Question

It is estimated that 20% of the members of a health club have high blood pressure. A sample of 4 club members are selected at random. Let r be the number with high blood pressure. Find P(r) for 0, 1, 2, 3, and 4. Round to four decimals. What is the expected number of members in the sample who will have high blood pressure?  What is the standard deviation? Could this situation be approximated with a normal model?  If we changed the sample size to 150 members, could we then use the normal model to approximate the binomial distribution? Now use the normal model to estimate the probability that the number of members with high blood pressure will be less than 30? Round to 4 decimals.

Answer 1

Probability of high blood pressure = 0.2

This is a binomial random variable with

probability of success, p=0.2; probability of failure, q = 0.8; n=4

(a) P(r=0) =

*0.2°0.84-0 = 0.4096 (0) * 0.20

P(r=1) =

(1) + 0.2'0.84–1 = 0.4096

P(r=2) =

(4) + 0.2?0.84-2 = 0.1536

P(r=3) =

(4) +0.280.84-3 = 0.0256

P(r=4) =

(4) + 0.20.84-4 = 0.0016

expected number of members in the sample who will have high blood pressure = n*p = 4*0.2 = 0.8

standard deviation = SQRT(n*p*q) = 0.8

This can not be approximated with a normal model as the sample size (n=4) is small.

For normal approximation to a binomial distribution, n*p >= 5 and n*q >= 5

For n=150, we have n*p = 150*0.2 = 30 and n*q = 150*0.8 = 120. Hence we can use the normal approximation.

Probability that the number of members with high blood pressure will be less than 30:

The normal distribution has mean = 150*0.2 = 30 and standard deviation = SQRT(n*p*q) = 4.899

Since mean iis 30, area under the normal curve lying to the left of the mean = 0.5

Probability that the number of members with high blood pressure will be less than 30 = 0.5

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