In this question you should state clearly the values of the parameters of any normal distributions that you use. The masses, in grams, of apples and pears have the independent distributions $N(260, 15^{2})$ and $N(210, 10^{2})$ respectively. A certain recipe requires 5 apples and 6 pears. The recipe requires the apples and pears to be prepared by removing the cores. This process reduces the mass of each apple by 5% and the mass of each pear by 15%.

1. State the distribution for the mass of an apple, clearly identifying the parameters.

Let $X$ be the mass of a randomly chosen apple in grams.
$X \sim \mathrm{N}(260, 15^2)$

2. Evaluate the probability that the mass is at least 250 grams.

$P(X \ge 250) = 0.747507\dots$
$P(X \ge 250) \approx 0.748 \text{ (to 3 s.f.)}$

1. Find the distribution of the difference in mass between a randomly chosen apple and a randomly chosen pear.

Let $X$ and $Y$ be the masses of a randomly chosen apple and pear respectively.
$E(X - Y) = E(X) - E(Y)$
$E(X - Y) = 260 - 210 = 50$
Since the masses are independent:
$\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)$
$\text{Var}(X - Y) = 15^2 + 10^2 = 225 + 100 = 325$
Therefore, $X - Y \sim \text{N}(50, 325)$.

2. Set up the modulus probability condition and evaluate it using a graphing calculator.

We require the probability that the mass differs by more than $55$ grams, which translates to $P(|X - Y| > 55)$.
$P(|X - Y| > 55) = P(X - Y < -55) + P(X - Y > 55)$
Using a graphing calculator:
$P(|X - Y| > 55) \approx 0.390757$
$P(|X - Y| > 55) \approx 0.391 \text{ (to 3 s.f.)}$

1. Find the expected total mass of the 5 prepared apples and 6 prepared pears.

Let $X$ and $Y$ be the mass of an apple and a pear respectively, so $X \sim \mathrm{N}(260, 15^2)$ and $Y \sim \mathrm{N}(210, 10^2)$.
Let $T$ be the total prepared mass.
$T = 0.95(X_1 + \dots + X_5) + 0.85(Y_1 + \dots + Y_6)$
$\mathrm{E}(T) = 0.95(5 \times 260) + 0.85(6 \times 210)$
$\mathrm{E}(T) = 1235 + 1071 = 2306$

2. Find the variance of the total mass of the prepared apples and pears.

Since the items are independently chosen, we sum their variances before multiplying by the scale factor squared:
$\mathrm{Var}(T) = 0.95^2(5 \times 15^2) + 0.85^2(6 \times 10^2)$
$\mathrm{Var}(T) = 0.9025(1125) + 0.7225(600)$
$\mathrm{Var}(T) = 1015.3125 + 433.5 = 1448.8125$

3. State the distribution parameters and calculate the probability that the total mass is at least 2300 grams.

Therefore, the total prepared mass follows the normal distribution $T \sim \mathrm{N}(2306, 1448.8125)$.
We need to find $\mathrm{P}(T \ge 2300)$.
Using a graphing calculator,
$\mathrm{P}(T \ge 2300) \approx 0.56262$
$\mathrm{P}(T \ge 2300) = 0.563 \text{ (to 3 s.f.)}$

1. Define the combined variable and find its mean

Let $Y=3(0.7W)-0.8(V_1+V_2)=2.1W-0.8V_1-0.8V_2$.
$E(Y)=2.1(260)-0.8(210)-0.8(210)=546-168-168=210.$

2. Find the variance (same apple scaled, two distinct pears)

Since $3(0.7W)=2.1W$ involves the SAME apple, its variance uses the factor $2.1^2$:
$Var(Y)=2.1^2(15^2)+0.8^2(10^2)+0.8^2(10^2)$
$=4.41(225)+0.64(100)+0.64(100)=992.25+64+64=1120.25.$
Hence $Y\sim N(210,\,1120.25)$.

3. Standardise and compute the probability

$P(Y>200)=P\!\left(Z>\frac{200-210}{\sqrt{1120.25}}\right)=P(Z>-0.2988)\approx 0.617.$

4. Interpret the probability in context

The value $0.617$ is the probability that three times 70% of the mass of one randomly chosen apple is more than 200 grams greater than 80% of the combined mass of two randomly chosen pears.