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 $28 and the estimated standard deviation is about $8.

(a) Consider a random sample of n = 40 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?
(b) What is the probability that x is between $38 and $42?
(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $38 and $42?

The central limit theorem explains that a sufficiently large enough sample size follows the same probability distribution as the population probability distribution.

For the mean, the central limit theorem explains further that, given a sufficiently large sample size from a population with a finite level of variance, the mean of all samples from the same population will be approximately equal to the mean of the population. That is,

μₓ = μ = $28

And the standard deviation of the distribution of sample means will be related to the standard deviation of the population through

σₓ = (σ/√n)

σ = standard deviation of the population = $8

n = sample size = 40

σₓ = (8/√40) = $1.265

The sampling distribution can easily be assumed to be normal with a mean of $28 and a standard error of $1.265.

It is more logical for a distribution with this mean and standard deviation to approximate a normal distribution as -3 and +3 standard deviations from the mean is still very logical.

b) Probability of the average of sample distribution x, being between $38 and $42.

We assume that the distribution of sample means approximate a normal distribution.

P(38 < x < 42)

We first standardize 38 and 42

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For $38

z = (x - μ)/σ = (38 - 28)/1.265 = 7.91

For $42

z = (x - μ)/σ = (42 - 28)/1.265 = 11.07

The required probability,

P(38 < x < 42) = P(7.91 < z < 11.07)

We'll use data from the normal probability table for these probabilities

P(38 < x < 42) = P(7.91 < z < 11.07)

= P(z < 11.07) - P(z < 7.91)

= 1 - 1 = 0.00

c) For this part, we evaluate the same probability for the population distribution

Our population distribution is assumed to be a normal distribution.

P(38 < x < 42)

We first standardize 38 and 42

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For $38

z = (x - μ)/σ = (38 - 28)/8 = 1.25

For $42

z = (x - μ)/σ = (42 - 28)/1.265 = 1.75

The required probability,

P(38 < x < 42) = P(1.25 < z < 1.75)

We'll use data from the normal probability table for these probabilities

P(38 < x < 42) = P(1.25 < z < 1.75)

= P(z < 1.75) - P(z < 1.25)

= 0.960 - 0.894 = 0.0066

Hope this Helps!!!