美国国家航空航天局（NASA）收集了航天飞机发射的数据并发布在其网站上。下表显示了 1981 年 4 月至 2011 年 7 月期间 135 次任务中每一次的机组人员人数的频数分布。

1. Construct the probability distribution table for $C$.

The total number of missions is $135$.
The probability distribution is found by calculating $P(C = c) = \frac{\text{Frequency of } c}{135}$.

2. Calculate the expected value $E(C)$.

$E(C) = \sum c \, P(C=c)$
$E(C) = 2\left(\frac{4}{135}\right) + 4\left(\frac{3}{135}\right) + 5\left(\frac{36}{135}\right) + 6\left(\frac{24}{135}\right) + 7\left(\frac{48}{135}\right) + 8\left(\frac{14}{135}\right) + 10\left(\frac{6}{135}\right)$
$E(C) = \frac{8 + 12 + 180 + 144 + 336 + 112 + 60}{135}$
$E(C) = \frac{852}{135}$
$E(C) = \frac{284}{45} \approx 6.31 \text{ (to 3 s.f.)}$

1. Identify all value pairs (with non-zero probability) that sum to 8

From part (i), the non-zero values of $C$ are $2,4,5,6,7,8,10$ with $P(C=2)=\tfrac{4}{135}$, $P(C=4)=\tfrac{3}{135}$, $P(C=6)=\tfrac{24}{135}$ (and $P(C=3)=P(C=9)=0$).

Let $C_1,C_2$ be the reports of the two students. The ordered pairs giving $C_1+C_2=8$ with non-zero probability are:
$(2,6),\ (6,2),\ (4,4).$
(Pairs such as $(3,5)$ or $(5,3)$ contribute $0$ since $P(C=3)=0$.)

2. Combine the cases using independence and addition

Since the two students are chosen independently,
$P(C_1+C_2=8)=2\,P(C=2)P(C=6)+\big(P(C=4)\big)^2$
$=2\cdot\frac{4}{135}\cdot\frac{24}{135}+\left(\frac{3}{135}\right)^2=\frac{192}{18225}+\frac{9}{18225}=\frac{201}{18225}\approx 0.0110.$

1. Set up the complement

Let $C_1,C_2$ be independent copies of $C$. The smallest possible total is $2+2=4$, so
$P(C_1+C_2>8)=1-P(C_1+C_2\le 8).$
Using the probabilities from (i) (out of $135$): $P(2)=\tfrac{4}{135},\,P(4)=\tfrac{3}{135},\,P(5)=\tfrac{36}{135},\,P(6)=\tfrac{24}{135},\,P(7)=\tfrac{48}{135},\,P(8)=\tfrac{14}{135},\,P(10)=\tfrac{6}{135}$, with $P(3)=P(9)=0$.

2. Enumerate the ordered pairs giving each total $\le 8$

Working in units of $\tfrac{1}{135^2}=\tfrac{1}{18225}$ (numerators are products of frequencies):

• Total $4$: $(2,2)\to 4\times4=16$
• Total $5$: $(2,3),(3,2)\to 0$ (since $P(3)=0$)
• Total $6$: $(2,4),(4,2)\to 2(4\times3)=24$; $(3,3)\to0$
• Total $7$: $(2,5),(5,2)\to 2(4\times36)=288$; $(3,4)\to0$
• Total $8$: $(2,6),(6,2)\to 2(4\times24)=192$; $(4,4)\to 3\times3=9$; $(3,5)\to0$

Total numerator $=16+0+24+288+(192+9)=529$.

3. Combine and conclude

$P(C_1+C_2\le 8)=\frac{529}{18225}=0.029026\ldots$
$P(C_1+C_2>8)=1-0.029026\ldots=0.97097\ldots=0.971\ (3\text{ s.f.})\;\blacksquare$

1. Identify the distribution of $T$ and its probability of success from part (iii).

From part (iii), the probability that a mentor needs to grade more than 8 reports is $0.971$.
$T \sim \text{B}(n, 0.971)$

2. Use the formula for the expected value of a binomial distribution to solve for $n$.

$\text{E}(T) = np$
$0.971n = 20.391$
$n = \frac{20.391}{0.971} = 21$

1. Find the total number of mentors, $n$, using the expected value given in part (iv).

From part (iii), the probability that a mentor needs to grade more than 8 reports (i.e. the 2 assigned students wrote more than 8 reports) is $p = 0.971$.
Let $T$ be the number of mentors who need to grade more than $8$ reports. We have $T \sim \text{B}(n, 0.971)$.
From part (iv), $\text{E}(T) = 20.391$.
Using the formula for the expected value of a binomial distribution $\text{E}(T) = np$:
$n(0.971) = 20.391$
$n = \frac{20.391}{0.971} = 21$

2. Calculate the probability that more than 19 mentors need to grade more than 8 reports.

We need to find $\text{P}(T > 19)$. Since $T$ is discrete, this is equivalent to $\text{P}(T \ge 20)$.
$\text{P}(T > 19) = \text{P}(T = 20) + \text{P}(T = 21)$
Using the probability mass function for $T \sim \text{B}(21, 0.971)$:
$\text{P}(T = 20) = \binom{21}{20}(0.971)^{20}(1 - 0.971)^1 = 21(0.971)^{20}(0.029) \approx 0.33779$
$\text{P}(T = 21) = \binom{21}{21}(0.971)^{21}(1 - 0.971)^0 = (0.971)^{21} \approx 0.53858$
$\text{P}(T > 19) \approx 0.33779 + 0.53858 = 0.87637$
Rounding to 3 significant figures, we get $0.876$.

1. Set up the inequality for $T=25$ to be the most probable value by comparing it to $T=24$.

From part (iii), the probability of a mentor grading more than 8 reports is $p = 0.971$. Thus, $T \sim \text{B}(n, 0.971)$.
If $T = 25$ is the most probable value (the mode), then $P(T=25) \ge P(T=24)$.
$\frac{P(T=25)}{P(T=24)} \ge 1$
$\frac{\binom{n}{25} (0.971)^{25} (1-0.971)^{n-25}}{\binom{n}{24} (0.971)^{24} (1-0.971)^{n-24}} \ge 1$
$\frac{n-24}{25} \left( \frac{0.971}{0.029} \right) \ge 1$

2. Solve the inequality to find the lower bound for $n$.

$n - 24 \ge \frac{25 \times 0.029}{0.971}$
$n - 24 \ge 0.7466$
$n \ge 24.747$

3. Set up the inequality comparing $T=25$ with $T=26$ to find the upper bound for $n$, and conclude the integer value.

For $T=25$ to be the maximum, we also require $P(T=26) \le P(T=25)$ (if $n \ge 26$).
$\frac{P(T=26)}{P(T=25)} \le 1$
$\frac{n-25}{26} \left( \frac{0.971}{0.029} \right) \le 1$
$n - 25 \le \frac{26 \times 0.029}{0.971}$
$n \le 25 + 0.7765$
$n \le 25.777$
Since $n$ must be an integer satisfying $24.747 \le n \le 25.777$, we conclude that $n = 25$.