Question

# A drug tester claims that a drug cures a rare skin disease 66​% of the time....

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.

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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