Question

The manager of a donut store believes that 35% of the customers are first-time customers. A random sample of 150 customers will be used to estimate the proportion of first-time customers. Assuming this belief is correct, what is the probability that the sample proportion will be between 0.2 and 0.4

Answer 1

0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

The manager of a donut store believes that 35% of the customers are first-time customers.

This means that
`p = 0.35`

Sample of 150 customers

This means that
`n = 150`

Mean and standard deviation:

`\mu = p = 0.35`

`s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.35*0.65)/(150)} = 0.0389`

What is the probability that the sample proportion will be between 0.2 and 0.4?

p-value of Z when X = 0.4 subtracted by the p-value of Z when X = 0.2.

X = 0.4

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (0.4 - 0.35)/(0.0389)`

`Z = 1.28`

`Z = 1.28` has a p-value of 0.8997

X = 0.2

`Z = (X - \mu)/(s)`

`Z = (0.2 - 0.35)/(0.0389)`

`Z = -3.85`

`Z = -3.85` has a p-value of 0.0001

0.8997 - 0.0001 = 0.8996

0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4