在本题中，你应该说明你所使用的任何分布的参数。 某客户服务呼叫中心将接入的电话进行分类，以便后续解决问题。对接入电话进行分类所需的时间 $T$（分钟）服从正态分布 $\text{N}(2, 0.2^2)$。接入的电话被分类为常规电话或复杂电话。解决常规电话所需的时间 $X$（分钟）服从正态分布 $\text{N}(5, k)$，而解决复杂电话所需的时间 $Y$（分钟）服从正态分布 $\text{N}(20, 6^2)$。你可以假设任何接入的电话都是先被分类为常规或复杂，然后立即进行后续问题的解决。通话的总时长定义为对电话进行分类的时间与解决问题的时间之和。

1. Sketch the normal distribution curve for $T$

Since $T \sim \text{N}(2, 0.2^2)$, the distribution is a symmetric, bell-shaped curve centred at the mean $\mu = 2$.
Draw a bell-shaped curve above a horizontal $t$-axis (or $T$-axis) and mark the value $2$ directly below the peak of the curve.

2. Shade the required probability area

The probability $\mathrm{P}(1.2 < T < 2.8)$ corresponds to the area under the normal curve between $t = 1.2$ and $t = 2.8$. Mark $1.2$ to the left of $2$, and $2.8$ to the right of $2$ (both equidistant from $2$, as they are $4$ standard deviations away), and shade the area bounded by the curve, the horizontal axis, and the vertical lines at $t = 1.2$ and $t = 2.8$.

1. Define the random variable for the total duration of a routine call and find its mean and variance.

Let $D$ be the duration of a routine call. Since it is the sum of the time taken to categorise and resolve, $D = T + X$.
$\text{E}(D) = \text{E}(T) + \text{E}(X) = 2 + 5 = 7$
$\text{Var}(D) = \text{Var}(T) + \text{Var}(X) = 0.2^2 + k = k + 0.04$
$D \sim \text{N}(7, k + 0.04)$

2. Set up the probability equation and standardise the variable.

Given that $\text{P}(D > 8) = 0.254$:
$\text{P}\left(Z > \frac{8 - 7}{\sqrt{k + 0.04}}\right) = 0.254$
$\text{P}\left(Z > \frac{1}{\sqrt{k + 0.04}}\right) = 0.254$

3. Use the inverse normal function to find the corresponding $z$-value.

Since $\text{P}(Z > z) = 0.254$, we have $\text{P}(Z \le z) = 1 - 0.254 = 0.746$.
Using the graphing calculator:
$\frac{1}{\sqrt{k + 0.04}} = \text{invNorm}(0.746) \approx 0.66197$

4. Solve for $k$ and round to 2 decimal places.

$\sqrt{k + 0.04} \approx \frac{1}{0.66197} \approx 1.51065$
$k + 0.04 \approx 1.51065^2 \approx 2.28206$
$k \approx 2.24206$
$k = 2.24 \text{ (shown to 2 d.p.)}$

1. State the distribution of the sample mean $\bar{Y}$ for $n$ complex calls.

Let $\bar{Y}$ be the mean time taken to resolve $n$ complex calls.
Since the time taken for a single complex call is $Y \sim \text{N}(20, 6^2)$, the sample mean follows the distribution $\bar{Y} \sim \text{N}\left(20, \frac{6^2}{n}\right)$.

2. Formulate the probability inequality and standardise the variable.

We are given that $\text{P}(\bar{Y} > 24) \le 0.01$.
Standardising to the standard normal variable $Z \sim \text{N}(0, 1)$:
$\text{P}\left(Z > \frac{24 - 20}{6/\sqrt{n}}\right) \le 0.01$

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

Using the inverse standard normal function, the critical value for the 99th percentile is $2.3263$ (i.e. $\text{P}(Z > 2.3263) \approx 0.01$).
$\frac{4}{6/\sqrt{n}} \ge 2.3263$
$\frac{2\sqrt{n}}{3} \ge 2.3263$
$\sqrt{n} \ge 3.4895$
$n \ge 12.177$

4. Determine the final set of possible values for $n$ based on the context constraints.

Since $n$ represents the number of calls, it must be an integer, which implies $n \ge 13$.
Given that there are exactly 20 incoming calls in total on that morning, $n \le 20$.
Therefore, $13 \le n \le 20$, and the set of possible values is $\{13, 14, 15, 16, 17, 18, 19, 20\}$.

1. Find the parameters of the new distribution for the time taken to resolve a complex call.

Let $C$ be the time taken to resolve a complex call after the review.
Since the time is reduced by $20\%$, $C = 0.8Y$.
$\text{E}(C) = 0.8 \times 20 = 16$
$\text{Var}(C) = 0.8^2 \times 6^2 = 23.04$
Therefore, $C \sim \text{N}(16, 23.04)$.

2. Formulate the required probability and define the linear combination of random variables.

We need the probability that the time taken to resolve 2 complex calls ($C_1 + C_2$) is more than twice the time taken to resolve a routine call ($2X$).
This is equivalent to finding $\text{P}(C_1 + C_2 > 2X)$, which can be rewritten as $\text{P}(C_1 + C_2 - 2X > 0)$.
Let $W = C_1 + C_2 - 2X$.

3. Calculate the mean and variance of the combined random variable $W$.

$\text{E}(W) = \text{E}(C_1) + \text{E}(C_2) - 2\text{E}(X)$
$\text{E}(W) = 16 + 16 - 2(5) = 22$

Using the given value $k = 2.24$ from part (b):
$\text{Var}(W) = \text{Var}(C_1) + \text{Var}(C_2) + (-2)^2\text{Var}(X)$
$\text{Var}(W) = 23.04 + 23.04 + 4(2.24) = 55.04$

Thus, we state the parameters of the combined distribution: $W \sim \text{N}(22, 55.04)$.

4. Calculate the final probability using the normal distribution.

$\text{P}(W > 0) = \text{P}\left(Z > \frac{0 - 22}{\sqrt{55.04}}\right)$
$\text{P}(W > 0) \approx 0.99849$
$\text{P}(W > 0) = 0.998$ (to 3 s.f.)