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.

Find the middle 80% for the total amount of time 64 batteries last.

Short answer

The middle $\mathit{80}\mathit{\%}$for the total amount of time $\mathit{64}$batteries last is $\mathit{21}\mathit{.}\mathit{98}$

Step-by-step solution

Given Information

It is given that a particular battery's lasting time follow exponential distribution with a mean of ten months.

We have to find the middle$80\%$ for the total amount of time.

Explanation

We know that the decay parameter $m$is $\frac{1}{10}=0.1$

Therefore, the amount of time $64$batteries last follows exponential distribution as$0.10e^{-0.10x}$

The difference between the ${90}^{th}$and ${10}^{th}$percentiles can be used to find the middle $80$percent.

The role="math" localid="1649330524181" ${\mathit{90}}^{\mathit{t}\mathit{h}}$percentile is calculated as follows:

role="math" localid="1649330516315" $\mathit{0}\mathit{.}\mathit{90}\mathit{=}\mathit{1}\mathit{-}{e}^{\mathit{-}\mathit{0}\mathit{.}\mathit{10}\mathit{x}{\mathit{90}}^{\mathit{t}\mathit{h}}\mathit{P}\mathit{e}\mathit{r}\mathit{c}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{l}\mathit{e}}$

role="math" localid="1649330508713" $\mathit{1}\mathit{-}\mathit{0}\mathit{.}\mathit{90}\mathit{=}{e}^{\mathit{-}\mathit{0}\mathit{.}\mathit{10}\mathit{x}\mathit{90}\mathit{t}\mathit{h}\mathit{p}\mathit{e}\mathit{r}\mathit{c}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{l}\mathit{e}}$

${\mathit{90}}^{\mathit{t}\mathit{h}}Percentile\mathit{=}\frac{\mathit{I}\mathit{n}\mathit{}\mathit{0}\mathit{.}\mathit{10}}{\mathit{-}\mathit{0}\mathit{.}\mathit{10}}$

$\mathit{=}\mathit{23}\mathit{.}\mathit{03}$

Let's find the ${10}^{th}Percentile$

role="math" localid="1649330637853" $\mathit{0}\mathit{.}\mathit{10}\mathit{=}\mathit{1}\mathit{-}{e}^{\mathit{-}\mathit{0}\mathit{.}\mathit{10}\mathit{x}\mathit{10}\mathit{t}\mathit{h}\mathit{p}\mathit{e}\mathit{r}\mathit{c}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{l}\mathit{e}}$

$\mathit{1}\mathit{-}\mathit{0}\mathit{.}\mathit{10}\mathit{=}{e}^{\mathit{-}\mathit{0}\mathit{.}\mathit{10}\mathit{x}\mathit{10}\mathit{t}\mathit{h}\mathit{p}\mathit{e}\mathit{r}\mathit{c}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{l}\mathit{e}}$

${\mathit{10}}^{\mathit{t}\mathit{h}}Percentile\mathit{=}\frac{\mathit{I}\mathit{n}\mathit{}\mathit{0}\mathit{.}\mathit{90}}{\mathit{-}\mathit{0}\mathit{.}\mathit{10}}$

$\mathit{=}\mathit{1}\mathit{.}\mathit{05}$

Hence, the middle $80\%$for the total amount of $64$batteries last is calculated as :

${\mathit{90}}^{\mathit{t}\mathit{h}}Percentile\mathit{-}{\mathit{10}}^{\mathit{t}\mathit{h}}Percentile$

$\mathit{=}\mathit{23}\mathit{.}\mathit{03}\mathit{-}\mathit{1}\mathit{.}\mathit{05}$

$\mathit{=}\mathit{21}\mathit{.}\mathit{98}$