Question

the following table displays a frequency distribution for the number of crew members on each shuttle mission over a 10 - year period. let x denote the crew size on a randomly selected shuttle from this time period.
crew size
frequency
4
5
5
2
6
39
7
20
8
4

1. 5, 6, 7, 8

a. what are the possible values of the random variable x?
b. use random - variable notation to represent the event that the shuttle mission obtained has a crew size of 7.
\\( \\{ x = 7 \\} \\)
c. find \\( p ( x = 5 ) \\).
\\( p ( x = 5 ) =.029 \\)
(round to three decimal places as needed.)
d. obtain the probability distribution of x.
(round to three decimal places as needed.)
\

$$\begin{tabular}{|c|c|} \\hline x & p(x) \\\\ \\hline 4 & \\\\ \\hline 5 & \\\\ \\hline 6 & \\\\ \\hline 7 & \\\\ \\hline 8 & \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Calculate total frequency

Total frequency \(n=4 + 5+39+20+4=72\)

Step2: Calculate \(P(X = 4)\)

\(P(X = 4)=\frac{4}{72}\approx0.056\)

Step3: Calculate \(P(X = 5)\)

\(P(X = 5)=\frac{5}{72}\approx0.069\)

Step4: Calculate \(P(X = 6)\)

\(P(X = 6)=\frac{39}{72}\approx0.542\)

Step5: Calculate \(P(X = 7)\)

\(P(X = 7)=\frac{20}{72}\approx0.278\)

Step6: Calculate \(P(X = 8)\)

\(P(X = 8)=\frac{4}{72}\approx0.056\)

Answer:

\(X\)	\(P(X)\)
\(5\)	\(0.069\)
\(6\)	\(0.542\)
\(7\)	\(0.278\)
\(8\)	\(0.056\)