Question

a drug tester claims that a drug cures a rare skin disease 67% of the time. the claim is checked by testing the drug on 100 patients. if at least 59 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: Check binomial - normal approximation conditions

For a binomial distribution \(X\sim B(n,p)\), we can use normal approximation \(X\sim N(np,np(1 - p))\) if \(np\geq5\) and \(n(1 - p)\geq5\).
Here, \(n = 100\), \(p=0.67\). Then \(np=100\times0.67 = 67\geq5\) and \(n(1 - p)=100\times(1 - 0.67)=33\geq5\).
The mean of the normal distribution \(\mu=np = 67\), and the standard deviation \(\sigma=\sqrt{np(1 - p)}=\sqrt{100\times0.67\times(1 - 0.67)}=\sqrt{22.11}\approx4.702\).

Step2: Continuity correction

The claim is rejected if \(X\lt59\). Using continuity correction for the binomial - to - normal approximation, we consider \(X\lt58.5\) (since we are approximating a discrete distribution with a continuous one).
We calculate the \(z\) - score: \(z=\frac{x-\mu}{\sigma}\), where \(x = 58.5\), \(\mu = 67\), \(\sigma\approx4.702\).
\(z=\frac{58.5 - 67}{4.702}=\frac{- 8.5}{4.702}\approx - 1.81\)

Step3: Find the probability using the standard normal table

We want to find \(P(X\lt58.5)\), which is equivalent to \(P(Z\lt - 1.81)\) in the standard normal distribution \(Z\sim N(0,1)\).
From the standard normal table (or using a calculator with a normal - distribution function), \(P(Z\lt - 1.81)=0.0351\)

Answer:

\(0.0351\)