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?

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

Step-by-step 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:

The standard deviation of the binomial distribution is:

Normal probability distribution

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

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 , .

In this problem, we have that:

So

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

Using continuity correction, this is , which is 1 subtracted by the pvalue of Z when X = 24.5

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

the first one, all those points on the line just make a strait line.

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step-by-step explanation: