91Ó°ÊÓ

$\begin{array}{c}\mathrm{μ}=48\\ \mathrm{σ}=8.2\\ \mathrm{n}=8\\ \mathrm{x}=42.2\end{array}$

The following formula was used:

$\mathrm{z}=\frac{\mathrm{x}−{\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}}$

The sampling distribution of the sample mean is normal because the population distribution is normal $\stackrel{¯}{\mathrm{x}}$is also typical.

z-score is

$\mathrm{z}=\frac{\mathrm{x}−{\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}}=\frac{\stackrel{¯}{\mathrm{x}}−\mathrm{μ}}{\mathrm{σ}/\sqrt{\mathrm{n}}}=\frac{42.2−48}{8.2\sqrt{\stackrel{¯}{\mathrm{B}}}}=−2.00$

The normal probability is used to calculate the associating probability

$\mathrm{P}(\mathrm{Z}<-2.00)$is presented in the standard normal probability table in the row beginning with 2.0 and the column beginning with .00.

$\mathrm{P}(\stackrel{¯}{\mathrm{X}}<42.2)=\mathrm{P}(\mathrm{Z}<-2.00)=0.0228=2.28\%$

$\begin{array}{c}\mathrm{μ}=48\\ \mathrm{σ}=8.2\\ \mathrm{n}=8\\ \mathrm{x}=42.2\end{array}$

The following formula was used:

$z=\frac{x−{\mathrm{μ}}_{\stackrel{¯}{x}}}{{\mathrm{σ}}_{\stackrel{¯}{x}}}$

The sampling distribution of the sample mean is normal because the population distribution is normal $\stackrel{¯}{\mathrm{x}}$ is also typical.

The sample mean's sampling distribution $\stackrel{¯}{\mathrm{x}}$ has mean $\mathrm{μ}$as well as standard deviation$\frac{\mathrm{σ}}{\sqrt{n}}$

The z-score is the difference between the mean and the standard deviation:

$\begin{array}{r}\mathrm{z}=\frac{\mathrm{x}−{\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}}\\ =\frac{\stackrel{¯}{\mathrm{x}}−\mathrm{μ}}{\mathrm{σ}/\sqrt{\mathrm{n}}}\\ =\frac{42.2−48}{8.2/\sqrt{8}}\\ =−2.00\end{array}$

The normal probability is used to calculate the associating probabilit

$\mathrm{P}(\mathrm{Z}<-2.00)$is given in the first row, beginning with -2.0 in the column that begins with .00 in the appendix to the standard normal probability table

$\mathrm{P}(\stackrel{¯}{\mathrm{X}}<42.2)=\mathrm{P}(\mathrm{Z}<−2.00)=0.0228=2.28\%$

When the chance is less than 0.05, it is considered less.

The possibility of a sample mean of at most 42.2 months occurring by accident is negligible, therefore the event is unlikely to happen by random, and there is compelling proof that the corporation is overstating the average lifetime in its batteries.