Chapter 6.2, Problem 31E

Eat your cereal: A cereal manufacturer claims that the weight of a box of cereal labeled as weighing 12 ounces has a mean of 12.0 ounces and a standard deviation of 0.1 ounce. You sample 75 boxes and weigh them. Let $\bar{x}$ denote the mean weight of the 75 boxes.

a.    If the claim is true, what is $P(\bar{x}\le 11.99)$?

b.    Based on the answer to part (a), if the claim is true, is 11.99 ounces an unusually small mean weight for a sample of 75 boxes?

c.    If the mean weight of the boxes were 11.99 ounces, would you be convinced that the claim was false? Explain.

d.    If the claim is true, what is $P(\bar{x}\le 11.97)$?

e.    Based on the answer to part (d), if the claim is true, is 11.97 ounces an unusually small mean weight for a sample of 75 boxes?

f.    If the mean weight of the boxes were 11.97 ounces, would you be convinced that the claim was false? Explain.

To determine

Find the probability $P\left(\overline{x}\le 11.99\right)$ if the claim is true.

Answer to Problem 31E

The probability $P\left(\overline{x}\le 11.99\right)$ if the claim is true is 0.1923.

Explanation of Solution

Calculation:

The given information is that the manufacturer claims that the weight of a box of cereal labelled as weighing 12 ounces has mean 12.0 ounces and standard deviation of 0.1 ounce. The sample of boxes is 75. Also, $\overline{x}$ is the mean weight of the 75 boxes.

Central limit theorem:

If $\overline{x}$ be the mean of the simple random sample of large size $\left(n>30\right)$ is drawn from the population with mean $\mu$ and standard deviation $\sigma$ (or) the population follows normal distribution then $\overline{x}$ follows normal distribution with mean ${\mu }_{\overline{x}}=\mu$ and standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

From the given information, it is observed that the simple size n is 75 and it is greater than 30. Therefore, it is appropriate to use the normal distribution.

For mean ${\mu }_{\overline{x}}$:

From the given information, the mean of the population $\mu$ is 12.0.

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =12.0\end{array}$

For standard deviation ${\sigma }_{\overline{x}}$:

From the given information, the standard deviation of the population $\sigma$ is 0.1 and sample size n is 75.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{0.1}{\sqrt{75}}\\ =\frac{0.1}{8.6602}\\ =0.0115\end{array}$

The probability $P\left(\overline{x}\le 11.99\right)$ represents the area to the left of 11.99.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 12.0 and Standard deviation as 0.0115.
• Click the Shaded Area tab.
• Choose X value and Left Tail for the region of the curve to shade.
• Enter the X value as 11.99.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability $P\left(\overline{x}\le 11.99\right)$ if the claim is true is 0.1923.

To determine

Check whether 11.99 ounces is an unusually small mean weight for a sample of 75 boxes based on the answer to part (a).

Answer to Problem 31E

No, the 11.99 ounces is not unusually small mean weight for a sample of 75 boxes based on the answer to part (a).

Explanation of Solution

Unusual:

If the probability of an event is less than 0.05 then the event is called unusual.

From part (a), the probability $P\left(\overline{x}\le 11.99\right)$ is 0.1932.

Here, the probability $P\left(\overline{x}\le 11.99\right)$ is greater than 0.05. That is, $0.1932>0.05$. Thus, the 11.99 ounces is not unusually small mean weight for a sample of 75 boxes based on the answer to part (a).

To determine

Explain whether the person can be convinced that the claim is false if the mean weigh of the boxes were 11.99 ounces.

Answer to Problem 31E

No, the person cannot be convinced that the claim is false if the mean weight of the boxes were 11.99 ounces because the result is not unusual if the claim is true.

Explanation of Solution

From part (b), it can be observed that the 11.99 ounces is not unusually small mean weight for a sample of 75 boxes based on the answer to part (a). Therefore, the person cannot be convinced that the claim is false if the mean weigh of the boxes were 11.99 ounces because the result is not unusual if the claim is true.

To determine

Find the probability $P\left(\overline{x}\le 11.97\right)$ if the claim is true.

Answer to Problem 31E

The probability $P\left(\overline{x}\le 11.97\right)$ if the claim is true is 0.0047.

Explanation of Solution

Calculation:

The probability $P\left(\overline{x}\le 11.97\right)$ represents the area to the left of 11.97.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 12.0 and Standard deviation as 0.0115.
• Click the Shaded Area tab.
• Choose X value and Left Tail for the region of the curve to shade.
• Enter the X value as 11.97.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability $P\left(\overline{x}\le 11.97\right)$ if the claim is true is approximately 0.0047.

To determine

Check whether 11.97 ounces is an unusually small mean weight for a sample of 75 boxes based on the answer to part (d).

Answer to Problem 31E

Yes, the 11.97 ounces is unusually small mean weight for a sample of 75 boxes based on the answer to part (d).

Explanation of Solution

Calculation:

From part (d), the probability $P\left(\overline{x}\le 11.97\right)$ is 0.0047.

Here, the probability $P\left(\overline{x}\le 11.97\right)$ is greater than 0.05. That is, $0.0047>0.05$. Thus, the 11.97 ounces is unusually small mean weight for a sample of 75 boxes based on the answer to part (d).

To determine

Explain whether the person can be convinced that the claim is false if the mean weight of the boxes were 11.97 ounces.

Answer to Problem 31E

Yes, the person can be convinced that the claim is false if the mean weigh of the boxes were 11.97 ounces because the result is unusual if the claim is true.

Explanation of Solution

From part (e), it can be observed that the 11.97 ounces is unusually small mean weight for a sample of 75 boxes based on the answer to part (d). Therefore, the person can be convinced that the claim is false if the mean weigh of the boxes were 11.97 ounces because the result is unusual if the claim is true.