Question

Y~ Normal Distribution with mean 4 and variance 4.

a. We have P(Y< y)=.9 find the y

b. Find P(1<Y<2)

c. Find P(Y< -2)

d. Find P(Y>-2)

Answer 1

Solution:-

Given that,

mean = = 4

variance = ² = 4

standard deviation = = ² = 4 = 2

a) Using standard normal table,

P(Z < z) =0.9

= P(Z < 1.28) = 0.9

z = 1.28

Using z-score formula,

y = z * +

y = 1.28 * 2 + 4

y = 6.56

b) P(1 < Y < 2) = P[(1 - 4)/ 2) < (Y - ) /  < (2 - 4) / 2) ]

= P( -1.5 < z < -1.0)

= P(z < -1.0) - P(z < -1.5)

Using z table,

= 0.1587 - 0.0668

= 0.0919

c) P(Y < -2 ) = P[(Y - ) / < (-2 - 4 ) / 2]

= P(z < -3.0 )

Using z table,

= 0.0013

d) P(Y > -2) = 1 - p( Y < -2)

=1- p P[(Y - ) / < (-2 - 4) / 2]

=1- P(z < -3.0)

Using z table,

= 1 - 0.0013

= 0.9987