In this question you should state the parameters of any distribution you use. A customer service call centre categorises incoming calls for follow-up resolution of the issues. The time taken to categorise an incoming call, $T$ minutes, follows a normal distribution $\text{N}(2, 0.2^2)$. Incoming calls are categorised as routine or complex calls. The time taken to resolve a routine call, $X$ minutes, follows a normal distribution $\text{N}(5, k)$ and the time taken to resolve a complex call, $Y$ minutes, follows a normal distribution $\text{N}(20, 6^2)$. You may assume that any incoming call is first categorised as either routine or complex and is immediately followed up with the resolution of the issue. The duration of the call is taken to be the sum of the time taken to categorise the call and then to resolve the issue.

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.)