The National Aeronautics and Space Administration (NASA) compiles data on space shuttle launches and publishes them on its website. The following table displays the frequency distribution for the number of crew members on each of the 135 missions from April 1981 to July 2011.

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 $n = 25$.