Question

A drug tester claims that a drug cures a rare skin disease 81% of the time. The claim is checked by testing the drug on 100 patients. If at least 78 patients are​ cured, the claim will be accepted. Find the probability that the claim will be rejected assuming that the​ manufacturer's claim is true. Use the normal distribution to approximate the binomial distribution if possible.

The probability is_______

​(Round to four decimal places as​ needed.)

Answer 1

Solution:

We are given n=100, p=0.81, q=1 p = 1 – 0.81 = 0.19

We have to find P(X<78)

We have to use normal approximation to binomial distribution.

Mean = np = 100*0.81 = 81

SD = sqrt(npq) = sqrt(100*0.81*0.19) = 3.923009

Z = (X – mean) / SD

Z = (78 - 81) / 3.923009

Z = -0.76472

P(Z<-0.76472) = P(X<78) = 0.222219

(by using z-table or excel)

Required probability = 0.2222

(Without using continuity correction factor)

By using continuity correction factor 0.5, we have to find P(X<78) = P(X<77.5)

Z = (77.5 - 81) / 3.923009

Z = -0.89217

P(Z<-0.89217) = P(X<77.5) = 0.18615

(by using z-table or excel)

Required probability = 0.1862

(Using continuity correction factor)