Question

A drug tester claims that a drug cures a rare skin disease 82​% of the time. The claim is checked by testing the drug on 100 patients. If at least 79 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.

Answer 1

Normal distribution to approximate the binomial distribution

so mean will be np and standard deviation will be given by npq

here n = 100

p = 82/100 = 0.82

q = 1-0.82 = 0.18

mean = np = 100*0.82 = 82

Standard deviation = npq =  100*0.82*0.18

= 14.76

= 3.841

Probability that the claim will be rejected assuming that the manufacturer's claim is true P(z-score> (x-mean)/standard deviation)

= P(Z-score > (79-82)/ 3.841)

= P(z-score > -3/3.841)

= P(z-score > -0.781)

= 1 - 0.2177 (the reason we subtracted 0.2177 from 1 is because in the z-table attached, it gives the area to the left of the z-score but we want the area above it, since here we are looking for a value greater than -0.781)

= 0.7823

The Z-table is attached below