In this question you should state clearly the parameters of any normal distributions you use. The masses in kilograms of chickens have the distribution $N(1.5,\sigma^2)$ such that 7.65% of them have a mass greater than 1.6 kg.

1. Express the given probability using the standard normal variable $Z$.

Let $C$ be the mass of a chicken, so $C \sim \mathrm{N}(1.5, \sigma^2)$.
We are given that $\mathrm{P}(C > 1.6) = 0.0765$.
Standardising, we have:
$\mathrm{P}\left(Z > \frac{1.6 - 1.5}{\sigma}\right) = 0.0765$
$\mathrm{P}\left(Z \le \frac{0.1}{\sigma}\right) = 1 - 0.0765 = 0.9235$

2. Use the inverse normal function to solve for $\sigma$.

Using the inverse normal distribution function on a graphing calculator:
$\frac{0.1}{\sigma} \approx 1.42898$
$\sigma = \frac{0.1}{1.42898} \approx 0.06998$
$\sigma \approx 0.0700 \text{ (to 3 s.f.)}$

1. Find the probability that a chosen chicken has a mass less than 1.6 kg.

Let $C$ be the mass of a chicken in kg.
Since $7.65\%$ of chickens have a mass greater than $1.6\text{ kg}$, the probability of being less than $1.6\text{ kg}$ is:
$\text{P}(C < 1.6) = 1 - 0.0765 = 0.9235$

2. Find the probability that a chosen duck has a mass more than 1.9 kg.

Let $D$ be the mass of a duck in kg.
Given the distribution $D \sim N(2.0, 0.08^2)$,
$\text{P}(D > 1.9) = 0.89435$

3. Calculate the combined probability of both independent events occurring.

Since the choice of a chicken and a duck are independent events, we multiply their respective probabilities:
$\text{Required probability} = \text{P}(C < 1.6) \times \text{P}(D > 1.9)$
$= 0.9235 \times 0.89435$
$= 0.82593$
$\approx 0.826 \text{ (to 3 s.f.)}$

1. Calculate the standard deviation $\sigma$ from the given probability (from part a).

From part (a), $C \sim \mathrm{N}(1.5, \sigma^2)$ and $\mathrm{P}(C > 1.6) = 0.0765$.
Standardising the variable:
$\mathrm{P}\left(Z > \frac{1.6 - 1.5}{\sigma}\right) = 0.0765$
Using the inverse normal distribution, $z \approx 1.4290$.
$\frac{0.1}{\sigma} = 1.4290 \implies \sigma \approx 0.069977 \text{ kg}$

2. Define the expected selling prices for the chickens and the duck.

Let $S_C = 11(C_1 + C_2 + C_3 + C_4)$ be the selling price of 4 randomly chosen chickens.
Let $S_D = 2 \times 18D = 36D$ be twice the selling price of 1 randomly chosen duck.
Let $W = S_D - S_C$ be the difference in selling prices.
$\mathrm{E}(S_C) = 11(4)(1.5) = 66$
$\mathrm{E}(S_D) = 36(2.0) = 72$
$\mathrm{E}(W) = 72 - 66 = 6$

3. Calculate the variance of the difference in selling prices.

$\mathrm{Var}(S_C) = 11^2(4)(\sigma^2) = 484(0.069977)^2 \approx 2.3701$
$\mathrm{Var}(S_D) = 36^2(0.08^2) = 8.2944$
Since the chickens and the duck are chosen independently,
$\mathrm{Var}(W) = \mathrm{Var}(S_D) + \mathrm{Var}(S_C) = 8.2944 + 2.3701 = 10.6645$
Thus, the difference $W$ follows a normal distribution: $W \sim \mathrm{N}(6, 10.6645)$.

4. Calculate the probability that the difference is at least \$5.

We require the probability that the difference is at least $5, which is $\mathrm{P}(|W| \ge 5)$.
$\mathrm{P}(|W| \ge 5) = \mathrm{P}(W \ge 5) + \mathrm{P}(W \le -5)$
Using a graphing calculator:
$\mathrm{P}(W \ge 5) \approx 0.62029$
$\mathrm{P}(W \le -5) \approx 0.00038$
$\mathrm{P}(|W| \ge 5) = 0.62029 + 0.00038 = 0.62067 \approx 0.621 \text{ (to 3 s.f.)}$