Question

suppose that a new treatment is successful in curing a common ailment 75% of the time. if the treatment is tried on a random sample of 95 patients, approximate the probability that more than 73 will be cured. use the normal approximation to the binomial with a correction for continuity. round your answer to at least three decimal places. do not round any intermediate steps. (if necessary, consult a list of formulas.)

Explanation:

Step1: Identify binomial parameters

Let $n = 95$ (number of trials/sample size) and $p=0.75$ (probability of success). The mean of the binomial distribution is $\mu = np$ and the standard - deviation is $\sigma=\sqrt{np(1 - p)}$.
$\mu=np=95\times0.75 = 71.25$
$\sigma=\sqrt{np(1 - p)}=\sqrt{95\times0.75\times(1 - 0.75)}=\sqrt{95\times0.75\times0.25}=\sqrt{17.8125}\approx4.22$.

Step2: Apply continuity correction

We want to find $P(X>73)$. With continuity correction for $P(X > k)$ in a binomial - normal approximation, we find $P(X>k + 0.5)$. So we want to find $P(X>73.5)$.

Step3: Standardize the value

We use the formula $z=\frac{x-\mu}{\sigma}$. Substituting $x = 73.5$, $\mu = 71.25$ and $\sigma\approx4.22$, we get $z=\frac{73.5 - 71.25}{4.22}=\frac{2.25}{4.22}\approx0.53$.

Step4: Find the probability

We want $P(Z>0.53)$. Since $P(Z>z)=1 - P(Z\leq z)$ and from the standard normal table $P(Z\leq0.53)=0.7019$, then $P(Z>0.53)=1 - 0.7019 = 0.2981$.

Answer:

$0.298$