Question

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 $84th$and $16th$percentiles for the sum of the sample. Interpret these values in context.

Short answer

(a)The probability that the sum of samples lies between$7$hours and $10$hours is $0.9991$.

(b)$84th$and $16th$percentile for the sums of sample are respectively $582.31$and $565.68$minutes.

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. Suppose the standard deviation is one minute. Take a sample size of$70$.

Explanation(part a)

According to the provided facts, the mean number of minutes for app engagement by a tablet user is $8.2$minutes, the standard deviation is $1$minute and the sample size is$70$.

The mean and standard deviation will follow the normal distribution. So, the distribution of sum of ages of tablet users is given as:

$\mathrm{\Sigma }X~N\left(n{\mu }_X,\sqrt{n}{\sigma }_X\right)$

$\sum X~N(70\times 8.2,\sqrt{70}\times 1)$

$\sum X~N(574,8.366)$

Probability

Consider $7$hours as $420$minutes and the $10$hours as $600$minutes. To find the probability that the sum of samples lies between $7$hours and$10$hours, use the Ti-83 calculator. For this, click on $2^{\text{nd}}$, then DISTR, and then scroll down to the normcdf option and enter the provided details. After this, click on ENTER button on the calculator to have the desired result. The screenshot is given below:

Therefore, the probability that the sum of samples lies between$7$hours and$10$hours is$0.9991$.

Final answer(part a)

The probability that the sum of samples lies between $7$hours and $10$hours is$0.9991$.

Given information(part b)

Given in the question that, 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 $70$.

Explanation(part b)

According to the provided details, the mean number of minutes for app engagement by a tablet user is $8.2$minutes, standard deviation is $1$minute and the sample size is $70$.

The mean and standard deviation will follow the normal distribution. So, the distribution of sum of ages of tablet users is given as:

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

$\sum X~N(70\times 8.2,\sqrt{70}\times 1)$

$\sum X~N(574,8.366)$

Calculate 84th  and 16th  percentile for the sums of sample.

To calculate ${84}^{\text{th}}$and ${16}^{\text{th}}$percentile for the sums of sample, use Ti-83 calculator. For this, click on $2^{\text{nd}}$, then DISTR, and then scroll down to the invnorm option and enter the provided details. After this, click on ENTER button on the calculator to have the desired result. The screenshot is given below:

Therefore,$84th$and $16th$ percentile for the sums of sample are respectively $582.31$ and $565.68$ minutes.

Final answer(part b)

$84th$ and $16th$ percentile for the sums of sample are respectively $582.31$ and $565.68$ minutes.