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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 :

Answer:

Option c) 761.0 is correct

Therefore [tex]\sigma^2=761[/tex]

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)        [tex]\frac{1}{40}[/tex]     [tex]\frac{1}{20}[/tex]      [tex]\frac{1}{10}[/tex]       [tex]\frac{33}{40}[/tex]

To calculate the variance of this sampling distribution :

X        P(X)     XP(X)       [tex]X^2P(X)[/tex]

0            [tex]\frac{1}{40}[/tex]          0                     0

10           [tex]\frac{1}{20}[/tex]           [tex]\frac{1}{2}[/tex]                     5

50          [tex]\frac{1}{10}[/tex]          5                  250

100        [tex]\frac{33}{40}[/tex]          [tex]\frac{165}{2}[/tex]               8250

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                 [tex]\sum XP(X)=88[/tex]    [tex]\sum X^2P(X)=8505[/tex]

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Variance [tex]\sigma^2=\sum X^2P(X)-(\sum XP(X))^2[/tex]

Now substitute the values in the above formula we get

  • [tex]\sigma^2=8505-88^2[/tex]
  • [tex]=8505-7744[/tex]
  • [tex]=761[/tex]
  • Therefore [tex]\sigma^2=761[/tex]
  • Therefore Variance is 761
  • Option c) 761.0 is correct

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