    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 0.5Given, 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)=1√2πσe−12⎛⎜⎝X−¯Xσ⎞⎟⎠2 Putting the value of ¯X and σ we get P(X)=12√2πe−(X−1)28 at X = 0 P(0)=12√2πe−−18 P(0)=0.176 at X = -1 P(−1)=12√2πe−(−1−1)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=0−12=−12 So, P(x<0)=P(Zx<−12)=P(Zx>12) =12−P(0≤Zx≤12)<0.5  Suggest Corrections  1      Similar questions  Explore more