Question

question number 6. (10.00 points) in testing a certain kind of missile, target accuracy is measured by the average distance x (from the target) at which the missile explodes. the distance x is measured in miles and the sampling distribution of x is given by:

x	p(x)
10	1/6
50	1/3
100	5/12

calculate the variance of this sampling distribution.
1416.7
472.2
60.0
2800.0
37.6
none of the above

Explanation:

Step1: Recall variance formula for discrete - random variable

The formula for the variance of a discrete - random variable $X$ is $Var(X)=\sum_{i}(x_{i}-\mu)^{2}P(x_{i})$, where $\mu = E(X)=\sum_{i}x_{i}P(x_{i})$. First, find the expected value $\mu$.
\[

$$\begin{align*} E(X)&=\sum_{i}x_{i}P(x_{i})\\ &=0\times\frac{1}{12}+ 10\times\frac{1}{6}+50\times\frac{1}{3}+100\times\frac{5}{12}\\ &=0+\frac{10}{6}+\frac{50}{3}+\frac{500}{12}\\ &=\frac{20 + 200+500}{12}\\ &=\frac{720}{12}\\ & = 60 \end{align*}$$

\]

Step2: Calculate variance

\[

$$\begin{align*} Var(X)&=(0 - 60)^{2}\times\frac{1}{12}+(10 - 60)^{2}\times\frac{1}{6}+(50 - 60)^{2}\times\frac{1}{3}+(100 - 60)^{2}\times\frac{5}{12}\\ &=3600\times\frac{1}{12}+2500\times\frac{1}{6}+100\times\frac{1}{3}+1600\times\frac{5}{12}\\ &=300+\frac{1250}{3}+\frac{100}{3}+\frac{2000}{3}\\ &=300+\frac{1250 + 100+2000}{3}\\ &=300+\frac{3350}{3}\\ &=\frac{900+3350}{3}\\ &=\frac{4250}{3}\approx1416.7 \end{align*}$$

\]

Answer:

1416.7