Q.54 - The Practice of Statistics for AP

Question

Stop the carl A car company has found that the lifetime of its disc brake pads varies from car to car according to a Normal distrilustion with mean $\mu =55000$miles and standard deviation$\sigma =4500$ miles. The company installs a new brand of brake pads on an SRS of $8$ cars.

(a) If the new brand has the same lifetime distribution as the previous bye of brake pad, what is the sampling distribution of the mean lifetime $\overline{x}$ ?

(b) The average life of the pads an these $8$ cars turns out to be $\overline{x}=51800$ miles. Find the probability that the sample mean lifetime is $51800$ miles or less if the lifetime distribution is unchanged, What conclusion would you draw?

Step-by-Step Solution

Answer

(a)The sampling distribution is normally with mean $=55000$ and standard deviation $=1590.99$

(b)The probability is $0.0222$

Part (a) Step-1 Given Information

Given in the question that 

Population mean $\left(\mu \right)=55000$

Population standard deviation $\left(\sigma \right)=4500$

Sample size $\left(n\right)=8$

We have to find that  the sampling distribution of the mean lifetime $\overline{x}$. 

 

Part (b) Step-2 Explanation

The sample distribution of $\overline{x}$ is written as :

$\overline{x}~N \left({\mu }_{\overline{X}},{\sigma }_{\overline{X}}\right)$

$\overline{x}~N \left(\mu ,\frac{\sigma }{\sqrt{n}}\right)$

$\overline{x}~N \left(55000,\frac{4500}{\sqrt{8}}\right)$

$\overline{x}~N (55000,1590.99)$

 Thus, the sampling distribution is normally distributed with mean $=55000$ and standard deviation $=1590.99$

Part (a) Step-1 Given Information

Given in the question that he average life of the pads an these $8$ cars turns out to be $\overline{x}=51800$ mile we have to find the probability that the sample mean lifetime is $51800$ miles or less and also if the lifetime distribution is unchanged, What conclusion would you draw.

Part (b) Step-2 Explanation

The probability that  the sample mean is $51800$ or less is calculated as follows:

 $P\left(\overline{x}<191\right)=P\left(\frac{{\displaystyle \frac{\overline{x}-\mu }{\sigma }}}{\sqrt{n}}<\frac{{\displaystyle \frac{51800-\mu }{\sigma }}}{\sqrt{n}}\right)$

                    $=P\left(Z,\frac{{\displaystyle \frac{51800-55000}{4500}}}{\sqrt{8}}\right)$

                     $=P(Z<-2.01)$

$=P\left(Z<-2.01\right)$

                    $=0.0222$

Thus, the required probability is $0.5346$

The computed probability is below $0.05$.thus, there is low chances for the occurance of the event.