ECBC 银行呼叫中心话务员平均每小时接听的电话次数正在被评估。

1. State the assumption of independence in the given context

Calls are received by the telephone operator independently of one another.

2. State the assumption of a constant average rate in the given context

The average rate at which calls are received by the telephone operator is constant over time.

1. Define the distributions for the number of calls in 1 hour and 4 hours.

Let $X$ be the number of credit card queries in 1 hour, and $Y$ be the number of credit card queries in 4 hours.
$X \sim \text{Po}(\mu)$
$Y \sim \text{Po}(4\mu)$

2. Translate the given relationship into a probability equation.

$\text{P}(X = 2) = 8 \times \text{P}(Y = 2)$
$\frac{e^{-\mu} \mu^2}{2!} = 8 \left( \frac{e^{-4\mu} (4\mu)^2}{2!} \right)$

3. Solve the equation for $\mu$.

$\frac{e^{-\mu} \mu^2}{2} = 8 \left( \frac{e^{-4\mu} 16\mu^2}{2} \right)$
$e^{-\mu} \mu^2 = 128 e^{-4\mu} \mu^2$

Since $\mu > 0$, we can divide both sides by $\mu^2$:
$e^{-\mu} = 128 e^{-4\mu}$
$e^{3\mu} = 128$
$3\mu = \ln 128$
$3\mu = \ln(2^7)$
$3\mu = 7 \ln 2$
$\mu = \frac{7}{3} \ln 2$

1. Determine the mean parameters and Poisson distributions for the 2-hour period.

Using $\mu = \frac{7}{3}\ln 2$ obtained from part (ii):
For a 2-hour period (1200 to 1400 hours), the number of credit card calls $C_2 \sim \text{Po}\left(\frac{14}{3}\ln 2\right)$, so $C_2 \sim \text{Po}(3.2343)$.
The number of other calls $X_2 = B_2 + P_2 \sim \text{Po}(12 + 14) = \text{Po}(26)$.
The total number of calls $T = C_2 + X_2 \sim \text{Po}\left(26 + \frac{14}{3}\ln 2\right)$, so $T \sim \text{Po}(29.2343)$.

2. Express the required conditional probability in terms of independent events.

We are looking for $P(C_2 = 0 \mid T > 50)$.
$P(C_2 = 0 \mid T > 50) = \frac{P(C_2 = 0 \text{ and } T > 50)}{P(T > 50)}$
Since $T = C_2 + X_2$, the intersection $(C_2 = 0 \text{ and } T > 50)$ simplifies to $(C_2 = 0 \text{ and } X_2 > 50)$.
As $C_2$ and $X_2$ are independent Poisson streams:
$P(C_2 = 0 \text{ and } X_2 > 50) = P(C_2 = 0) P(X_2 > 50)$

3. Evaluate the probabilities using a graphing calculator and find the final answer.

$P(C_2 = 0) = e^{-\frac{14}{3}\ln 2} = 2^{-14/3} \approx 0.039373$
$P(X_2 > 50) = 1 - P(X_2 \le 50) \approx 1.81156 \times 10^{-5}$
$P(T > 50) = 1 - P(T \le 50) \approx 3.97449 \times 10^{-4}$
Therefore,
$P(C_2 = 0 \mid T > 50) = \frac{0.039373 \times 1.81156 \times 10^{-5}}{3.97449 \times 10^{-4}} \approx 0.0017946$
To 3 significant figures, the probability is $0.00179$.

1. Establish the distributions for the number of calls in a two-hour period and their Normal approximations.

Let $B_2$ and $P_2$ be the number of business banking queries and personal banking queries respectively in a 2-hour period.
$B_2 \sim \text{Po}(2 \times 6) \implies B_2 \sim \text{Po}(12)$
$P_2 \sim \text{Po}(2 \times 7) \implies P_2 \sim \text{Po}(14)$

Since $\lambda > 10$ for both distributions, they can be suitably approximated by Normal distributions:
$B_2 \sim \text{N}(12, 12)$ approximately
$P_2 \sim \text{N}(14, 14)$ approximately

2. Determine the distribution of the difference $B_2 - P_2$.

Let $W = B_2 - P_2$.
$\text{E}(W) = \text{E}(B_2) - \text{E}(P_2) = 12 - 14 = -2$
$\text{Var}(W) = \text{Var}(B_2) + \text{Var}(P_2) = 12 + 14 = 26$

Therefore, $W \sim \text{N}(-2, 26)$ approximately.

3. Apply a continuity correction and calculate the probability that $B_2 > P_2$.

We want to find $\text{P}(B_2 > P_2)$, which is equivalent to $\text{P}(B_2 - P_2 > 0)$.
Since the number of calls are discrete integers, $B_2 - P_2 > 0 \implies B_2 - P_2 \ge 1$.

Applying the continuity correction for the continuous Normal variable $W$:
$\text{P}(B_2 - P_2 \ge 1) \approx \text{P}(W > 0.5)$

Calculating the standard normal value:
$\text{P}\left(Z > \frac{0.5 - (-2)}{\sqrt{26}}\right) = \text{P}\left(Z > \frac{2.5}{\sqrt{26}}\right)$
$\approx \text{P}(Z > 0.49029)$
$\approx 0.31196$
$\approx 0.312 \text{ (to 3 s.f.)}$