Chapter 6.1, Problem 6E

a.

To determine

To show: The probability distribution is a legitimate probability distribution.

Explanation of Solution

Given:

The probability distribution is:

X	0	1	2	3	4	5	6	7
Probability	0.68	0.05	0.07	0.08	0.05	0.04	0.01	0.02

Calculation:

The sum of probabilities is:

  $\begin{array}{c}\text{Sum of probabilities = 0}\text{.68+0}\text{.05+0}\text{.08+0}\text{.05+0}\text{.01+0}\text{.02}\\ \text{=1}\end{array}$

The probabilities are between 0 and 1 and the addition of probabilities is equal to 1. Thus, the provided probability distribution is a legitimate probability distribution.

b.

To determine

To construct: The graph for the probability distribution and interpret it.

Explanation of Solution

Graph:

The graph could be constructed as:

  

Interpretation:

The most of the data is on the right side of the graph. Thus, the distribution is skewed to the right.

c.

To determine

To describe: The $Y<7$ in words. Also, compute the $P\left(Y<7\right)$

Answer to Problem 6E

The probability is 0.98

Explanation of Solution

Calculation:

  $X<7$ implies that number of days that people workout is less than 7.

  $P\left(X<7\right)$ can be calculated as:

  $\begin{array}{c}P\left(X<7\right)=P\left(X=1\right)+P\left(X=2\right)+\mathrm{.....}+P\left(X=6\right)\\ =0.68+0.05+\mathrm{.....}+0.01\\ =0.98\end{array}$

Thus, the required probability is 0.98

d.

To determine

To express: The event “workout at least once” in terms of Y and compute it.

Answer to Problem 6E

The probability is 0.32.

Explanation of Solution

Calculation:

The event “workout at least once” in terms of Y could be written as $Y\ge 1$ .

  $P\left(Y\ge 1\right)$ can be calculated as:

  $\begin{array}{c}P\left(Y\ge 1\right)=1-P\left(Y<0\right)\\ =1-0.68\\ =0.32\end{array}$

Thus, the required probability is 0.32.