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 $18 and the estimated standard deviation is about $9. (a) Consider a random sample of n = 70 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 = 18 and standard error σx = $1.08. The sampling distribution of x is not normal. The sampling distribution of x is approximately normal with mean μx = 18 and standard error σx = $9. The sampling distribution of x is approximately normal with mean μx = 18 and standard error σx = $0.13. Is it necessary to make any assumption about the shape of the x distribution? Explain your answer. It is not necessary to make any assumption about the x distribution because both μ and σ are given and they are very large. It is necessary to assume that the shape of the x distribution is approximately Normal even though it is not stated. It is not necessary to make any assumption about the x distribution since the sample size, n, is large. It is necessary to assume since the population size was not given. (b) What is the probability that x is between $16 and $20? (Round your answer to four decimal places.)

Answer 1

A) = 18

   = = 9/ = 1.08

The sampling distribution of is approximately normal with mean and standard error = 1.08

It is not necessary to make any assumption about the x distribution since the sample size n is large.

B) P(16 < < 20)

= P((16 - )/()) < ( - )/()) < (20 - )/())

= P((16 - 18)/(9/)) < Z < (20 - 18)/(9/))

= P(-1.86 < Z < 1.86)

= P(Z < 1.86) - P(Z < -1.86)

= 0.9686 - 0.0314

= 0.9372