Question

A random sample of 36 observations is drawn from a population with a mean equal to 51 and a standard deviation equal to 15.

1. What is the mean and the standard deviation of the sampling distribution of x̄?

1. Calculate the z-score corresponding to a value of x̄ = 45.5.
   
   
2. Calculate the z-score corresponding to a value of x̄ = 46.5.
   
   
3. Find P(x̄ ≥ 45.5) (to 4 decimals)
   
   
4. Find P(x̄ < 46.5) (to 4 decimals)
   
   
5. Find P(45.5 ≤ x̄ ≤ 46.5) (to 4 decimals)
   
   
6. There is a 60% chance that the value of x̄ is above  (to 4 decimals).

Answer 1

= 51

= 15

n = 36

a) For sampling distribution of ,

mean, = = 51

standard deviation, =

=

= 2.5

a) Z = (X - mean)/standard deviation

For = 45.5,

Z = (45.5 - 51)/2.5

= -2.2

b) For = 46.5,

Z = (46.5 - 51)/2.5

= -1.8

c) P( 45.5) = 1 - P( < 45.5)

= 1 - P(Z < -2.2)

= 1 - 0.0139

= 0.9861

d) P( < 46.5) = P(Z < -1.8)

= 0.0359

e) P(45.5 46.5) = P( < 46.5) - P( < 45.5)

= 0.0139 - 0.0359

= -0.0220

f) Let the value for which there is 60% chance that is above it be K

P( > K) = 0.60

P( < K) = 1 - 0.60

P(Z < (K - 51)/2.5) = 0.40

Take Z value corresponding to 0.40 from standard normal distribution table.

(K - 51)/2.5 = -0.25

K = 44.75