Chapter 7.3, Problem 61E

(a)

To determine

To Calculate: the probability that the sample mean lifetime is 42.2 months or less if the company’s claim is true.

Answer to Problem 61E

2.28%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =48\\ \sigma =8.2\\ n=8\\ x=42.2\end{array}$

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

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}\\ =\frac{42.2-48}{8.2/\sqrt{8}}\\ =-2.00\end{array}$

The associating probability using the normal probability

  $P\left(Z<-2.00\right)$ is given in the row starting with $-2.0$ and in the column starting with $.00$ of the standard normal probability

  $P\left(\overline{X}<42.2\right)=P\left(Z<-2.00\right)=0.0228=2.28\%$

(b)

To determine

To Explain: that there is convincing evidence that the company is overstating the average lifetime of its batteries.

Answer to Problem 61E

Yes

Explanation of Solution

Given:

  $\begin{array}{c}\mu =48\\ \sigma =8.2\\ n=8\\ x=42.2\end{array}$

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 sampling distribution of the sample mean $\overline{x}$ has mean $\mu$ and standard deviation $\frac{\sigma }{\sqrt{n}}\cdot$

The z-score is the value decreased by the mean, divided by the standard deviation:

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}\\ =\frac{42.2-48}{8.2/\sqrt{8}}\\ =-2.00\end{array}$

The associating probability using the normal probability

  $P\left(Z<-2.00\right)$ is given in the row starting with $-2.0$ and in the column starting with $.00$ of the standard normal probability table in the appendix.

  $P\left(\overline{X}<42.2\right)=P\left(Z<-2.00\right)=0.0228=2.28\%$

The probability is considered to be smaller, when it is less than $0.05$ .

It is observed that the probability of a sample mean of at most $42.2$ months is small, therefore the event is un usual to occur by chance and therefore there is convincing evidence that the company is over starting the average lifetime in its batteries.