Question

38% of people in the United States have type O+blood. You randomly select 50 people in the United States and record their blood type. a. What is the probability that exactly 20 have type O+blood? b. Find the mean and standard deviation for this probability distribution. c. If you only select 8 people, what is the probability that at least one has type O+blood?

Answer 1

Answer:

a) 10.98%

b) Mean = 19, Standard deviation = 3.4322

c) 97.82%

Explanation:

This can be modeled with a Binomial Distribution with probability p of “success” (selecting a person O+) is 38% (0.38) and the likelihood q of “failure” (selecting a person with blood type other than O+) is 62% (0.62).

As we know, the probability of having exactly k “successes” in a sample of size n is given by

$\large P(k;n)=\binom{n}{k}p^kq^{n-k}$

where

$\large \binom{n}{k}$ are the combination of n elements taken k at a time

a. What is the probability that exactly 20 have type O+blood?

Since the sample size is 50, we are looking for P(20;50)

$\large P(20;50)=\binom{50}{20}0.38^{20}0.62^{30}=0.1098=10.98\%$

b. Find the mean and standard deviation for this probability distribution.

The mean for a Binomial distribution in a sample of size n is given by

$\large \bar x=np$

and the standard deviation is  

$\large s=\sqrt{npq}$

So,  

$\large \bar x = 50*0.38=19$

$\large s=\sqrt{50*0.38*0.62}=3.4322$

c. If you only select 8 people, what is the probability that at least one has type O+blood?

If we only select 8 people, the sample size is now 8 and we are looking for P(1;8)+P(2;8)+...+P(8;8)

But P is a probability function, so the sum above equals

1 - P(0;8) (1 minus the probability that none of them is O+)

So, the likelihood we are looking for is

$\large 1-P(0;8)=1-\binom{8}{0}0.38^00.62^8=1-0.021834=0.97816\approx 0.9782=97.82\%$