On a very hot summer day, 5 percent of the production employees at Highlander Steel are dehydrated. The production employees are randomly selected for a special in-depth study on dehydration. What is the probability of randomly selecting 8 production employees on a hot summer day and finding that none of them are dehydrated

P(X=0) = 0.0332

Step-by-step explanation:

5% of the employees are dehydrated. I.e. p = 5/100

p = 0.05

The percentage of employees that are not dehydrated = 100-5 = 95%

q = 1 - p = 1 - 0.05

q = 0.95

A total of 8 employees were randomly selected. n = 8

This is a binomial distribution question

P(X =r) = nCr (p^r)q^(n-r)

n = 8, r = 0

n -r = 8-0 = 8

P(X = 0) = 8C0 (0.05^0)(0.95^8)

8C0 = 1

P(X=0) = 1 * 1* 0.6634

P(X=0) = 0.6634

what is the question here

step-by-step explanation: