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?

Respuesta :

Answer:

[tex]Mean = 0.36[/tex]

[tex]SD = 0.5817[/tex]

[tex]P(x=3) = 0.003588[/tex]

Step-by-step explanation:

Given

Let

A = Event of being a universal donor.

So:

[tex]P(A) = 0.06[/tex]

[tex]n = 6[/tex]

Solving (a): Mean and Standard deviation.

The mean is:

[tex]Mean = np[/tex]

[tex]Mean = 6 * 0.06[/tex]

[tex]Mean = 0.36[/tex]

The standard deviation is:

[tex]SD = \sqrt{np(1-p)}[/tex]

[tex]SD = \sqrt{6*0.06*(1-0.06)}[/tex]

[tex]SD = \sqrt{0.3384}[/tex]

[tex]SD = 0.5817[/tex]

Solving (b): P(x = 3)

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

[tex]P(x) = ^nC_x * p^x * (1-p)^{n-x}[/tex]

So, we have:

[tex]P(x=3) = ^6C_3 * 0.06^3 * (1-0.06)^{6-3}[/tex]

[tex]P(x=3) = ^6C_3 * 0.06^3 * (1-0.06)^3[/tex]

[tex]P(x=3) = 20 * 0.06^3 * (1-0.06)^3[/tex]

[tex]P(x=3) = 0.003588[/tex]

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