Question

One of the problems encountered by corporations in America is finding an adequate number of employees who want to move into management. Recent surveys of workers in America taken by the Department of Labor in Washington D. C. revealed that only 20% of employees would like to move into management and be the boss. Suppose that a random sample of 75 U.S. workers was taken and each person was asked whether or not they would like to move into management. Find the probability that at least 18 of the 75 sampled employees would like to move into management.

Answer 1

Answer:

0.2358 = 23.58% probability that at least 18 of the 75 sampled employees would like to move into management.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes 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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions 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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

20% of employees would like to move into management and be the boss.

This means that $p = 0.2$

Sample of 75:

This means that $n = 75$

Mean and standard deviation:

$\mu = E(X) = np = 75(0.2) = 15$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{75*0.2*0.8} = 3.4641$

Find the probability that at least 18 of the 75 sampled employees would like to move into management.

Using continuity correction, this is $P(X \geq 18 - 0.5) = P(X \geq 17.5)$, which is 1 subtracted by the p-value of X = 17.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17.5 - 15}{3.4641}$

$Z = 0.72$

$Z = 0.72$ has a p-value of 0.7642.

1 - 0.7642 = 0.2358

0.2358 = 23.58% probability that at least 18 of the 75 sampled employees would like to move into management.