7.7 The mean number of minutes for app engagement by a table use is $8.2$minutes. Suppose the standard deviation is one minute. Take a sample size of $70$ .
a. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem?
b. Find the ${84}^{\text{th}}$ and ${16}^{\text{th}}$ percentiles for the sum of the sample. Interpret these values in context.

Given in the question that, the mean number of minutes for app engagement by a table use is $8.2$minutes.

We have to find the probability that the sum of the sample is between seven hours and ten hours, if the standard deviation is one minute.

According to the information, we have :$\mu \mathrm{\Sigma }x=n\mu x=70\times \left(8.2\right)=574$minutes

$\sigma \mathrm{\Sigma }x=\left(\sqrt{n}\right)\left({\sigma }_x\right)=\left(\sqrt{70}\right)\times \left(1\right)=8.37$minutes.

Let's find the sum of the sample between$7\mathrm{h}(420\min )$and $10\mathrm{h}(600\min )$
Therefore,

We can write the probability as follow:

$\begin{array}{r}\mathrm{Pr}(420<\mathrm{\Sigma }x<600)=\mathrm{Pr}\left(\frac{420-574}{8.37}<Z<\frac{600-574}{8.37}\right)\end{array}$

$=\mathrm{Pr}(-18.399<Z<3.1063)$

$=\mathrm{Pr}(Z<3.1063)-\mathrm{Pr}(Z<-18.399)$

$=0.991-0$

$=0.9991$

The ${84}^{\text{th}}$percentiles for the sum be $k_1$and${16}^{\text{th}}$percentiles for the sum be $k_2$.

Let $k_1={84}^{\text{th}}$percentile.

${\mathrm{k}}_1=\mathrm{invNorm}(0.84,574,8.37)=582.3236$
Let $k_2={16}^{\text{th}}$percentile.
${\mathrm{k}}_2=\mathrm{invNorm}(0.16,574,8.37)=565.6764$