[对于本题，你应该清楚地写出你所使用的任何正态分布的参数值。] 每天晚饭后，班尼迪克（Benedict）会在晚上 8 点开始复习前给自己留出一些放松的时间。他在晚上 7 点过 $D$ 分钟时吃完晚饭，其中 $D$ 服从分布 $\text{N}(10, 2^2)$。晚饭后，他通常会花 $M$ 分钟在社交媒体上，其中 $M$ 服从分布 $\text{N}(k, 12^2)$。

1. Define the distribution of the total time taken before starting revision.

Let $T$ be the total time in minutes past 7 pm before Benedict starts his revision. Then $T = D + M$.
Assuming $D$ and $M$ are independent:
$\text{E}(T) = \text{E}(D) + \text{E}(M) = 10 + k$
$\text{Var}(T) = \text{Var}(D) + \text{Var}(M) = 2^2 + 12^2 = 148$
$T \sim \text{N}(10 + k, 148)$

2. Set up the probability equation for starting revision late.

Benedict plans to start his revision at 8 pm, which is 60 minutes past 7 pm. He starts late if $T > 60$.
Given that he is late 12.5% of the time:
$\text{P}(T > 60) = 0.125$
$\text{P}(T \le 60) = 1 - 0.125 = 0.875$

3. Standardise the normal variable and solve for $k$.

Standardising the normal variable $T$ to the standard normal variable $Z$:
$\text{P}\left(Z \le \frac{60 - (10+k)}{\sqrt{148}}\right) = 0.875$
Using the inverse normal function (from GC or tables), the $z$-value for an area of 0.875 is approximately $1.1503$.
$\frac{50 - k}{\sqrt{148}} \approx 1.1503$
$50 - k \approx 1.1503 \times \sqrt{148}$
$50 - k \approx 13.994$
$k \approx 50 - 13.994$
$k \approx 36.006$
$k = 36.0 \text{ (shown to 1 d.p.)}$

1. Find the distribution of the total time spent on dinner and social media.

Let $T$ be the total time spent, so $T = D + M$.
Assuming $D$ and $M$ are independent:
$\text{E}(T) = \text{E}(D) + \text{E}(M) = 10 + 36 = 46$
$\text{Var}(T) = \text{Var}(D) + \text{Var}(M) = 2^2 + 12^2 = 148$
So, $T \sim \text{N}(46, 148)$.

2. Sketch the normal distribution curve for $T$.

Draw a normal distribution curve with the following features:

• A horizontal axis labelled $t$ (time in minutes).
• A bell-shaped, symmetrical curve entirely above the horizontal axis.
• The peak (maximum point) of the curve is aligned with $t = 46$.
• The spread can be indicated by points of inflexion at $\mu \pm \sigma$, i.e., $46 \pm \sqrt{148}$ ($33.8$ and $58.2$), though marking the central mean of $46$ is the primary requirement.
• The tails of the curve asymptotically approach the horizontal axis.

1. Define the start times for the two days and find the mean of their difference.

Let $T_1 = D_1 + M$ be the revision start time on the social media day, and $T_2 = D_2 + G$ be the revision start time on the gaming day.
We are looking for the difference $T_1 - T_2 = D_1 + M - D_2 - G$.
$\text{E}(T_1 - T_2) = \text{E}(D_1) + \text{E}(M) - \text{E}(D_2) - \text{E}(G)$
$\text{E}(T_1 - T_2) = 10 + 36 - 10 - 45 = -9$

2. Find the variance of the difference and state its normal distribution.

Since the events occur on different days, $D_1$ and $D_2$ are independent instances. We sum the individual variances:
$\text{Var}(T_1 - T_2) = \text{Var}(D_1) + \text{Var}(M) + \text{Var}(D_2) + \text{Var}(G)$
$\text{Var}(T_1 - T_2) = 2^2 + 12^2 + 2^2 + 10^2 = 4 + 144 + 4 + 100 = 252$
Thus, the difference follows a normal distribution: $T_1 - T_2 \sim \text{N}(-9, 252)$.

3. Calculate the probability that the difference is at most 5 minutes.

We require $\text{P}(|T_1 - T_2| \le 5)$, which is equivalent to $\text{P}(-5 \le T_1 - T_2 \le 5)$.
Using a graphing calculator with $\mu = -9$ and $\sigma = \sqrt{252}$:
$\text{P}(-5 \le T_1 - T_2 \le 5) \approx 0.21160$
$\approx 0.212 \text{ (to 3 s.f.)}$

1. Find the distribution of the total time taken for dinner and online games before revision.

Let $T_O = D + G$ be the total time spent from 7 pm to the start of revision if he plays online games.
Assuming $D$ and $G$ are independent,
$\text{E}(T_O) = \text{E}(D) + \text{E}(G) = 10 + 45 = 55$
$\text{Var}(T_O) = \text{Var}(D) + \text{Var}(G) = 2^2 + 10^2 = 104$
$T_O \sim \text{N}(55, 104)$

2. Calculate the conditional probability that he is late given he played online games.

He is late if the total time exceeds 60 minutes (from 7 pm to 8 pm).
$P(\text{Late} \mid \text{Online games}) = P(T_O > 60)$
Using the graphing calculator:
$P(T_O > 60) \approx 0.311999$

3. Determine the total probability that Benedict starts his revision late.

Let $S$ be the event he uses social media and $O$ be the event he plays online games.
Since he only does one or the other, $P(O) = 0.2$ means $P(S) = 1 - 0.2 = 0.8$.
From part (i), we know $P(\text{Late} \mid S) = 0.125$.
$P(\text{Late}) = P(S)P(\text{Late} \mid S) + P(O)P(\text{Late} \mid O)$
$P(\text{Late}) = 0.8(0.125) + 0.2(0.311999)$
$P(\text{Late}) = 0.1 + 0.062400 = 0.162400$

4. Find the conditional probability that he played online games given he started late.

Using Bayes' Theorem:
$P(O \mid \text{Late}) = \frac{P(O \cap \text{Late})}{P(\text{Late})}$
$P(O \mid \text{Late}) = \frac{0.062400}{0.162400}$
$P(O \mid \text{Late}) \approx 0.38423$
$P(O \mid \text{Late}) \approx 0.384 \text{ (3 s.f.)}$

1. State the necessary assumption of independence between the variables.

To calculate the variance of the combined times ($D + M$ and $D + G$) by simply adding their individual variances, the underlying random variables must be independent of each other.