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.
sketch the graph of the normal distribution with the indicated probability shaded.
a. $mu = 24.18$
b. $mu = 24.18$
c. $mu = 24.18$
d. $mu = 24.18$
e. the normal distribution cannot be used.

Explanation:

Step1: Check normal - approximation conditions

For a binomial distribution \(X\sim B(n,p)\), we check \(np\geq5\) and \(n(1 - p)\geq5\). Here, \(n = 39\) and \(p=0.62\). So, \(np=39\times0.62 = 24.18\geq5\) and \(n(1 - p)=39\times(1 - 0.62)=39\times0.38 = 14.82\geq5\). We can use the normal - approximation to the binomial, where \(\mu=np = 24.18\) and \(\sigma=\sqrt{np(1 - p)}=\sqrt{24.18\times0.38}\approx3.03\).

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

When using normal - approximation for \(P(X = k)\) in binomial, we find \(P(23.5\lt X\lt24.5)\) for the normal distribution \(N(\mu = 24.18,\sigma\approx3.03)\). First, find the z - scores: \(z_1=\frac{23.5 - 24.18}{3.03}\approx - 0.22\) and \(z_2=\frac{24.5 - 24.18}{3.03}\approx0.11\). Then \(P(-0.22\lt Z\lt0.11)=\Phi(0.11)-\Phi(-0.22)\), where \(\Phi(z)\) is the standard - normal cumulative distribution function. \(\Phi(0.11) = 0.5438\), \(\Phi(-0.22)=0.4129\), so \(P(-0.22\lt Z\lt0.11)=0.5438 - 0.4129 = 0.1309\).

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

For \(P(X\geq29)\) in binomial, we find \(P(X\gt28.5)\) for the normal distribution. The z - score is \(z=\frac{28.5 - 24.18}{3.03}=\frac{4.32}{3.03}\approx1.43\). Then \(P(Z\gt1.43)=1-\Phi(1.43)=1 - 0.9236 = 0.0764\).

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

For \(P(X\leq23)\) in binomial, we find \(P(X\lt23.5)\) for the normal distribution. The z - score is \(z=\frac{23.5 - 24.18}{3.03}\approx - 0.22\). Then \(P(Z\lt - 0.22)=0.4129\).

Step5: Identify unusual events

An event is considered unusual if \(P(X)\lt0.05\). The event \(P(X\geq29)\approx0.0764\gt0.05\) and \(P(X\leq23)\approx0.4129\gt0.05\) and \(P(X = 24)\approx0.1309\gt0.05\), so there are no unusual events.

Answer:

(a) \(P(X = 24)\approx0.1309\)
(b) \(P(X\geq29)\approx0.0764\)
(c) \(P(X\leq23)\approx0.4129\)
No unusual events.