Chapter 8.3, Problem 79E

(a)

To determine

To calculate: The probability for day X equal to 7.

Answer to Problem 79E

The probability when X is equal to 7 is 0.57.

Explanation of Solution

Given information:

Probability of sample is:

Days X	Probabilities
0	0.04
1	0.03
2	0.06
3	0.08
4	0.09
5	0.08
6	0.05

Calculation:

Add all probabilities are:

  $\begin{array}{c}P\left(X\le 6\right)=0.04+0.03+0.06+0.08+0.09+0.08+0.05\\ =0.43\end{array}$

The sum of the all probabilities should be equal to 1.

Now, find the missing probabilities.

  $\begin{array}{c}P\left(X=7\right)=1-P\left(X\le 6\right)\\ =1-0.43\\ =0.57\end{array}$

Hence, the probability when X is equal to 7 is 0.57.

(b)

To determine

To calculate: The mean of the random variable X.

Answer to Problem 79E

The mean of the random variable X is 5.44.

Explanation of Solution

Given information:

Probability of sample is:

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

Formula used:

The sample mean $\mu ={\displaystyle \sum xP\left(x\right)}$

Calculation:

Find the mean, use the formula $\mu ={\displaystyle \sum xP\left(x\right)}$ .

  $\begin{array}{c}\mu ={\displaystyle \sum xP\left(x\right)}\\ =0\times 0.04+1\times 0.03+2\times 0.06+3\times 0.08+4\times 0.09+5\times 0.08+6\times 0.05+7\times 0.57\\ =5.44\end{array}$

Hence, the mean of the random variable X is 5.44.

(c)

To determine

To calculate: The standard deviation of the random variable X.

Answer to Problem 79E

The sample mean of 4.96 is within standard deviation of the mean 2.1369.

Explanation of Solution

Given information:

Probability of sample is:

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

Sample mean $\mu =5.44$

Formula used:

The standard deviation $\sigma =\sqrt{{\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}}$

Calculation:

Find the standard deviation, use the formula $\sigma =\sqrt{{\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}}$ .

  $\begin{array}{c}\sigma =\sqrt{{\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}}\\ =\sqrt{\left[\begin{array}{c}{\left(0-5.44\right)}^2\times 0.04+{\left(1-5.44\right)}^2\times 0.03+{\left(2-5.44\right)}^2\times 0.06\\ +{\left(3-5.44\right)}^2\times 0.08+{\left(4-5.44\right)}^2\times 0.09+{\left(5-5.44\right)}^2\times 0.08\\ +{\left(6-5.44\right)}^2\times 0.05+{\left(7-5.44\right)}^2\times 0.57\end{array}\right]}\\ =2.1369\end{array}$

Hence, the sample mean of 4.96 is within standard deviation of the mean 2.1369.