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

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Accepted Answer

# 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

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QUESTION IMAGE

Question

Explanation:

Step1: Assume binomial distribution

Step2: Probability of at least 10 graduating

Step3: Probability of exactly 8 graduating

Step4: Probability of at most 8 graduating

Answer: