Question

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 = 0.7796 × if 11 of the students from the special programs are randomly selected, find the probability that exactly 8 of them graduated. prob = 0.0417 × 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 = 0.0527 × would it be unusual to randomly select 11 students from

Explanation:

Step1: Assume binomial distribution

Let \(n = 11\) (number of trials, i.e., number of students selected). Let \(p\) be the probability of a single - student graduating. We need to use the binomial probability formula \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Probability of at least 10 graduating

\(P(X\geq10)=P(X = 10)+P(X = 11)\)
\[

$$\begin{align*} P(X = 10)&=C(11,10)\times p^{10}\times(1 - p)^{11- 10}=\frac{11!}{10!(11 - 10)!}\times p^{10}\times(1 - p)^{1}=11\times p^{10}\times(1 - p)\\ P(X = 11)&=C(11,11)\times p^{11}\times(1 - p)^{11-11}=p^{11} \end{align*}$$

\]

Step3: Probability of exactly 8 graduating

\[

$$\begin{align*} P(X = 8)&=C(11,8)\times p^{8}\times(1 - p)^{11 - 8}\\ &=\frac{11!}{8!(11 - 8)!}\times p^{8}\times(1 - p)^{3}\\ &=\frac{11\times10\times9}{3\times2\times1}\times p^{8}\times(1 - p)^{3}=165\times p^{8}\times(1 - p)^{3} \end{align*}$$

\]
A result is considered unusual if \(P(X)\leq0.05\). Since \(P(X = 8)=165\times p^{8}\times(1 - p)^{3}\), if \(P(X = 8)\leq0.05\), it is unusual.

Step4: Probability of at most 8 graduating

\(P(X\leq8)=1-(P(X = 9)+P(X = 10)+P(X = 11))\)
\[

$$\begin{align*} P(X = 9)&=C(11,9)\times p^{9}\times(1 - p)^{11 - 9}=\frac{11!}{9!(11 - 9)!}\times p^{9}\times(1 - p)^{2}=55\times p^{9}\times(1 - p)^{2} \end{align*}$$

\]

Assume \(p = 0.8\) (since it is not given in the problem, we assume a reasonable graduation - rate for illustration purposes).
\[

$$\begin{align*} P(X\geq10)&=P(X = 10)+P(X = 11)\\ &=11\times0.8^{10}\times0.2+0.8^{11}\\ &=11\times0.1074\times0.2 + 0.0859\\ &=0.2363+0.0859\\ &=0.3222 \end{align*}$$

\]
\[

$$\begin{align*} P(X = 8)&=165\times0.8^{8}\times0.2^{3}\\ &=165\times0.1678\times0.008\\ &=0.2128 \end{align*}$$

\]
\[

$$\begin{align*} P(X\leq8)&=1-(P(X = 9)+P(X = 10)+P(X = 11))\\ P(X = 9)&=55\times0.8^{9}\times0.2^{2}=55\times0.1342\times0.04 = 0.2952\\ P(X\geq9)&=P(X = 9)+P(X = 10)+P(X = 11)=0.2952 + 0.2363+0.0859 = 0.6174\\ P(X\leq8)&=1 - 0.6174=0.3826 \end{align*}$$

\]

Answer:

1. For the probability that at least 10 of 11 students graduated: 0.3222
2. For the probability that exactly 8 of 11 students graduated: 0.2128
3. For whether it is unusual to get exactly 8 graduates: no, it is not unusual
4. For the probability that at most 8 of 11 students graduated: 0.3826