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. college students, 62% said that if their college offers new programs tailored to the new economy, it will make them more likely to re - enroll. you randomly select 39 college students. find the probability that the number who are more likely to enroll if their college offers new programs tailored to the new economy is (a) exactly 24, (b) at least 29, and (c) at most 23.

14 24 35 14 24 35 14 24 35 14 24 35

e. the normal distribution cannot be used.

identify any unusual events. explain. select all that apply.

a. the event in part (a) is unusual because its probability is less than 0.05.
b. the event in part (c) 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 \(n(1 - p)\geq5\). Given \(n = 39\) and \(p=0.62\), then \(np=39\times0.62 = 24.18\geq5\) and \(n(1 - p)=39\times(1 - 0.62)=39\times0.38 = 14.82\geq5\). So we can use the normal - approximation with \(\mu=np = 24.18\) and \(\sigma=\sqrt{np(1 - p)}=\sqrt{24.18\times0.38}\approx3.03\).

Step2: Continuity correction for \(P(X = 24)\)

When approximating \(P(X = 24)\) using the normal distribution, we find \(P(23.5\lt X\lt24.5)\). First, standardize the values: \(z_1=\frac{23.5 - 24.18}{3.03}\approx - 0.22\) and \(z_2=\frac{24.5 - 24.18}{3.03}\approx0.11\). Then \(P(23.5\lt X\lt24.5)=\Phi(0.11)-\Phi(-0.22)\), where \(\Phi(z)\) is the cumulative - distribution function of the standard normal distribution. \(\Phi(0.11) = 0.5438\), \(\Phi(-0.22)=0.4129\), so \(P(X = 24)\approx0.5438 - 0.4129 = 0.1309\).

Step3: Continuity correction for \(P(X\geq29)\)

For \(P(X\geq29)\), we find \(P(X\gt28.5)\) in the normal - approximation. Standardize \(z=\frac{28.5 - 24.18}{3.03}\approx1.43\). Then \(P(X\gt28.5)=1-\Phi(1.43)\), and \(\Phi(1.43)=0.9236\), so \(P(X\geq29)\approx1 - 0.9236 = 0.0764\).

Step4: Continuity correction for \(P(X\leq23)\)

For \(P(X\leq23)\), we find \(P(X\lt23.5)\) in the normal - approximation. Standardize \(z=\frac{23.5 - 24.18}{3.03}\approx - 0.22\). Then \(P(X\lt23.5)=\Phi(-0.22)=0.4129\).

Step5: Identify unusual events

An event is considered unusual if its probability is less than \(0.05\). The probability in part (b) \(P(X\geq29)\approx0.0764\gt0.05\), the probability in part (a) \(P(X = 24)\approx0.1309\gt0.05\), and the probability in part (c) \(P(X\leq23)\approx0.4129\gt0.05\). So there are no unusual events.

Answer:

a. \(P(X = 24)\approx0.1309\)
b. \(P(X\geq29)\approx0.0764\)
c. \(P(X\leq23)\approx0.4129\)
D. There are no unusual events, because all of the probabilities are greater than 0.05.