Question

In: Statistics and Probability

A drug tester claims that a drug cures a rare skin disease 82​% of the time....

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.

Solutions

Expert Solution

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


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