Chapter 7, Problem 11CQ

The running time for videos submitted to YouTube in a given week is normally distributed with $\mu =390$ seconds and standard deviation $\sigma =148$ seconds.

1. If a single video is randomly selected, what is the probability that the running time of the video exceeds 6 minutes (360 seconds)?
2. Suppose that a sample of 40 videos is selected. What is the probability that the mean running time of the sample exceeds 6 minutes?

To determine

To find:the probability that the running time of the video exceeds 6 minutes ( $360$ seconds),

If a single video is randomly selected,

Answer to Problem 11CQ

The required answer is $0.58$ .

Explanation of Solution

Given,

If a single video is randomly selected then the mean of $\overline{x}\text{ is }{\mu }_{\overline{x}}=\mu =390$

Calculation Required:

The required probability that $\overline{x}$ is greater than $360$ is

  $\begin{array}{l}P\left(\overline{x}>360\right)\\ =1-P\left(\overline{x}\le 360\right)\\ =1-P(Z\le \frac{360-390}{148})\\ =1-P(Z\le -2027)\end{array}$

Using normal table,

  $P\left(\overline{x}>360\right)=0.58$

Calculation Steps:

Given that a normal distribution has mean of $390$ seconds and a standard deviation of $148$ seconds.

If a single video is randomly selected then the mean of $\overline{x}\text{ is }{\mu }_{\overline{x}}=\mu =390$

The standard deviation is,

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{148}{\sqrt{1}}\\ =148\end{array}$

To compute the probability that $\overline{x}$ will be greater than 6 minutes or $6$ seconds using Minitab follow the steps given below.

1. Click CaIc, then select Probability distributions and then go to Normal.

2. Select the Cumulative option.

3. Enter $390$ in the Mean field and enter $148$ in the standard deviation field.

4. Enter $360$ in tile input constant field.

  

5. Click Ok. The final output is given below.

Cumulative distribution function.

Normal with mean $=390$ and standard deviation $=148$

  $\begin{array}{l}xP\left(X\le x\right)\\ 3600.419684\end{array}$

The required probability that $\overline{x}$ is greater than $360$ is

  $\begin{array}{l}P\left(\overline{x}>360\right)\\ =10-P\left(\overline{x}<360\right)\\ =1-0.419684\\ =0.58\end{array}$

The required answer is $0.58$

To determine

To find: the probability that the mean running time of the sample exceeds $6$ minutes.

Answer to Problem 11CQ

The required answer is $0.901$

Explanation of Solution

Given,

If $40$ videos are randomly selected then the mean of $\overline{x}\text{ is }{\mu }_{\overline{x}}=\mu =390$

Calculation Required:

  $\begin{array}{l}P\left(\overline{x}>360\right)=1-P\left(\overline{x}\le 360\right)\\ P\left(\overline{x}>360\right)=1-P\left(\frac{\overline{x}-390}{23.4}\le \frac{360-390}{23.4}\right)\\ P\left(\overline{x}>360\right)=1-P\left(Z\le -1.282\right)\end{array}$

Using normal table,

  $\begin{array}{l}P\left(\overline{x}>360\right)=1-0.099\\ P\left(\overline{x}>360\right)=0.901\end{array}$

Calculation Steps :

If $40$ videos are randomly selected then the mean of $\overline{x}\text{ is }{\mu }_{\overline{x}}=\mu =390$

The standard deviation is,

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{148}{\sqrt{40}}\\ =23.4\end{array}$

The final output is given below.

The final output is given below.

Cumulative distribution function

Normal with mean $=390$ and standard deviation $=23.4$ .

  $\begin{array}{l}xP\left(X\le x\right)\\ 3600.0999123\end{array}$

The required probability that $\overline{x}$ is greater than $360$ is

  $\begin{array}{l}P\left(\overline{x}>360\right)=1-P\left(\overline{x}\le 360\right)\\ P\left(\overline{x}>360\right)=1-P\left(\frac{\overline{x}-390}{23.4}\le \frac{360-390}{23.4}\right)\\ P\left(\overline{x}>360\right)=1-P\left(Z\le -1.282\right)\\ P\left(\overline{x}>360\right)=1-0.099\\ P\left(\overline{x}>360\right)=0.901\end{array}$

The required answer is $0.901$