Question

Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $48 and the estimated standard deviation is about $7.

(a) Consider a random sample of n = 60 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution?

The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.12. The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $7. The sampling distribution of x is not normal. The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.90.


Is it necessary to make any assumption about the x distribution? Explain your answer.

It is necessary to assume that x has an approximately normal distribution. It is not necessary to make any assumption about the x distribution because μ is large. It is necessary to assume that x has a large distribution. It is not necessary to make any assumption about the x distribution because n is large.
(b) What is the probability that x is between $46 and $50? (Round your answer to four decimal places.)
(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $46 and $50? (Round your answer to four decimal places.)
(d) In part (b), we used x, the average amount spent, computed for 60 customers. In part (c), we used x, the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen?

Answer 1

(a) σx = σ/√n = 7/√60 = 0.90

The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.90

It is necessary to assume that x has an approximately normal distribution.

(b) x1-bar = 46 and x2-bar = 50

z1 = (x1-bar - μ)/σx = ((46 - 48)/0.90 = -2.2131 and z2 = (x2-bar - μ)/σx = (50 - 48)/0.90 = 2.2131

P(46 < x-bar < 50) = P(-2.2131 < z < 2.2131) = 0.9731

(c) x1 = 46 and x2 = 50

z1 = (x1 - μ)/σ = (46 - 50)/7 = -0.2857 and z2 = (x2 - μ)/σ = (50 - 48)/7 = 0.2857

P(46 < x < 50) = P(-0.2857 < z < 0.2857) = 0.2249

(d) The x-bar distribution is approximately normal while the x distribution is not normal.

(e) The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer.

Explanation:

(a) σx = σ/√n = 7/√60 = 0.90

The sampling distribution of x is approximately normal with mean μx = 48 and standard error σx = $0.90

It is necessary to assume that x has an approximately normal distribution.

(b) x1-bar = 46 and x2-bar = 50

z1 = (x1-bar - μ)/σx = ((46 - 48)/0.90 = -2.2131 and z2 = (x2-bar - μ)/σx = (50 - 48)/0.90 = 2.2131

P(46 < x-bar < 50) = P(-2.2131 < z < 2.2131) = 0.9731

(c) x1 = 46 and x2 = 50

z1 = (x1 - μ)/σ = (46 - 50)/7 = -0.2857 and z2 = (x2 - μ)/σ = (50 - 48)/7 = 0.2857

P(46 < x < 50) = P(-0.2857 < z < 0.2857) = 0.2249

(d) The x-bar distribution is approximately normal while the x distribution is not normal.

(e) The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer.