Question

Given a Normal distribution with a mean (µ) of 60 and standard deviation (ᵟ) of 8, what is the probability that the sample mean (Xbar) is:

1. Less than 57?
2. Between 57and 62.5?
3. Above 65?
4. There is a 38% chance that the Xbar is above what value?

Answer 1

Mean = 60

S.D. = 8

a)

P( X < 57)

Z = (X - μ) / σ
Z = (57 - 60) / 8
Z = -0.375

Feom Z score table

P( X < 57)= P( Z < -0.375)

P( X < 57) = 0.3538

b)

P( 57< X < 62.5)

Z = (X - μ) / σ
Z = (57 - 60) / 8
Z = -0.375

Z= (X - μ) / σ
Z = (62.5 - 60) / 8
Z = 0.3125

Feom Z score table

P( 57< X < 62.5) = P( Z < 0.3125) - P( Z < -0.375)

0.6227-0.3538 = 0.2688

c)

P( X > 65)

Z = (X - μ) / σ
Z = (65 - 60) / 8
Z = 0.625

Feom Z score table

P( X > 65)= P( Z > 0.625)

P( X > 65) = 0.266

d)

P( X > x) =0.38

Z = 0.30548

Z = (X - μ) / σ
0.30548 = (X - 60) / 8
X = 60 + 2.44384

X=62.4438