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.
a. μ = 36.1 graph with shaded region from 30 to right side, x - axis from 18 to 54
b. μ = 36.1 graph with a thin - shaded region near 30, x - axis from 18 to 54
c. μ = 36.1 graph with shaded region from left side to 30, x - axis from 18 to 54
d. μ = 36.1 graph with shaded region from 30 to 54
e. the normal distribution cannot be used.
(b) find the probability that the number who have participated in a clinical trial is more than 38.
(round to four decimal places as needed.)

Explanation:

Step1: Check normal - approximation conditions

For a binomial distribution $B(n,p)$, we can approximate it with a normal distribution $N(np,np(1 - p))$ if $np\geq5$ and $n(1 - p)\geq5$. Here, $n = 190$ and $p=0.19$. So, $np=190\times0.19 = 36.1\geq5$ and $n(1 - p)=190\times(1 - 0.19)=190\times0.81 = 153.9\geq5$. The mean of the normal - approximation is $\mu=np = 36.1$ and the standard deviation is $\sigma=\sqrt{np(1 - p)}=\sqrt{36.1\times0.81}\approx\sqrt{29.241}\approx5.4075$.

Step2: Standardize for part (b)

We want to find $P(X>38)$ for the binomial distribution, which is approximated by $P(X > 38.5)$ (continuity correction) for the normal distribution. The z - score is calculated as $z=\frac{x-\mu}{\sigma}$. Substituting $x = 38.5$, $\mu = 36.1$, and $\sigma\approx5.4075$, we get $z=\frac{38.5 - 36.1}{5.4075}=\frac{2.4}{5.4075}\approx0.44$.

Step3: Find the probability

Using the standard normal table, $P(Z>0.44)=1 - P(Z\leq0.44)$. From the standard - normal table, $P(Z\leq0.44) = 0.6700$. So, $P(Z>0.44)=1 - 0.6700=0.3300$.

Answer:

0.3300