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Suppose X is a normally distributed random variable with mean μ = 10 and variance σ2...

Suppose X is a normally distributed random variable with mean μ = 10 and variance σ2 = 4. Find P(9 < X < 12).

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Answer #1

Solution :

Given that ,

mean =   = 10

standard deviation = = 4=2  

P(9< x <12 ) = P[(9-10) / 2< (x - ) / < (12-10) /2 )]

= P( -0.5< Z < 1)

= P(Z <1 ) - P(Z < -0.5)

Using z table   

= 0.8413-0.3085

probability= 0.5328

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