605$

The formula to convert the $\overline{x}$ value into z score is,

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

Substitute 112 for $\overline{x}$, 100 for ${\mu }_{\overline{x}}$ and 8.6605 for ${\sigma }_{\overline{x}}$

That is,

$\begin{array}{c}z=\frac{112-100}{8.6605}\\ =\frac{12}{8.6605}\\ =1.39\end{array}$

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

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

Procedure:

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

That is, $P\left(z<1.39\right)=0.9177$

The area to the right of 1.39 is,

$\begin{array}{c}P\left(z>1.39\right)=1-P\left(z<1.39\right)\\ =1-0.9177\\ =0.0823\end{array}$

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

Interpretation:

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

To determine

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

Answer to Problem 1UA

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

Explanation of Solution

Given info:

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

Calculation:

The mean of the sampling distribution of sample means is, ${\mu }_{\overline{x}}=\mu$.

Here, the population mean is, $\mu =100$. Therefore, the sampling distribution of the sample mean is, ${\mu }_{\overline{x}}=100$.

The formula to find the standard deviation of the sample means $\left({\sigma }_{\overline{x}}\right)$ is,

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Substitute 15 for $\sigma$ and 75 for n

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{15}{\sqrt{75}}\\ =\frac{15}{8.6602}\\ =1.7321\end{array}$

Thus, the standard deviation of the sampling distribution of sample means is ${\sigma }_{\overline{x}}=1.7321$

The formula to convert the $\overline{x}$ value into z score is,

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

Substitute 105 for $\overline{x}$, 100 for ${\mu }_{\overline{x}}$ and 1.7321 for ${\sigma }_{\overline{x}}$

That is,

$\begin{array}{c}z=\frac{105-100}{1.7321}\\ =\frac{5}{1.7321}\\ =2.89\end{array}$

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

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

Procedure:

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

That is, $P\left(z<2.89\right)=0.9981$

The area to the right of 2.89 is,

$\begin{array}{c}P\left(z>2.89\right)=1-P\left(z<2.89\right)\\ =1-0.9981\\ =0.0019\end{array}$

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

Interpretation:

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