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the random variable x is the crew size of a randomly selected shuttle m…

Question

the random variable x is the crew size of a randomly selected shuttle mission. its probability distribution is shown below. complete parts a through c.

x2345678
--------
p(x = x)0.0460.0340.0390.3020.1920.3050.082

a. as the number of observations, n, decreases, the mean of the observations will approach the mean of the random variable.
b. as the number of observations, n, increases, the mean of the observations will approach the mean of the random variable.
c. the observed value of the random variable will be equal to the mean of the random variable in most observations.
d. the observed value of the random variable will be less than the mean of the random variable in most observations.
b. obtain the standard deviation of the random variable.
σ = (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the mean of a discrete - random variable

The mean $\mu$ of a discrete random variable $X$ is given by $\mu=\sum_{x}x\cdot P(X = x)$.
\[

$$\begin{align*} \mu&=2\times0.046 + 3\times0.034+4\times0.039 + 5\times0.302+6\times0.192+7\times0.305+8\times0.082\\ &=0.092+0.102 + 0.156+1.51+1.152+2.135+0.656\\ &=5.803 \end{align*}$$

\]

Step2: Recall the formula for the variance of a discrete - random variable

The variance $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(X = x)$.
\[

$$\begin{align*} &(2 - 5.803)^{2}\times0.046+(3 - 5.803)^{2}\times0.034+(4 - 5.803)^{2}\times0.039+(5 - 5.803)^{2}\times0.302+(6 - 5.803)^{2}\times0.192+(7 - 5.803)^{2}\times0.305+(8 - 5.803)^{2}\times0.082\\ &=(- 3.803)^{2}\times0.046+(-2.803)^{2}\times0.034+(-1.803)^{2}\times0.039+(-0.803)^{2}\times0.302+(0.197)^{2}\times0.192+(1.197)^{2}\times0.305+(2.197)^{2}\times0.082\\ &=14.462809\times0.046 + 7.856809\times0.034+3.250809\times0.039 + 0.644809\times0.302+0.038809\times0.192+1.432809\times0.305+4.826809\times0.082\\ &=0.665289+0.267132+0.126782+0.194732+0.007451+0.436007+0.395808\\ &=2.103191 \end{align*}$$

\]

Step3: Recall the relationship between standard deviation and variance

The standard deviation $\sigma=\sqrt{\sigma^{2}}$.
\[
\sigma=\sqrt{2.103191}\approx1.450
\]

Answer:

1.450