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 distribution for the number of calls reporting a lost credit card in a one-hour period.

Let $X$ be the number of calls reporting a lost credit card in a one-hour period.
Since there are $3$ calls on average in a $10$-minute period, the average number of calls in $60$ minutes is $3 \times 6 = 18$.
$X \sim \text{Po}(18)$

2. Set up and evaluate the required probability using the complement rule.

We want to find the probability of receiving more than $12$ calls.
$\text{P}(X > 12) = 1 - \text{P}(X \le 12)$
Using a graphing calculator,
$\text{P}(X > 12) = 1 - 0.091669 \approx 0.90833$
$\text{P}(X > 12) \approx 0.908 \text{ (to 3 s.f.)}$

1. Determine the mean number of calls for each category in a 15-minute interval.

Let $X_1$, $X_2$, and $X_3$ be the number of lost card calls, credit limit calls, and nuisance calls received in a 15-minute interval, respectively.

The mean rates for 15 minutes are:
$\lambda_1 = 3 \times \frac{15}{10} = 4.5$
$\lambda_2 = 2 \times \frac{15}{10} = 3$
$\lambda_3 = 6$

2. Identify the probability distribution for the total number of calls.

Assuming the different types of calls are independent, the total number of calls $W = X_1 + X_2 + X_3$ also follows a Poisson distribution.

$\lambda_W = 4.5 + 3 + 6 = 13.5$
Thus, $W \sim \text{Po}(13.5)$.

3. Calculate the probability that at most 12 calls are received.

We need to find $P(W \le 12)$.
Using a graphing calculator:
$P(W \le 12) \approx 0.40932$
$P(W \le 12) = 0.409 \text{ (to 3 s.f.)}$

1. Recalculate the probability from part (i) that more than 12 lost card calls are received in a 1-hour period.

Let $X$ be the number of calls reporting a lost credit card in a 1-hour period.
Since the average is 3 calls per 10 minutes, the mean for 60 minutes is $3 \times 6 = 18$.
$X \sim \text{Po}(18)$
$P(X > 12) = 1 - P(X \le 12) \approx 0.90833 \text{ (to 5 s.f.)}$

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

Let $Y$ be the number of 1-hour periods out of 24 where more than 12 such calls are received.
Assuming the 1-hour periods are independent, $Y$ follows a binomial distribution:
$Y \sim \text{B}(24, 0.90833)$

3. Calculate the probability that there are at least 20 such periods.

We require $P(Y \ge 20)$.
$P(Y \ge 20) = 1 - P(Y \le 19)$
Using a graphing calculator,
$P(Y \ge 20) \approx 0.93667$
$P(Y \ge 20) = 0.937 \text{ (to 3 s.f.)}$

1. Calculate the expected number of calls for the nuisance category and the combined total of the other two categories in a one-hour period.

Let $N$ be the number of nuisance calls in a one-hour period.
Let $T$ be the total number of calls reporting a lost card or requesting a credit limit increase in a one-hour period.
$\lambda_N = 6 \times \frac{60}{15} = 24$
$\lambda_T = \left(3 \times \frac{60}{10}\right) + \left(2 \times \frac{60}{10}\right) = 18 + 12 = 30$

2. State the Normal approximations for the distributions of $N$ and $T$.

Since the Poisson means $\lambda_N = 24$ and $\lambda_T = 30$ are both greater than $10$, we can approximate them using the Normal distribution:
$N \sim \text{N}(24, 24) \text{ approximately}$
$T \sim \text{N}(30, 30) \text{ approximately}$

3. Determine the distribution of the linear combination $N - T$.

Let $W = N - T$. Since $N$ and $T$ are independent:
$\text{E}(W) = \text{E}(N) - \text{E}(T) = 24 - 30 = -6$
$\text{Var}(W) = \text{Var}(N) + \text{Var}(T) = 24 + 30 = 54$
Therefore, $W \sim \text{N}(-6, 54)$ approximately.

4. Apply the continuity correction to find the required probability expression.

We need the probability that $N$ exceeds $T$, which is $\text{P}(N - T > 0)$.
Since the original variables are discrete, $N - T > 0 \implies N - T \ge 1$.
Applying the continuity correction:
$\text{P}(N - T \ge 1) \approx \text{P}(W > 0.5)$

5. Compute the final probability.

$\text{P}(W > 0.5) = \text{P}\left(Z > \frac{0.5 - (-6)}{\sqrt{54}}\right)$
$= \text{P}\left(Z > \frac{6.5}{\sqrt{54}}\right)$
$\approx \text{P}(Z > 0.8845)$
$\approx 0.18821$
$\approx 0.188 \text{ (to 3 s.f.)}$

1. State the first assumption (independence).

The occurrences of the different types of calls (reporting a lost card, requesting a credit limit increase, and "nuisance" calls) are independent of one another (or calls are received independently of one another).

2. State the second assumption (constant rate).

The average rate of receiving each type of call remains constant throughout the time intervals considered.