The manufacturer of a certain brand of aluminum foil claims that the amount of foil on each roll follows a Normal distribution with a mean of $250$ square feet (ft² ) and a standard deviation of $2$ ft² . To test this claim, a restaurant randomly selects 10 rolls of this aluminum foil and carefully measures the mean area to be$\overline{x}=249.6f{t}^2$.

a. Find the probability that the sample mean area is $249.6f{t}^2$or less if the manufacturer’s claim is true.

b. Based on your answer to part (a), is there convincing evidence that the company is overstating the average area of its aluminum foil rolls?

Given:

Mean, $\mu =250$

Standard deviation,$\sigma =2$

$n=10$

$\overline{x}=249.6f{t}^2$

The sampling distribution of the sample mean is also normal because the population distribution is normal.

$$\begin{gathered} z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}} \\ =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}} \\ =\frac{249.6-250}{2/\sqrt{10}}=\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{63}\mathbf{}\mathbf{} \end{gathered}$$

is the $z$-score.

In the row beginning with $-0.6$and in the column beginning with $.03$of the standard normal probability, the corresponding probability using the normal probability $P(Z<-0.63)$is given :

$P(\overline{X}<249.6)=P(Z<-0.63)=0.2643=\mathbf{26}\mathbf{.}\mathbf{43}\%$

Given :

Mean,$\mu =250$

Standard deviation, $\sigma =2$

$n=10$

$\overline{x}=249.6f{t}^2$

When the likelihood is less than $0.05$, the probability is deemed modest. The likelihood is large, implying that a sample mean area of at most $249.6$ foot square is likely to occur, and hence there is no persuasive proof that the corporation is exaggerating the average area of its aluminum foil rolls.