Question

On a very hot summer day, 5% of the production employees at Midland States Steel are absent from work. The production employees are randomly selected for a special in-depth study on absenteeism. What is the probability of randomly selecting 10 production employees on a hot summer day and finding that none of them are absent

Answer 1

Answer:

59.87% probability of randomly selecting 10 production employees on a hot summer day and finding that none of them are absent

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they are absent, or they are not. The probability of an employee being absent is independent from other employees. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

5% of the production employees at Midland States Steel are absent from work.

This means that $p = 0.05$

What is the probability of randomly selecting 10 production employees on a hot summer day and finding that none of them are absent

This is P(X = 0) when n = 10. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.05)^{0}.(0.95)^{10} = 0.5987$

59.87% probability of randomly selecting 10 production employees on a hot summer day and finding that none of them are absent