Question

Suppose that 60% of employees at a particular corporation participate in the optional retirement plan. If a random sample of 50 employees is selected, what is the probability that at least 25 employees in the sample will participate in the optional retirement plan?

Answer 1

94.41% probability that at least 25 employees in the sample will participate in the optional retirement plan

Explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`n = 50, p = 0.6`

So

`\mu = E(X) = np = 50*0.6 = 30`

`\sigma = √(V(X)) = √(np(1-p)) = √(50*0.6*0.4) = 3.46`

What is the probability that at least 25 employees in the sample will participate in the optional retirement plan?

Using continuity correction, this is
`P(X \geq 25 - 0.5) = P(X \geq 24.5)`, which is 1 subtracted by the pvalue of Z when X = 24.5

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

`Z = (24.5 - 30)/(3.46)`

`Z = -1.59`

`Z = -1.59` has a pvalue of 0.0559

1 - 0.0559 = 0.9441

94.41% probability that at least 25 employees in the sample will participate in the optional retirement plan