91Ó°ÊÓ

Given in the question that

Population mean$\left(\mathrm{μ}\right)=55000$

Population standard deviation$\left(\mathrm{σ}\right)=4500$

Sample size$\left(n\right)=8$

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

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

$\overline{x}~N\left({\mathrm{μ}}_{\stackrel{¯}{X}},{\mathrm{σ}}_{\stackrel{¯}{X}}\right)$

$\overline{x}~N\left(\mathrm{μ},\frac{\mathrm{σ}}{\sqrt{n}}\right)$

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

$\stackrel{¯}{x}~N(55000,1590.99)$

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

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.

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

$P\left(\overline{x}<191\right)=P\left(\frac{{\displaystyle \frac{\stackrel{¯}{x}-\mathrm{μ}}{\mathrm{σ}}}}{\sqrt{n}}<\frac{{\displaystyle \frac{51800-\mathrm{μ}}{\mathrm{σ}}}}{\sqrt{n}}\right)$

localid="1649200029249" $=P\left(Z,\frac{{\displaystyle \frac{51800-55000}{4500}}}{\sqrt{8}}\right)$

$=P(Z<-2.01)$

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

$=0.0222$

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.