Question

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 0 10 50 100
P(X) 1⁄40 1/20 1⁄10 33⁄40

Calculate the variance of this sampling distribution.

a) 27.6

b) 5138.7

c) 761.0

d) 253.7

e) 88.0

f) None of the above

Answer 1

Answer:

Option c) 761.0 is correct

Therefore $\sigma^2=761$

Therefore Variance is 761

Step-by-step explanation:

Given that 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             0     10     50     100

P(X)        $\frac{1}{40}$     $\frac{1}{20}$      $\frac{1}{10}$       $\frac{33}{40}$

To calculate the variance of this sampling distribution :

X        P(X)     XP(X)       $X^2P(X)$

0            $\frac{1}{40}$          0                     0

10           $\frac{1}{20}$           $\frac{1}{2}$                     5

50          $\frac{1}{10}$          5                  250

100        $\frac{33}{40}$          $\frac{165}{2}$               8250

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                 $\sum XP(X)=88$    $\sum X^2P(X)=8505$

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Variance $\sigma^2=\sum X^2P(X)-(\sum XP(X))^2$

Now substitute the values in the above formula we get

• $\sigma^2=8505-88^2$
• $=8505-7744$
• $=761$
• Therefore $\sigma^2=761$
• Therefore Variance is 761
• Option c) 761.0 is correct