Question

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.

What is the distribution for the total length of time 64 batteries last?

Short answer

The distribution for the total length of time 64 batteries last$.10$month.

Step-by-step solution

:Given Information

According to the given details,

The mean of the distribution is $\mu =10$months.

The sample size is$\mathit{64}\mathit{.}$

Explanation

From the information, we know that the probability density function of the exponential distribution

$X\mathit{~}\mathit{Exp}\mathit{\left(}m\mathit{\right)}$

where $m$is the decay parameter.

Therefore,

localid="1649332271197" $f\mathit{\left(}X\mathit{\right)}\mathit{=}m{e}^{\mathit{(}\mathit{-}\mathit{m}\mathit{x}\mathit{)}}$

localid="1649332290236" $\text{Where,}X\mathit{\ge }\mathit{0}\text{and}m\mathit{>}\mathit{0}\text{.}$

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

The distribution of $\overline{X}$is the mean length time of $64$batteries last is given as follow:

$\overline{X}-N\left({\mu }_X,{\sigma }_X/\sqrt{n}\right)$

$\overline{X}-N(10,10/\sqrt{64})$

$\overline{X}~N(10,10/8)$

Therefore, the distribution for total length of $\mathit{64}$batteries last is:

$m=\frac{\mathit{1}}{\mathit{\mu }}$

$\mathit{=}\mathit{.}\mathit{10}$month