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 $32 and the estimated standard deviation is about $7.

(a) Consider a random sample of n = 80 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 μ_x = 32 and standard error σ_x = $0.09. The sampling distribution of x is not normal.     The sampling distribution of x is approximately normal with mean μ_x = 32 and standard error σ_x = $7. The sampling distribution of x is approximately normal with mean μ_x = 32 and standard error σ_x = $0.78.

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 $30 and $34? (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 $30 and $34? (Round your answer to four decimal places.)
(d) In part (b), we used x, the average amount spent, computed for 80 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?

The sample size is smaller for the x distribution than it is for the x distribution. The mean is larger for the x distribution than it is for the x distribution.     The standard deviation is smaller for the x distribution than it is for the x distribution. The x distribution is approximately normal while the x distribution is not normal. The standard deviation is larger for the x distribution than it is for the x distribution.

Answer 1

Solution :

Given that,

mean = = $ 32

standard deviation = = $ 7

n = 80

a) = = $ 32

= / n = 7/ 80 = $ 0.78

The sampling distribution of x is approximately normal with mean μ_x = 32 and standard error σ_x = $0.78

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

b) P(30 < < 34)  

= P[(30 - 32) /0.78 < ( - ) / < (34 - 32) / 0.78)]

= P(-2.56 < Z < 2.56)

= P(Z < 2.56) - P(Z < -2.56)

Using z table,  

= 0.9948 - 0.0052  

= 0.9896

c) P( 30 < x < 34) = P[(30 - 32 ) / 7 ) < (x - ) /  < (34 - 32) / 7) ]

= P( -0.29 < z < 0.29)

= P(z < 0.29 ) - P(z < -0.29)

Using z table,

= 0.6141 - 0.3859

= 0.2282

d) The standard deviation is smaller for the    distribution than it is for the x distribution.