Question

A drug tester claims that a drug cures a rare skin disease 66% of the time. The claim is checked by testing the drug on 100 patients. If at least 61 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

Answer)

N = 100

P = 0.66

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 66

N*(1-p) = 34

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/s.d

Mean = n*p = 66

S.d = √{n*p*(1-p)} = 4.73708771293

We need to find P(x<61) (because then only claim will be rejected)

By continuity correction

P(x<60.5)

Z = (60.5 - 66)/4.73708771293 = -1.16

From z table, P(z<-1.16) = 0.1230

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