Question

The mean amount purchased by each customer at Churchill’s Grocery Store is $22 with a standard deviation of $7. The population is positively skewed. For a sample of 50 customers, answer the following questions:

a. What is the likelihood the sample mean is at least $23? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)

Sample mean            

b. What is the likelihood the sample mean is greater than $20 but less than $23? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)

Sample mean            

c. Within what limits will 95% of the sample means occur? (Round the final answers to 2 decimal places.)

Sample mean             and

Answer 1

a)

µ = 22, σ = 7, n = 50
P(X̅ > 23) =
= P( (X̅-μ)/(σ/√n) > (23-22)/(7/√50) )
= P(z > 1.01)
= 1 - P(z < 1.01)
Using excel function:
= 1 - NORM.S.DIST(1.01, 1)
= 0.1562

b)

P(20 < X̅ < 23) =
= P( (20-22)/(7/√50) < (X-µ)/(σ/√n) < (23-22)/(7/√50) )
= P(-2.02 < z < 1.01)
= P(z < 1.01) - P(z < -2.02)
Using excel function:
= NORM.S.DIST(1.01, 1) - NORM.S.DIST(-2.02, 1)
= 0.8221

c)

Proportion in the middle = 0.95
Proportion on left and right side of normal curve = 1-0.95 = 0.025
Z score at p = 0.025 using excel = NORM.S.INV(0.025) = 1.96
Value of X1 = µ - z*(σ/√n) = 22 - (1.96)*7/√50 = 20.06
Value of X2 = µ + z*(σ/√n) = 22 + (1.96)*7/√50 = 23.94