Question

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

Answer 1

The probability of picking 8 employee and none of them is dehydrated is 0.66

Explanation:

P(employee is dehydrated) = 5%=5/100=1/20= 0.05

P(employee not dehydrated) = 1 - 0.05= 0.95

The probability of picking 8 employee and they are not dehydrated is:

= (0.95)^8

= 0.6634204312890625

= 0.66

Answer 2

P(X=0) = 0.0332

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