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A pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed.

The exponential function is,

$m=\frac{1}{渭}$

$=\frac{1}{18}$

$=0.055556$

The following is an exponential distribution of how long running shoes last:

$X~\mathrm{Exp}鈦�\left(0.055556\right)$

Here,

$X=\text{Random variable for the length of time running shoes last}$

It is evident that the exponential distribution's cumulative distribution function is:

$P(X<x)=1鈭�e^{鈭�mx}$

The probability that the running will endure longer than 15 months can be computed as follows:

$P(x>15)=1鈭�P(x<15)$

$=1鈭�\left(1鈭�e^{鈭�0.055556脳15}\right)$

$=e^{鈭�0.83334}$

$=0.4346$

Thus, the value of the running last in more than $15$months is $0.4346$.

It is known that $1$pair of running shoes on an average last in,

$=18\text{months}$

So, $6$pairs of running shoes on an average will last in,

$=18脳6\text{months}$

$=108\text{months}$

Here, you must compute the time it takes for $80$percent of the running shoes to last, which implies you must calculate the data's ${80}^{th}$percentile So, calculate it as below:

$P(x<k)=1鈭�e^{鈭�0.055556k}$

$0.80=1鈭�e^{鈭�0.055556k}$

$e^{鈭�0.055556k}=1鈭�0.80$

Take logarithmic on both the sides. Then,

$\ln 鈦�\left(e^{鈭�0.055556k}\right)=\ln 鈦�0.20$

$鈭�0.055556k=鈭�1.6094$

$k=\frac{1.6094}{0.055556}$

$k鈮�28.97$