91Ó°ÊÓ

Given:

Mean, $\mathrm{μ}=250$

Standard deviation,$\mathrm{σ}=2$

$n=10$

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

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

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

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,$\mathrm{μ}=250$

Standard deviation, $\mathrm{σ}=2$

$n=10$

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

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.