Section B: Statistics [60 marks] A manager claims that the time spent by each customer at his supermarket follows a normal distribution with mean 35 minutes and standard deviation 30 minutes. A statistician comments that the distribution $\text{N}(35, 30^2)$ will not provide an adequate model.

1. Determine the probability of a customer spending a negative amount of time under the proposed model.

Let $T$ be the time spent by a customer. Under the manager's model, $T \sim \text{N}(35, 30^2)$.
The probability of a customer spending a negative amount of time is:
$P(T < 0) = P\left(Z < \frac{0 - 35}{30}\right)$
$P(T < 0) = P(Z < -1.167) \approx 0.1216$

2. Relate the probability to the physical constraints of the context to draw a conclusion.

Yes, I agree with the statistician.
Since time spent must be non-negative ($T \ge 0$), a model that predicts a relatively large probability (about $12.2\%$) for an impossible outcome (negative time) is not an adequate model.

1. Define the appropriate linear combination and calculate its expected value.

Let $X$ be the time spent by a randomly chosen customer, so $X \sim \text{N}(35, 10^2)$.
Let $W = X_1 + X_2 - 3X_3$ represent the difference between the total time of 2 customers and 3 times the time of a 3rd customer.
$\text{E}(W) = \text{E}(X_1) + \text{E}(X_2) - 3\text{E}(X_3)$
$\text{E}(W) = 35 + 35 - 3(35) = -35$

2. Calculate the variance of the linear combination.

$\text{Var}(W) = \text{Var}(X_1) + \text{Var}(X_2) + (-3)^2\text{Var}(X_3)$
$\text{Var}(W) = 10^2 + 10^2 + 9(10^2) = 100 + 100 + 900 = 1100$
Therefore, $W \sim \text{N}(-35, 1100)$.

3. Find the required probability using the normal distribution.

We require the probability that the total time spent by 2 customers is more than 3 times that of another customer, which is $\text{P}(X_1 + X_2 > 3X_3)$ or $\text{P}(W > 0)$.
$\text{P}(W > 0) \approx 0.1456488$
$\text{P}(W > 0) \approx 0.146 \text{ (to 3 s.f.)}$