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Question

Let X be a normal random variable with mean 1 and variance 4. The probability P(X < 0) is

A
0.5
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B
greater than zero and less than 0.5
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C
greater than 0.5 and less than 1.0
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D
1.0
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Solution

The correct option is B greater than zero and less than 0.5
Given, a normal random variable, X with

mean (¯X)=1

variance (σ2)=4

Standard deviation=σ2=4=2

We know that probability density function of normal random variable is

P(X)=12πσe12X¯Xσ2

Putting the value of ¯X and σ we get

P(X)=122πe(X1)28

at X = 0

P(0)=122πe18

P(0)=0.176

at X = -1

P(1)=122πe(11)28=0.1209<0.5

Hence increasing the value of X in negative direction, P(X) will decrease and tend to zero.

P(X < 0) always lies in the range 0 < P(X < 0) < 0.5.

Alternative Solution

Mean (¯X)=1, σ=2

So, Zx=x¯xσ=012=12

So, P(x<0)=P(Zx<12)=P(Zx>12)

=12P(0Zx12)<0.5

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