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Question 2 (a) Identify the mean and variance of a standard normal random variable Z. Determine the following probabilities for the standard normal random variable Z: (i) P(Z1.52) (iii) P(Z> -2.15) (iv) P-2.34Z176) (4 marks) (b) Solve for the value of k if the probability P(-1.24 <Z-0.8? (3 marks) (c) Assume that X is normally distributed with a mean of 10 and a variance of 4. Determine the following probabilities: (ii) P(2<X<4) (ii) P-2<X<8) (5 marks) Assume that Y is normally distributed with a mean of 0 and a standard deviation of 0.45. What is the value of m if the probability P(-mY<m)-0.99? (d) (4 marks)

Answer 1

Part a)

i)

X ~ N ( µ = 0 , σ = 1 )
P ( X < 1.52 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1.52 - 0 ) / 1
Z = 1.52
P ( ( X - µ ) / σ ) < ( 1.52 - 0 ) / 1 )
P ( X < 1.52 ) = P ( Z < 1.52 )
P ( X < 1.52 ) = 0.9357

ii)

X ~ N ( µ = 0 , σ = 1 )
P ( X > 1.44 ) = 1 - P ( X < 1.44 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1.44 - 0 ) / 1
Z = 1.44
P ( ( X - µ ) / σ ) > ( 1.44 - 0 ) / 1 )
P ( Z > 1.44 )
P ( X > 1.44 ) = 1 - P ( Z < 1.44 )
P ( X > 1.44 ) = 1 - 0.9251
P ( X > 1.44 ) = 0.0749

iii)

X ~ N ( µ = 0 , σ = 1 )
P ( X > -2.15 ) = 1 - P ( X < -2.15 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( -2.15 - 0 ) / 1
Z = -2.15
P ( ( X - µ ) / σ ) > ( -2.15 - 0 ) / 1 )
P ( Z > -2.15 )
P ( X > -2.15 ) = 1 - P ( Z < -2.15 )
P ( X > -2.15 ) = 1 - 0.0158
P ( X > -2.15 ) = 0.9842

iv)

X ~ N ( µ = 0 , σ = 1 )
P ( -2.34 < X < 1.76 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( -2.34 - 0 ) / 1
Z = -2.34
Z = ( 1.76 - 0 ) / 1
Z = 1.76
P ( -2.34 < Z < 1.76 )
P ( -2.34 < X < 1.76 ) = P ( Z < 1.76 ) - P ( Z < -2.34 )
P ( -2.34 < X < 1.76 ) = 0.9608 - 0.0096
P ( -2.34 < X < 1.76 ) = 0.9512

Part b)

P ( -1.24 < Z < k )
P ( -1.24 < X < k ) = P ( Z < k ) - P ( Z < -1.24 )
0.8 = P ( Z < k) - 0.1075

P ( Z < k) = 0.8 + 0.1075

P ( Z < k) = 0.9075

Looking for value 0.9075 in standard normal table to find the Z value.

Z = 1.33 i.e k = 1.33
P ( -1.24 < X < 1.33 ) = 0.80

Part c)

i)

X ~ N ( µ = 10 , σ = 2 )
P ( X > 9 ) = 1 - P ( X < 9 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 9 - 10 ) / 2
Z = -0.5
P ( ( X - µ ) / σ ) > ( 9 - 10 ) / 2 )
P ( Z > -0.5 )
P ( X > 9 ) = 1 - P ( Z < -0.5 )
P ( X > 9 ) = 1 - 0.3085
P ( X > 9 ) = 0.6915

ii)

X ~ N ( µ = 10 , σ = 2 )
P ( 2 < X < 4 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 2 - 10 ) / 2
Z = -4
Z = ( 4 - 10 ) / 2
Z = -3
P ( -4 < Z < -3 )
P ( 2 < X < 4 ) = P ( Z < -3 ) - P ( Z < -4 )
P ( 2 < X < 4 ) = 0.0013 - 0
P ( 2 < X < 4 ) = 0.0013

iii)

X ~ N ( µ = 10 , σ = 2 )
P ( -2 < X < 8 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( -2 - 10 ) / 2
Z = -6
Z = ( 8 - 10 ) / 2
Z = -1
P ( -6 < Z < -1 )
P ( -2 < X < 8 ) = P ( Z < -1 ) - P ( Z < -6 )
P ( -2 < X < 8 ) = 0.1587 - 0
P ( -2 < X < 8 ) = 0.1587

Part d)

X ~ N ( µ = 0 , σ = 0.45 )
P ( a < X < b ) = 0.99
Dividing the area 0.99 in two parts we get 0.99/2 = 0.495
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.495
Area above the mean is b = 0.5 + 0.495
Looking for the probability 0.005 in standard normal table to calculate critical value Z = -2.58
Looking for the probability 0.995 in standard normal table to calculate critical value Z = 2.58
Z = ( Y - µ ) / σ
-2.58 = ( Y - 0 ) / 0.45
a = -1.16
2.58 = ( Y - 0 ) / 0.45
b = 1.16
P ( -1.16 < Y < 1.16 ) = 0.99
i.e m = $\pm$ 1.16