Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is \$23.50 with a...

 The mean amount purchased by a typical customer at Churchill's Grocery Store is \$23.50 with a standard deviation of \$6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 52 customers, answer the following questions.
 a. What is the likelihood the sample mean is at least \$25.50? (Round z value to 2 decimal places and final answer to 4 decimal places.)
 Probability
 b. What is the likelihood the sample mean is greater than \$22.50 but less than \$25.50? (Round z value to 2 decimal places and final answer to 4 decimal places.)
 Probability
 c. Within what limits will 95 percent of the sample means occur? (Round your answers to 2 decimal places.)
 Sample mean and

a)

To give that the likelihood the sample mean is at least \$25.50

P(xbar >= 25.5)

= P((xbar - mean)/(s /vn) >(25.5 - 23.5)/(6 /sqrt(52)))

= P(Z > 2.40)

P(xbar >= 25.5) = 0.0082 [since from standard normal distribution table]

b)

To give that  the likelihood the sample mean is greater than \$22.50 but less than \$25.50

P(22.5 < X < 25.5)

= P((22.5 - 23.5)/(6 /sqrt(52)) < Z < (25.5 - 23.5)/(6/sqrt(52)))

= P(-1.2 < Z < 2.40)

P(22.5 < X < 25.5) = 0.8767

c)

To determine the limits for 95% confidence interval

a = 0.05,

Z(0.025) = 1.96 (since from standard normal table)

Here 95% confidence interval is xbar +/- Z*s/vn

substitute values then we get limits

= 23.5 +/- 1.96*(6/sqrt(52)

Limits are = (21.87, 25.13)