Question

a study was conducted to determine whether there were significant differences between college students admitted through special programs (such as retention incentive and guaranteed placement programs) and college students admitted through the regular admissions criteria. it was found that the graduation rate was 91.5% for the college students admitted through special programs. round answers to 4 decimal places. if 11 of the students from the special programs are randomly selected, find the probability that at least 10 of them graduated. prob = if 11 of the students from the special programs are randomly selected, find the probability that exactly 8 of them graduated. prob = would it be unusual to randomly select 11 students from the special programs and get exactly 8 that graduate? no, it is not unusual yes, it is unusual if 11 of the students from the special programs are randomly selected, find the probability that at most 8 of them graduated. prob = would it be unusual to randomly select 11 students from the special programs and get at most 8 that graduate? no, it is not unusual yes, it is unusual

Explanation:

Step1: Identify the binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 11$ and $p=0.9153$.

Step2: Probability that at least 10 graduated

$P(X\geq10)=P(X = 10)+P(X = 11)$.
$C(11,10)=\frac{11!}{10!(11 - 10)!}=\frac{11!}{10!1!}=11$, $P(X = 10)=C(11,10)\times(0.9153)^{10}\times(1 - 0.9153)^{11 - 10}=11\times(0.9153)^{10}\times0.0847\approx11\times0.4049\times0.0847\approx0.3754$.
$C(11,11)=\frac{11!}{11!(11 - 11)!}=1$, $P(X = 11)=C(11,11)\times(0.9153)^{11}\times(1 - 0.9153)^{11 - 11}=(0.9153)^{11}\approx0.3407$.
$P(X\geq10)=0.3754 + 0.3407=0.7161$.

Step3: Probability that exactly 8 graduated

$C(11,8)=\frac{11!}{8!(11 - 8)!}=\frac{11\times10\times9}{3\times2\times1}=165$.
$P(X = 8)=C(11,8)\times(0.9153)^{8}\times(1 - 0.9153)^{11 - 8}=165\times(0.9153)^{8}\times(0.0847)^{3}$.
$(0.9153)^{8}\approx0.4747$, $(0.0847)^{3}\approx0.0006$.
$P(X = 8)=165\times0.4747\times0.0006\approx0.0470$.

Step4: Determine if getting exactly 8 is unusual

A result is considered unusual if $P(X = k)<0.05$. Since $P(X = 8)\approx0.0470<0.05$, it is unusual.

Step5: Probability that at most 8 graduated

$P(X\leq8)=1 - P(X\geq9)=1-(P(X = 9)+P(X\geq10))$.
$C(11,9)=\frac{11!}{9!(11 - 9)!}=\frac{11\times10}{2\times1}=55$.
$P(X = 9)=C(11,9)\times(0.9153)^{9}\times(1 - 0.9153)^{11 - 9}=55\times(0.9153)^{9}\times(0.0847)^{2}$.
$(0.9153)^{9}\approx0.4344$, $(0.0847)^{2}\approx0.0072$.
$P(X = 9)=55\times0.4344\times0.0072\approx0.1739$.
$P(X\geq9)=P(X = 9)+P(X\geq10)=0.1739 + 0.7161 = 0.89$.
$P(X\leq8)=1 - 0.89 = 0.11$.

Step6: Determine if getting at most 8 is unusual

Since $P(X\leq8)=0.11>0.05$, it is not unusual.

Answer:

prob (at least 10 graduated) $\approx0.7161$
prob (exactly 8 graduated) $\approx0.0470$
yes, it is unusual
prob (at most 8 graduated) $\approx0.1100$
no, it is not unusual