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.
0.0000 (round to four decimal places as needed.)
sketch the graph of the normal distribution with the indicated probability shaded.
a. μ = 36.1
b. μ = 36.1
c. μ = 36.1
d. μ = 36.1
e. the normal distribution cannot be used.

Explanation:

Step1: Check normal - approximation conditions

For a binomial distribution \(X\sim B(n,p)\) to be approximated by a normal distribution \(N(np,np(1 - p))\), we need \(np\geq5\) and \(n(1 - p)\geq5\). Here, \(n = 190\) and \(p=0.19\). So \(np=190\times0.19 = 36.1\geq5\) and \(n(1 - p)=190\times(1 - 0.19)=190\times0.81 = 153.9\geq5\). We can use the normal - approximation \(X\sim N(np = 36.1,np(1 - p)=36.1\times0.81\approx29.241)\), and the standard deviation \(\sigma=\sqrt{np(1 - p)}=\sqrt{29.241}\approx5.4075\).

Step2: Standardize for part (a)

The probability that the number is no fewer than 30, \(P(X\geq30)\). Using the continuity correction for the normal approximation of the binomial, we find \(P(X\geq29.5)\). The z - score is \(z=\frac{29.5 - 36.1}{5.4075}=\frac{- 6.6}{5.4075}\approx - 1.22\). Then \(P(X\geq29.5)=1 - P(Z\lt - 1.22)=1-0.1112 = 0.8888\).

Step3: Standardize for part (b)

The probability that the number is more than 38, \(P(X > 38)\). Using the continuity correction, we find \(P(X\geq38.5)\). The z - score is \(z=\frac{38.5 - 36.1}{5.4075}=\frac{2.4}{5.4075}\approx0.44\). Then \(P(X\geq38.5)=1 - P(Z\lt0.44)=1 - 0.6700=0.3300\).

Step4: Standardize for part (c)

The probability that the number is between 33 and 43 inclusive, \(P(33\leq X\leq43)\). Using the continuity correction, we find \(P(32.5\leq X\leq43.5)\). The z - score for \(x = 32.5\) is \(z_1=\frac{32.5 - 36.1}{5.4075}=\frac{-3.6}{5.4075}\approx - 0.67\), and the z - score for \(x = 43.5\) is \(z_2=\frac{43.5 - 36.1}{5.4075}=\frac{7.4}{5.4075}\approx1.37\). Then \(P(32.5\leq X\leq43.5)=P(-0.67\leq Z\leq1.37)=P(Z\leq1.37)-P(Z\leq - 0.67)=0.9147-0.2514 = 0.6633\).

Unusual events are those with a probability of less than 0.05 or greater than 0.95. None of the events in (a), (b), and (c) are unusual.

For the graph of \(P(X\geq30)\) (part (a)), we shade the area to the right of \(x = 29.5\) (after continuity - correction) under the normal curve with \(\mu = 36.1\) and \(\sigma\approx5.4075\).

Answer:

(a) \(0.8888\)
(b) \(0.3300\)
(c) \(0.6633\)