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

Option c) 761.0 is correct

Therefore
`\sigma^2=761`

Therefore Variance is 761

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)
`(1)/(40)`
`(1)/(20)`
`(1)/(10)`
`(33)/(40)`

To calculate the variance of this sampling distribution :

X P(X) XP(X)
`X^2P(X)`

0
`(1)/(40)` 0 0

10
`(1)/(20)`
`(1)/(2)` 5

50
`(1)/(10)` 5 250

100
`(33)/(40)`
`(165)/(2)` 8250

________________________________

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

________________________________

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