Question

a drug tester claims that a drug cures a rare skin disease 76% of the time. the claim is checked by testing the drug on 100 patients. if at least 71 patients are cured, the claim will be accepted. find the probability that the claim will be rejected assuming that the manufacturers claim is true. use the normal distribution to approximate the binomial distribution if possible. the probability is (round to four decimal places as needed.)

Explanation:

Step1: Identify binomial parameters

Let \(n = 100\) (number of patients), \(p=0.76\). The mean of the binomial distribution is \(\mu = np\) and the standard - deviation is \(\sigma=\sqrt{np(1 - p)}\).
\(\mu=np=100\times0.76 = 76\)
\(\sigma=\sqrt{np(1 - p)}=\sqrt{100\times0.76\times(1 - 0.76)}=\sqrt{100\times0.76\times0.24}=\sqrt{18.24}\approx4.27\)

Step2: Determine the rejection condition

The claim is rejected if the number of cured patients \(X<71\). When using the normal approximation to the binomial, for \(X < 71\), we apply the continuity correction. We consider \(P(X < 70.5)\) (continuity correction for \(X<71\)).

Step3: Calculate the z - score

The z - score is calculated as \(z=\frac{x-\mu}{\sigma}\), where \(x = 70.5\), \(\mu = 76\), and \(\sigma\approx4.27\).
\(z=\frac{70.5 - 76}{4.27}=\frac{- 5.5}{4.27}\approx - 1.29\)

Step4: Find the probability using the standard normal table

We want to find \(P(Z < - 1.29)\). Looking up the value in the standard - normal table, we find that \(P(Z < - 1.29)=0.0985\)

Answer:

\(0.0985\)