Question

The probability of being a universal donor is 6% (O-negative-blood type). Suppose that 6 people come to a blood drive.' a) What are the mean and standard deviation of the number of universal donors among the 6 people? b) What is the probability that there are exactly three universal donors?

Answer 1

`Mean = 0.36`

`SD = 0.5817`

`P(x=3) = 0.003588`

Explanation:

Given

Let

A = Event of being a universal donor.

So:

`P(A) = 0.06`

`n = 6`

Solving (a): Mean and Standard deviation.

The mean is:

`Mean = np`

`Mean = 6 * 0.06`

`Mean = 0.36`

The standard deviation is:

`SD = √(np(1-p))`

`SD = √(6*0.06*(1-0.06))`

`SD = √(0.3384)`

`SD = 0.5817`

Solving (b): P(x = 3)

The event is a binomial event an dthe probability is calculated as:

`P(x) = ^nC_x * p^x * (1-p)^(n-x)`

So, we have:

`P(x=3) = ^6C_3 * 0.06^3 * (1-0.06)^(6-3)`

`P(x=3) = ^6C_3 * 0.06^3 * (1-0.06)^3`

`P(x=3) = 20 * 0.06^3 * (1-0.06)^3`

`P(x=3) = 0.003588`