Question

Space shuttles. The random variable X is the crew size of a randomly selected shuttle mission between April 12, 1981 and July 8, 2001. Its probability distribution is as follows.

Part (a) Find and interpret the mean of the random variable.

Part (b) Obtain the standard deviation of the random variable by using one of the formulas given in definition 5.10

Part (c) Construct a probability histogram for the random variable, locate the mean: and show one, two, and three standard deviation intervals.

Short answer

Part (a) 6.05

Part (b) 1.152

Part (c) Graph

Step-by-step solution

Part (a) Step 1. Given information. 

Take a look at the data table below. The crew size of a randomly selected crew mission is the random variable X.

Part (a) Step 2. The random variable's average.

$$\begin{gathered} \mu =\sum _{}x·(X=x) \\ \mu =2\left(0.03\right)+\left(4\times 0.022\right)+\left(5\times 0.267\right)+\left(6\times 0.207\right)+\left(7\times 0467\right)+8\left(0.007\right) \\ \mu =0.06+0.088+1.335+1.242+3.269+0.056 \\ \mu =6.05 \end{gathered}$$

The average number of people on a shuttle mission is 6.05

Part (b) Step 1.The random variable's standard deviation:

Standard deviation =$\sqrt{\sum _{}{x}^{2}P(X=x)-{\mu }^2}$

$$\begin{gathered} \sigma =\sqrt{\sum _{}{x}^{2}P(X=x)-{\mu }^2} \\ \sigma =\sqrt{4\left(0.03\right)+\left(16\times 0.022\right)+\left(25\times 0.267\right)+\left(36\times 0.207\right)+\left(49\times 0.467\right)-36.6} \\ \sigma =\sqrt{1.327} \\ \sigma =1.152 \end{gathered}$$

As a result, standard deviation = 1.152

Part (c) Step 1.  The mean, one, two, and three standard deviation intervals on the probability histogram graph of the random variable.

We have $\mu$and $\sigma$from sections a and b.

$$\begin{gathered} \left(\mu -\sigma ,\mu +\sigma \right)=\left(6.05-1.152,6.05+1.152\right) \\ =\left(4.9,7.2\right) \\ \left(\mu -2\sigma ,\mu +2\sigma \right)=\left(6.05-2\left(1.152\right),6.05+2\left(1.152\right)\right) \\ =(3.75,8.35) \\ \left(\mu -3\sigma ,\mu +3\sigma \right)=\left(6.05-3\left(1.152\right),6.05+3\left(1.152\right)\right) \\ =\left(2.6,9.5\right) \end{gathered}$$

Draw the probability histogram graph:

The probability histogram graph of the random variable is shown above, along with the mean and one, two, and three standard deviation intervals.