Question

In a large company, the proportion of employees who were promoted during the last year was 0.10. If 100

Answer 1

Answer:

0.0475

Step-by-step explanation:

We use the binomial approximation to the normal 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:

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

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

In a large company, the proportion of employees who were promoted during the last year was 0.10.

This means that $p = 0.1$

100 employees

This means that $n = 100$

Mean and standard deviation:

$\mu = E(X) = np = 100*0.1 = 10$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.1*0.9} = 3$

What is the probability that at least 15 of them were promoted during the last year?

This is $P(X \geq 15)$, which is 1 subtracted by the pvalue of Z when X = 15. So

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

$Z = \frac{15 - 10}{3}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of 0.9525

1 - 0.9525 = 0.0475.

0.0475 is the answer.