A credit card company has a special 24-hour telephone hotline for customers to either report a lost credit card or to request for an increase in their credit limit. It is found that in any 10-minute period, on average, three calls are received reporting on a lost card and two calls are received requesting for an increase in credit limit. In addition, six "nuisance" calls on average are received every 15 minutes for other customer services (i.e. those unrelated to reporting of a lost card or a credit limit increase). Assuming that the calls received may be modeled with Poisson distributions,

1. Determine the parameter of the Poisson distribution for a one-hour period.

Let $X$ be the number of calls reporting a lost credit card in a 1-hour (60 minutes) period.
Since the average is 3 calls per 10 minutes, the average for 60 minutes is proportional:
$\lambda = 3 \times \frac{60}{10} = 18$.
So, $X \sim \text{Po}(18)$.

2. Calculate the probability that more than 12 calls are received.

$\text{P}(X > 12) = 1 - \text{P}(X \le 12)$
Using a graphing calculator,
$\text{P}(X > 12) = 1 - 0.0916739 \dots$
$\text{P}(X > 12) = 0.908326 \dots \approx 0.908 \text{ (to 3 s.f.)}$

1. Calculate the average number of each type of call in a 15-minute interval.

Average lost card calls $= \frac{15}{10} \times 3 = 4.5$
Average credit limit calls $= \frac{15}{10} \times 2 = 3$
Average nuisance calls $= 6$

2. Determine the distribution of the total number of calls in 15 minutes.

Let $W$ be the total number of calls in a 15-minute interval.
Since the sum of independent Poisson random variables follows a Poisson distribution:
$\lambda = 4.5 + 3 + 6 = 13.5$
$W \sim \text{Po}(13.5)$

3. Calculate the probability of receiving at most 12 calls.

$\text{P}(W \le 12) = 0.40933... \approx 0.409$ (to 3 s.f.)

1. Calculate the probability of the condition being met in a single one-hour period (result from part i).

Let $L$ be the number of calls reporting a lost credit card in a one-hour period.
Given an average of 3 calls per 10 minutes, the average rate per hour is $\lambda = 3 \times 6 = 18$.
$L \sim \text{Po}(18)$
$P(L > 12) = 1 - P(L \le 12) \approx 0.90833$

2. Set up the binomial distribution for the 24 consecutive one-hour periods.

Let $W$ be the number of one-hour periods out of 24 where more than 12 calls are received.
Assuming calls follow a Poisson process, non-overlapping periods are independent. Thus, $W$ follows a binomial distribution with $n = 24$ and $p = 0.90833$.
$W \sim \text{B}(24, 0.90833)$

3. Evaluate the probability of at least 20 such periods using a graphing calculator.

We need to find the probability of at least 20 periods, which is $P(W \ge 20)$.
$P(W \ge 20) = 1 - P(W \le 19)$
Using a graphing calculator:
$P(W \ge 20) \approx 0.93820$
$\approx 0.938 \text{ (to 3 s.f.)}$

1. Find the expected number of each type of call in a one-hour period and define the discrete distributions.

Let $X$ be the total number of calls reporting a lost card or requesting a credit limit increase in a one-hour period.
Let $Y$ be the number of 'nuisance' calls in a one-hour period.

Scaling the rates to one hour (60 minutes):
Lost card: $3 \times 6 = 18$
Credit limit: $2 \times 6 = 12$
Nuisance: $6 \times 4 = 24$

Thus,
$X \sim \text{Po}(18 + 12) \implies X \sim \text{Po}(30)$
$Y \sim \text{Po}(24)$

2. Approximate the discrete Poisson distributions with continuous Normal distributions.

Since $\lambda_X = 30 > 10$ and $\lambda_Y = 24 > 10$, we can approximate both using the Normal distribution:
$X \sim \text{N}(30, 30)$ approximately
$Y \sim \text{N}(24, 24)$ approximately

3. Determine the distribution of the linear combination $Y - X$.

We want to find $P(Y > X)$, which is $P(Y - X > 0)$.
Let $W = Y - X$.
Since $X$ and $Y$ are independent,
$\text{E}(W) = \text{E}(Y) - \text{E}(X) = 24 - 30 = -6$
$\text{Var}(W) = \text{Var}(Y) + \text{Var}(X) = 24 + 30 = 54$

Therefore, $W \sim \text{N}(-6, 54)$ approximately.

4. Apply the continuity correction and calculate the required probability.

Since $X$ and $Y$ originally take integer values, $Y - X > 0$ implies $Y - X \ge 1$.
Applying the continuity correction, we calculate $P(W > 0.5)$:

$P(W > 0.5) = P\left(Z > \frac{0.5 - (-6)}{\sqrt{54}}\right)$
$= P\left(Z > \frac{6.5}{\sqrt{54}}\right)$
$= P(Z > 0.8845)$
$\approx 0.1882 \approx 0.188 \text{ (to 3 s.f.)}$

(Note: If calculated without continuity correction as is the convention in some recent syllabuses, $P(W > 0) = P\left(Z > \frac{6}{\sqrt{54}}\right) \approx 0.207$. However, traditional rigorous approximation historically includes it)

1. State two necessary assumptions for combining distributions and using the Binomial model in the earlier parts.

In part (ii) and (iv), different types of calls were combined, which requires assuming that the number of calls for lost credit cards, credit limit increases, and nuisance calls are independent of one another.

In part (iii), the 24 one-hour periods were modelled as independent trials (Binomial distribution), which requires assuming that the number of calls received in any one-hour period is independent of the number of calls received in other one-hour periods. (Alternatively, we must assume the average rate of calls remains constant throughout the 24-hour period).