Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50, with a standard deviation of $500. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions (a) What is the likelihood the sample mean is at least $25.00 decimal places and final answer to 4 decimal places.) Round z value to 2 Probability What is the likelihood the sample mean is greater than $2250 but less than $25.00% Round z value to 2 decimal places and final answer to 4 decimal places.) Probability wthin what limits will 90% of the sample means occur? Round your answers 10 2 The limits are and

Answer 1

Given the data,

The mean amount purchased by a typical customer μ is $23.50

standard deviation σ is $5.00

Sample size n is 50

a) we have a z-score given by:

Z=( X-μ)/(σ⁄√n)=(25-23.50)/(5⁄√50)=2.12

Using the Z-table, we get: P(Z>2.12)=0.5-0.4830=0.0170

Hence, the probability of a sample mean of at least $25.00 is 0.0170.

b) For $22.50, the z-score is:

Z=(X-μ))/(σ⁄√n)=(22.50-23.50)/(5⁄√50)=-1.41

Using the Z-table, we get: P(-1.41>Z>2.12)=1-0.0793-0.0170=0.9037

Hence the probability of a sample mean between $22.50 and $25.00 is 0.9037.

c) We use the Z-table to find the Z-score that gives us 0.45 (0.45 * 2 = 0.90).

We find that the Z-score is 1.645. So, the limits correspond to Z-score = -1.645 and Z-score = 1.645.

Using the definition of Z-score, we get:

X ̅=(Z*σ)/√n+μ=(1.645*5)/√50+23.50=24.67
and:
X ̅=(Z*σ)/√n+μ=(-1.645*5)/√50+23.50=22.33

Hence the limits are $22.33 and $24.67.