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D Start e chegg Study | Guide G 1 Just Need (c). I Keel Ocean Connect WebAssign Assignm | 'Waiting for resp + × www wet assignnet vet Student/Assignment Responses sub itdepr 1806610SIQ14 points 1 Previous Answers BBUnderStat12650 My Notes Ask Your 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 $16 and estimated standard deviation is about $9. (a) Consider a random sample of n 130 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say abou probability distribution of x, the average amount spent by these customers due to impulise buying? what are the mean and standard deviation of the i distribution? O The sampling distribution of x is not normal O The sampling distribution of-is approximately normal with mean :-16 andetandard error . $9. The sampling distribution of x is approximately normal with mean :-16 and standard error oi-S0.79. The sampling distribution of x is approximately normal witn mean " 16 and standard error . SO07 Is it necessary to make any assumption about the x distribution? Explain wour answer. O It is not necessary to make any assumption about the x distribution because is large. O 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 It is necessary to assume that x has an approximately normal distribution. (b) What is the probability that x is between $14 and $187 (Round your answer to four decimal places.) 9886 (c) Let us assume that x has a distribution that is approximately normal. What is the probabiity that x is between $14 and si8 the amount spent by only one customer. The answers to parts (b) and (c) are (d) In part (b), we used x, the average amount spent, computed for 130 customers. In part (c), we used x, different. Why would this happen? 531 P O Type here to search

Explanation / Answer

Part b

We are given

Mean = 16, SE = 0.79, n = 130

We have to find P(14<Xbar<18)

P(14<Xbar<18) = P(Xbar<18) – P(Xbar<14)

First we have to find P(Xbar<18)

Z = (Xbar - µ) / [/sqrt(n)]

SE = [/sqrt(n)]

Z = (18 – 16) / [ 0.79] = 2.531646

P(Z<2.531646) = 0.994324

P(Xbar<18) = 0.994324

Now, we have to find P(Xbar<14)

Z = (14 – 16) / 0.79 = -2/0.79 = -2.53165

P(Z<-2.53165) = 0.005676

P(Xbar<14) = 0.005676

P(14<Xbar<18) = P(Xbar<18) – P(Xbar<14)

P(14<Xbar<18) = 0.994324 - 0.005676

P(14<Xbar<18) = 0.988647

Answer: 0.9886

Part c

We have to find P(14<X<18)

P(14<X<18) = P(X<18) – P(X<14)

Z = (X – mean) / SD

For X = 18

Z = (18 – 16) / 9

Z = 2/9 = 0.222222

P(Z< 0.222222) = 0.58793

For X = 14

Z = (14 – 16) / 9 = -2/9 = -0.22222

P(Z< -0.22222) = 0.41207

P(14<X<18) = P(X<18) – P(X<14)

P(14<X<18) = 0.58793 - 0.41207 = 0.175859

P(14<X<18) = 0.175859

Required probability = 0.1759

Part d

This would happen because in the first case we consider the sampling distribution of sample means, however in the second case we find out probability for individual score X which follows an approximate normal distribution (as described in part c).

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