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Chapter 5: Q.99 (page 357)

According to the American Red Cross, about one out of nine people in the U.S. have Type B blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B blood types that arrive roughly follows the Poisson distribution.

a. If 100 people arrive, how many on average would be expected to have Type B blood?

b. What is the probability that over 10people out of these 100 have type B blood?

c. What is the probability that more than 20 people arrive before a person with type B blood is found?

Short Answer

Expert verified

a). The average number of people is expected will be11.11.

b). The value of P(x>10)is 0.5532.

c). The value of P(x>20)is 0.1084.

Step by step solution

01

Part (a) Step 1: Given Information 

According to the American Red Cross, about one out of nine people in the U.S. have Type B blood.

02

Part (a) Step 2: Explanation 

The average number of people is expected will be:

=1009

=11.11

03

Part (b) Step 1: Given Information

According to the American Red Cross, about one out of nine people in the U.S. have Type B blood.

04

Part (b) Step 2: Explanation 

The required probability that is P(x>10)can be written as below:

P(x>10)=1-P(x鈮�10)

P(x>10)=1-0.4468

=0.5532

05

Part (c) Step 1: Given Information 

According to the American Red Cross, about one out of nine people in the U.S. have Type B blood.

06

Part (c) Step 2: Explanation

The required probability will follow exponential distribution with decay rate 1/9. So, the required probability can be calculated as below:

P(x>20)=1-P(x<20)

=1-1-e19-20

=e-19脳20

=0.1084

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