Question

Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping...

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 $44 and the estimated standard deviation is about $9.

(a) Consider a random sample of n = 120 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 not normal. The sampling distribution of x is approximately normal with mean μx = 44 and standard error σx = $9.     The sampling distribution of x is approximately normal with mean μx = 44 and standard error σx = $0.08. The sampling distribution of x is approximately normal with mean μx = 44 and standard error σx = $0.82.

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 necessary to assume that x has a large distribution.     It is not necessary to make any assumption about the x distribution because μ is large. 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 $42 and $46? (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 $42 and $46? (Round your answer to four decimal places.)
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 $44 and the estimated standard deviation is about $9.

Homework Answers

Answer #1

a) = 44

  

        

The sampling distribution of is approximately normal with mean = 44 and standard error = 0.82

b)

= P(-2.44 < Z < 2.44)

= P(Z < 2.44) - P(Z < -2.44)

= 0.9927 - 0.0073

= 0.9854

c) P(42 < X < 46)

= P(-0.22 < Z < 0.22)

= P(Z < 0.22) - P(Z < -0.22)

= 0.5871 - 0.4129

= 0.1742

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