Blood type A occurs in about 45% of the population. A clinic needs 3 pints of ty
ID: 3244129 • Letter: B
Question
Blood type A occurs in about 45% of the population. A clinic needs 3 pints of type A blood. A donor usually gives a pint of blood. Let n be a random variable representing the number of donors needed to provide 3 pints of type A blood.
(a) Explain why a negative binomial distribution is appropriate for the random variable n.
We have binomial trials for which the probability of success is p = 0.45 and failure is q = 0.55; k is a fixed whole number > 1.We have binomial trials for which the probability of success is p = 0.45 and failure is q = 0.55; k is a fixed whole number 1. We have poisson trials for which the probability of success is p = 0.45 and failure is q = 0.55; k is a fixed whole number 1.We have binomial trials for which the probability of success is p = 0.55 and failure is q = 0.45; k is a fixed whole number 1.
Write out the formula for P(n) in the context of this application.
P(n) =
(b) Compute P(n = 3), P(n = 4), P(n = 5), and P(n = 6). (Use 4 decimal places.)
(c) What is the probability that the clinic will need from three to six donors to obtain the needed 3 pints of type A blood? (Use 4 decimal places.)
(d) What is the probability that the clinic will need more than six donors to obtain 3 pints of type A blood? (Use 4 decimal places.)
(e) What are the expected value and standard deviation of the random variable n? (Use 2 decimal places.)
Interpret these values in the context of this application.
The expected number of donors in which the fourth pint of blood type A is acquired is , with a standard deviation of .The expected number of donors in which the second pint of blood type A is acquired is , with a standard deviation of . The expected number of donors in which the third pint of blood type A is acquired is , with a standard deviation of .The expected number of donors in which the third pint of blood type A is acquired is , with a standard deviation of .
P(3) P(4) P(5) P(6)Explanation / Answer
a) p = 0.45, and the probability of failure is q = 0.55. k = 3
P(n) = (n-1)C(k-1) * p^k * q^(n-k) = (n-1)C2 * 0.45^3 *0.55^(n-3)
b) P(3) = (0.45)^3 * (0.25 ^( 3- 3) = 0.091125
P(4) = 3 * (0.45)^3 * (0.25 ^(4 - 3) =0.0683437
P(5) = 6* (0.45)^3 * (0.25 ^( 5- 3) = 0.034171
P(6) =10 * (0.45)^3 * (0.25 ^(6- 3) = 0.01423828
c) P(3 n 6) = 0.091125+0.0683437+ 0.034171+0.01423828
=0.20787
d)P(n>=7) = 1- P(n<=6) = 1 -0.20787 = 0.79213 ,note that n can not be less than 3 , n>= k {here k =3}
e) The expected number of donors in which the third pint of blood type A is acquired is , with a standard deviation of .