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?

Respuesta :

Answer:

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:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 50, p = 0.6[/tex]

So

[tex]\mu = E(X) = np = 50*0.6 = 30[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.6*0.4} = 3.46[/tex]

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{24.5 - 30}{3.46}[/tex]

[tex]Z = -1.59[/tex]

[tex]Z = -1.59[/tex] 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