Question

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.

Short answer

(a) The probability that the sum of the sample between seven hours and ten hours is $0.9991$

(b) The ${84}^{th}$and ${16}^{th}$percentiles for the sum of the samples are respectively$582.3236,and565.6764$

Step-by-step solution

Given information Part (a)

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.

Explanation Part (a)

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:

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

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

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

$=0.991-0$

$=0.9991$

Given information Part (b)

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

Explanation Part (b)

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.
${\mathrm{k}}_2=\mathrm{invNorm}(0.16,574,8.37)=565.6764$