Chapter 7.3, Problem 62E

(a)

To determine

To find: the probability that the sample mean area is 249.6 foot square or less if the manufacturer’s claim is true.

Answer to Problem 62E

26.43%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =250\\ \sigma =2\\ n=10\\ \overline{x}=249.6\end{array}$

Formula used:

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

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean $\overline{x}$ is also normal.

The z-score is

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}\\ =\frac{249.6-250}{2/\sqrt{10}}\\ =-0.63\end{array}$

The corresponding probability using the normal probability

  $P\left(Z<-0.63\right)$ is given in the row starting with $-0.6$ and in the column starting with $.03$ of the standard normal probability

  $P\left(\overline{X}<249.6\right)=P\left(Z<-0.63\right)=0.2643=26.43\%$

(b)

To determine

To Explain: that there is convincing evidence that the company is overstating the average area of its aluminium foil rolls.

Answer to Problem 62E

No

Explanation of Solution

Given:

  $\begin{array}{c}\mu =250\\ \sigma =2\\ n=10\\ \overline{x}=249.6\end{array}$

Formula used:

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

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean $\overline{x}$ is also normal.

The z-score is

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}\\ =\frac{249.6-250}{2/\sqrt{10}}\\ =-0.63\end{array}$

The associating probability using the normal probability table

  $P\left(Z<-0.63\right)$ is given in the row starting with $-0.6$ and in the column starting with $.03$ of the standard normal probability

  $P\left(\overline{X}<249.6\right)=P\left(Z<-0.63\right)=0.2643=26.43\%$

The probability is considered to be small, when the probability is less than $0.05$ .

It is observed that the probability is large, which implies that it is likely to occur a sample mean are of at most $249.6$ foot square and thus there is no convincing evidence that the company is overstating the average are of its aluminium foil rolls.