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	1/9
50	2/9
100	11/18

calculate the variance of this sampling distribution.
3548.1
36.1
1300.0
433.3
73.3
none of the above

Explanation:

Step1: Recall variance formula

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, calculate the expected value $E(X)$.
$E(X)=\sum_{i}x_{i}P(x_{i})=0\times\frac{1}{18}+10\times\frac{1}{9}+50\times\frac{2}{9}+100\times\frac{11}{18}$
$=\frac{0 + 10\times2+50\times4 + 100\times11}{18}=\frac{0 + 20+200 + 1100}{18}=\frac{1320}{18}=\frac{220}{3}$.

Step2: Calculate variance

$Var(X)=(0 - \frac{220}{3})^{2}\times\frac{1}{18}+(10-\frac{220}{3})^{2}\times\frac{1}{9}+(50-\frac{220}{3})^{2}\times\frac{2}{9}+(100-\frac{220}{3})^{2}\times\frac{11}{18}$
$(0 - \frac{220}{3})^{2}\times\frac{1}{18}=\frac{220^{2}}{9}\times\frac{1}{18}=\frac{48400}{162}$
$(10-\frac{220}{3})^{2}\times\frac{1}{9}=(\frac{30 - 220}{3})^{2}\times\frac{1}{9}=(\frac{- 190}{3})^{2}\times\frac{1}{9}=\frac{36100}{81}$
$(50-\frac{220}{3})^{2}\times\frac{2}{9}=(\frac{150 - 220}{3})^{2}\times\frac{2}{9}=(\frac{-70}{3})^{2}\times\frac{2}{9}=\frac{4900\times2}{81}=\frac{9800}{81}$
$(100-\frac{220}{3})^{2}\times\frac{11}{18}=(\frac{300 - 220}{3})^{2}\times\frac{11}{18}=(\frac{80}{3})^{2}\times\frac{11}{18}=\frac{6400\times11}{162}=\frac{70400}{162}$
$Var(X)=\frac{48400+36100\times2 + 9800\times2+70400}{162}$
$=\frac{48400 + 72200+19600 + 70400}{162}=\frac{210600}{162}=1300.0$

Answer:

1300.0