Question

determine whether you can use a normal distribution to approximate the binomial distribution. if you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. if you cannot, explain why and use a binomial distribution to find the indicated probabilities. identify any unusual events. explain.
in a survey of u.s. adults, 19% said that they have participated in a clinical trial. you randomly select 190 u.s. adults. find the probability that the number who have participated in a clinical trial is (a) no fewer than 30, (b) more than 38, and (c) between 33 and 43 inclusive.
determine whether a normal distribution can be used to approximate the binomial distribution. choose the correct answer below.
a. no, because np < 5.
b. yes, because both np ≥ 5 and nq ≥ 5.
c. no, because nq < 5.
(a) find the probability that the number who have participated in a clinical trial is no fewer than 30.
(round to four decimal places as needed.)

Explanation:

Step1: Identify binomial parameters

Let $n = 190$ (number of trials), $p=0.19$, $q = 1 - p=0.81$.

Step2: Calculate mean and standard - deviation for normal approximation

The mean of the binomial distribution $\mu=np=190\times0.19 = 36.1$.
The standard deviation $\sigma=\sqrt{npq}=\sqrt{190\times0.19\times0.81}=\sqrt{36.1\times0.81}=\sqrt{29.241}\approx5.4075$.

Step3: Standardize for part (a)

For $x\geq30$, using the continuity correction, we consider $x = 29.5$. The z - score is $z=\frac{x-\mu}{\sigma}=\frac{29.5 - 36.1}{5.4075}=\frac{- 6.6}{5.4075}\approx - 1.22$.
$P(X\geq30)=P(Z\geq - 1.22)=1 - P(Z < - 1.22)$.
From the standard normal table, $P(Z < - 1.22)=0.1112$, so $P(X\geq30)=1 - 0.1112 = 0.8888$.

Answer:

(a) $0.8888$