Question

Find the indicated probability by using an appropriate normal model to approximate the binomial distribution

People with O-negative blood are called "universal donors" because O-negative blood can be given to anyone else regardless of the recipient's blood type. About 6% of people have type O-negative blood.
A clinic is running a blood drive. If at least 220 O-negative donors give blood, the clinic will have sufficient O-negative blood for the coming month. If 4000 donors come to the blood drive, what's the probability that the clinic will not have sufficient O-negative donors?

Group of answer choices

0.4641

0.1312

0.1093

0.0915

0.9085

Answer 1

Solution :

Let X be a random variable which represents the number of O-negative blood doners in 4000 doners.

Given that, About 6% of people have type O-negative blood. Hence, probability that a person has O-negative blood type is 6/100 = 0.06

Let us consider "people with O- negative blood type" as success. Hence, we have now only two mutually exclusive outcomes.

Probability of success (p) = 0.06

Number of trials (n) = 4000

Since, probability of success remains constant in each of the trials, number of trials are finite, we have only two mutually exclusive outcomes for each of the trials and outcomes are independent, therefore we can consider that X follows binomial distribution with parameters n = 4000 and p = 0.06.

If at least 220 O-negative blood doners give blood , the clinic will have sufficient O-negative blood. We have to obtain the probability that the clinic will not have sufficient O-negative donors. It means we have to obtain the probability that less than 220 O-negative blood doners give the blood.

i.e. We need to obtain P(X < 220).

To obtain the probability we shall use normal approximation to the binomial distribution.

If X is a binomial distributed random variable with parameters n and p and if np ≥ 5 and nq ≥ 5, then binomial distributed random variable follows approximately normal distribution with mean np and variance npq.

(Where, n is number of trials, p is probability of success and q = 1 - p)

We have, n = 4000, p = 0.06 and q = 1 - 0.06 = 0.94

np = 4000×0.06 = 240 which is greater than 5.

nq = 4000×0.94 = 3760 which is greater than 5.

Hence, we can use normal approximation to the binomial distribution. So, our random variable X is approximately normally distributed with mean np = 240 and variance npq = 4000×0.06×0.94 = 225.6.

i.e. X ~ N(240, 225.6)

μ = 240 and σ​​​​​​2 = 225.6

We need to obtain P(X < 220).

We know that if X ~ N(μ, σ​​​​​​2) then

Using "pnorm" function of R we get, P(Z < -1.3316) = 0.0915

Hence, the probability that the clinic will not have sufficient O-negative donors is 0.0915.

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