Chapter 8.3, Problem 85E

a)

To determine

To find $P\left(X=7\right)$

Answer to Problem 85E

  $P\left(X=7\right)=0.57$

Explanation of Solution

Given:

Days	0	1	2	3	4	5	6	7
Probability	0.04	0.03	0.06	0.08	0.09	0.08	0.05	?

Calculation:

We know for every probability distribution, the sum of the probabilities equal to 1.

Using this condition,

  $\begin{array}{l}{\displaystyle \sum P(X=x)=1}\\ 0.04+0.03+0.06+0.08+0.09+0.08+0.05+P(X=7)=1\\ P(X=7)=1-0.43=0.57\end{array}$

b)

To determine

To interpret mean and standard deviation.

Answer to Problem 85E

On average the number of days that a young person watched television is 5.44 days.

The number of days that a young person watched television varies on average by 2.14 days from the mean of 5.44 days.

Explanation of Solution

Given:

Days	0	1	2	3	4	5	6	7
Probability	0.04	0.03	0.06	0.08	0.09	0.08	0.05	0.57

  $\begin{array}{l}{\mu }_x=5.44\\ {\sigma }_x=2.14\end{array}$

Calculation:

First, need to understand mean and standard deviation.

Mean represents the expected value of the random variable and standard deviation represents the average deviation from the mean.

Therefore, on average the number of days that a young person watched television is 5.44 days.

The number of days that a young person watched television varies on average by 2.14 days from the mean of 5.44 days.

c)

To determine

To explain whether the mean of the 100 sample is 4.96.

Answer to Problem 85E

Not surprising.

Explanation of Solution

Given:

Days	0	1	2	3	4	5	6	7
Probability	0.04	0.03	0.06	0.08	0.09	0.08	0.05	0.57

  $\begin{array}{l}{\mu }_x=5.44\\ {\sigma }_x=2.14\end{array}$

Calculation:

The sample size n=100 and their sample mean = $\overline{x}=4.96$

We need check whether it contains in one standard deviation.

Therefore,

  $\begin{array}{l}{\mu }_x-{\sigma }_x=5.44-2.14=3.3\\ {\mu }_x+{\sigma }_x=5.44+2.14=7.58\end{array}$

Since $\overline{x}=4.96$ is contained in one standard deviation, it is not surprising.