91Ó°ÊÓ

Given in the question that the length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months.

A sample of $64$of these smartphones is taken.

According to the information, we observed that the mean of the distribution is:

$\mathrm{μ}\mathit{=}\mathit{10}$

The probability density function of exponential distribution $X\mathit{~}Exp\mathit{\left(}m\mathit{\right)}$

Where, $m$is the decay parameter.

Therefore,

$f\mathit{\left(}X\mathit{\right)}\mathit{=}m{e}^{\mathit{(}\mathit{-}\mathit{m}\mathit{X}\mathit{)}}$

Where, role="math" localid="1649331353581" $X\mathit{â‰\yen }\mathit{0}$and $m\mathit{>}\mathit{0}$

The standard deviation of the exponential distribution is$\mathrm{σ}\mathit{=}\mathrm{μ}\mathit{=}\mathit{10}$

Consider $\stackrel{\mathit{¯}}{\mathit{X}}$as the mean length time of $\mathit{64}$batteries last, then the distribution of mean length of $\mathit{64}$batteries will follow the normal distribution.

The distribution of $\stackrel{¯}{X}$is the mean length time of $\mathit{64}$batteries last is given as below:

$\stackrel{¯}{X}-N\left({\mathrm{μ}}_x,{\mathrm{σ}}_x/\sqrt{n}\right)$

$\stackrel{¯}{X}~N(10,10/\sqrt{64})$

$\stackrel{¯}{X}~N(10,10/8)$

Let's use Ti-83 calculator to find the probability :

For this, click on $2^{nd}$.

Then DISTR and then scroll down to the normalcdf option and enter the provided details. After this, click on ENTER button of calculator to get the result.

Hence, the value of $P(7<\stackrel{¯}{X}<11)$is approximately $\mathit{77}\mathit{.}\mathit{99}\mathit{\%}$.