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 lower than 30, (b) more than 38, and (c) between 33 and 43 inclusive.
the normal distribution cannot be used.
identify any unusual events. explain. select all that apply.
a. the event in part (c) is unusual because its probability is less than 0.05.
b. the event in part (a) is unusual because its probability is less than 0.05.
c. the event in part (b) is unusual because its probability is less than 0.05.
d. there are no unusual events, because all of the probabilities are greater than 0.05.

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 \(p=0.19\). Calculate \(np=190\times0.19 = 36.1\geq5\) and \(n(1 - p)=190\times(1 - 0.19)=190\times0.81 = 153.9\geq5\). So, we can use a normal - distribution approximation. The mean of the normal approximation is \(\mu=np = 36.1\) and the standard deviation is \(\sigma=\sqrt{np(1 - p)}=\sqrt{36.1\times0.81}\approx\sqrt{29.241}\approx5.41\).

Step2: Standardize the values for part (a)

We want \(P(X\geq30)\). Using the normal - approximation, we first standardize \(x = 30\) to \(z=\frac{x-\mu}{\sigma}=\frac{30 - 36.1}{5.41}\approx\frac{- 6.1}{5.41}\approx - 1.13\). Then \(P(X\geq30)=P(Z\geq - 1.13)=1 - P(Z\lt - 1.13)\). From the standard - normal table, \(P(Z\lt - 1.13)=0.1292\), so \(P(X\geq30)=1 - 0.1292 = 0.8708\).

Step3: Standardize the values for part (b)

We want \(P(X>38)\). Standardize \(x = 38\) to \(z=\frac{x-\mu}{\sigma}=\frac{38 - 36.1}{5.41}=\frac{1.9}{5.41}\approx0.35\). Then \(P(X>38)=P(Z>0.35)=1 - P(Z\leq0.35)\). From the standard - normal table, \(P(Z\leq0.35)=0.6368\), so \(P(X>38)=1 - 0.6368 = 0.3632\).

Step4: Standardize the values for part (c)

We want \(P(33\leq X\leq43)\). Standardize \(x_1 = 33\) to \(z_1=\frac{33 - 36.1}{5.41}\approx\frac{-3.1}{5.41}\approx - 0.57\) and \(x_2 = 43\) to \(z_2=\frac{43 - 36.1}{5.41}=\frac{6.9}{5.41}\approx1.28\). Then \(P(33\leq X\leq43)=P(-0.57\leq Z\leq1.28)=P(Z\leq1.28)-P(Z\leq - 0.57)\). From the standard - normal table, \(P(Z\leq1.28)=0.8997\) and \(P(Z\leq - 0.57)=0.2843\), so \(P(33\leq X\leq43)=0.8997 - 0.2843 = 0.6154\).

Step5: Identify unusual events

An event is considered unusual if its probability is less than \(0.05\). Since none of the probabilities \(P(X\geq30)=0.8708\), \(P(X>38)=0.3632\), and \(P(33\leq X\leq43)=0.6154\) is less than \(0.05\), there are no unusual events.

Answer:

D. There are no unusual events, because all of the probabilities are greater than 0.05