Question

It has been estimated that 17.0 % of mutual fund shareholders are retired persons. Assuming a simple random sample of 400 shareholders has been selected: a. What is the probability that at least 20 % of those in the sample will be retired? b. What is the probability that between 15 % and 21 % of those in the sample will be retired?

Answer 1

a) 0.0548 = 5.48% probability that at least 20 % of those in the sample will be retired.

b) 0.8388 = 83.88% probability that between 15 % and 21 % of those in the sample will be retired.

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)}`

17.0 % of mutual fund shareholders are retired persons.

This means that
`p = 0.17`

Assuming a simple random sample of 400 shareholders has been selected

This means that
`n = 400`

Mean and standard deviation:

`\mu = p = 0.17`

`s = \sqrt{(0.17*0.83)/(400)} = 0.0188`

a. What is the probability that at least 20 % of those in the sample will be retired?

This is the 1 p-value of Z when X = 0.2. So

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

By the Central Limit Theorem

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

`Z = (0.2 - 0.17)/(0.0188)`

`Z = 1.6`

`Z = 1.6` has a p-value of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability that at least 20 % of those in the sample will be retired.

b. What is the probability that between 15 % and 21 % of those in the sample will be retired?

This is the p-value of Z when X = 0.21 subtracted by the p-value of Z when X = 0.15.

X = 0.21

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

`Z = (0.21 - 0.17)/(0.0188)`

`Z = 2.13`

`Z = 2.13` has a p-value of 0.9834

X = 0.15

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

`Z = (0.15 - 0.17)/(0.0188)`

`Z = -1.06`

`Z = -1.06` has a p-value of 0.1446

0.9834 - 0.1446 = 0.8388

0.8388 = 83.88% probability that between 15 % and 21 % of those in the sample will be retired.