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 $33 and the estimated standard deviation is about $6.

(a) Consider a random sample of *n* = 50 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? (multiple
choice)

a.The sampling distribution of *x* is approximately
normal with mean *μ*_{x} = 33 and standard
error *σ*_{x} = $0.12.

b.The sampling distribution of *x* is not normal.

c. The sampling distribution of *x* is approximately
normal with mean *μ*_{x} = 33 and standard
error *σ*_{x} = $0.85.

d.The sampling distribution of *x* is approximately
normal with mean *μ*_{x} = 33 and standard
error *σ*_{x} = $6.

Is it necessary to make any assumption about the shape of the
*x* distribution? Explain your answer. (multiple choice)

a.It is not necessary to make any assumption about the
*x* distribution because both *μ* and *σ* are
given and they are very large.

b.It is necessary to assume that the shape of the *x*
distribution is approximately Normal even though it is not
stated.

c.It is necessary to assume since the population size was not given.

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

(b) What is the probability that *x* is between $31 and $35?
(Round your answer to four decimal places.)

Answer #1

Part a

We are given n=50, µ = 33, σ = 6

So, µ_{xbar} = 33, σ_{xbar} = σ/sqrt(n) =
6/sqrt(50) = 0.848528137 = 0.85 approximately

So, correct answer is given as below:

c. The sampling distribution of *x* is approximately
normal with mean *μ** _{xbar}* = 33 and
standard error

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

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

Explanation: We know that the sampling distribution of sample statistic follows an approximate normal distribution even though the population distribution does not follow a normal distribution. Also, n is given as 50 which is greater than 30, this means n is adequate or large.

(b) What is the probability that *x* is between $31 and
$35? (Round your answer to four decimal places.)

P(31<Xbar<35) = P(Xbar<35) – P(Xbar<31)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (35 – 33)/[6/sqrt(50)]

Z = 2.357022604

P(Z<2.357022604) = P(Xbar<35) = 0.990788937

(by using z-table)

Now find P(Xbar <31)

Z = (31 – 33)/[6/sqrt(50)]

Z = -2.3570226

P(Z< -2.3570226) = P(Xbar<31) = 0.009211063

(by using z-table)

P(31<Xbar<35) = P(Xbar<35) – P(Xbar<31)

P(31<Xbar<35) = 0.990788937 - 0.009211063

P(31<Xbar<35) = 0.9815779

Required probability = 0.9816

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