Question

4. X has a normal distribution with mean=100 and a variance=25. Evaluate the following.

(e) Pr( 101.7 < X < 112.3)

(f) Pr( 93.85 < X < 106.15)

(g) Pr( 90.15 < X < 94.65)

(h) The corresponding X0 value, given that Pr(X < X0)= 0.9115.

Answer 1

Mean = 100

S.D= =5

(e) Pr( 101.7 < X < 112.3)

Z score at 101.7

Z = (X - μ) / σ
Z = (101.7 - 100) / 5
Z = 0.34

Z score at 112.3

Z = (X - μ) / σ
Z = (112.3 - 100) / 5
Z = 2.46

Pr( 101.7 < X < 112.3) = P( 2.46 < Z ) - P(o.34< Z )

= 0.36

(f) Pr( 93.85 < X < 106.15)

at x=93.85

Z = (X - μ) / σ
Z = (93.85 - 100) / 5
Z = -1.23

at x=106.15

Z = (X - μ) / σ
Z = (106.15 - 100) / 5
Z = 1.23

P( -1.23 < Z < 1.23) = 0.7813

(g) Pr( 90.15 < X < 94.65)

at x= 90.15

Z = (X - μ) / σ
Z = (90.15 - 100) / 5
Z = -1.97

at x=94.65

Z = (X - μ) / σ
Z = (94.65 - 100) / 5
Z = -1.07

P( -1.97 < Z < -1.07) = 0.1179

h)

P( X < X0) = 0.9115

Z score at p value of 0.9115

Z = 1.35005

Z = (X - μ) / σ
1.35005 = (X0 - 100) / 5
X0= 100+6.75

X0= 106.75