165k views
5 votes
In a large company, the proportion of employees who were promoted during the last year was 0.10. If 100

employees were chosen from this company randomly, what is the probability that at least 15 of them were
promoted during the last year?
O 0.5
O 0.4525
O 0.0475
O 0.9525

User Sidon
by
4.5k points

1 Answer

1 vote

Answer:

0.0475

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:


√(V(X)) = √(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 = (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 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 = √(V(X)) = √(np(1-p)) = √(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 = (X - \mu)/(\sigma)


Z = (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.

User Denis Kokorin
by
3.7k points