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Question

Suppose X is a normal random variable with mean 0 and variance 4. Then the mean of the absolute value of X is

A
12π
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B
22π
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C
22π
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D
2π
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Solution

The correct option is B 22π
For normal distribution:
f(x)=1σ2πe(xμ)22σ2=122πex28

μmean, σstandard deviation & given μ=0, σ2=4

mean of absolute value of x = E(|x|) =
=|x|f(x)dx=20|x|f(x)dx
( even function)

Now put x28=y

xdx=4dy

=12π0ey(4dy)

=22π0ey dy

=22π

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