In this question, you should state clearly all the distributions that you use, together with the values of the appropriate parameters. A company produces two types of yarn balls with two machines. Machine $A$ produces balls of 2-ply yarn. Over a period of time, it is found that 80% of the balls of yarn from Machine $A$ have lengths less than 340 metres and 10% of the balls of yarn from Machine $A$ have lengths less than 280 metres.

1. Standardise the normal variable and use the inverse standard normal distribution to set up two simultaneous equations.

Let $X \sim \mathrm{N}(\mu, \sigma^2)$ be the length of yarn in balls from Machine $A$.
Given $\mathrm{P}(X < 340) = 0.80$ and $\mathrm{P}(X < 280) = 0.10$.
Standardising the variables, we get:
$\mathrm{P}\left(Z < \frac{340 - \mu}{\sigma}\right) = 0.80 \implies \frac{340 - \mu}{\sigma} \approx 0.84162$
$\mathrm{P}\left(Z < \frac{280 - \mu}{\sigma}\right) = 0.10 \implies \frac{280 - \mu}{\sigma} \approx -1.28155$

2. Solve the simultaneous equations to find and show the value of $\sigma$.

Rearranging the equations:
$340 - \mu = 0.84162\sigma \quad \text{--- (1)}$
$280 - \mu = -1.28155\sigma \quad \text{--- (2)}$
Subtracting (2) from (1):
$(340 - 280) = (0.84162 - (-1.28155))\sigma$
$60 = 2.12317\sigma$
$\sigma = 28.2596 \approx 28.3 \text{ (shown)}$

3. Substitute the precise value of $\sigma$ back into one of the equations to find $\mu$.

Substituting $\sigma = 28.2596$ into (1):
$340 - \mu = 0.84162(28.2596)$
$\mu = 340 - 23.784$
$\mu = 316.216 \approx 316 \text{ (to 3 s.f.)}$

1. State the distribution and find the probabilities of a ball falling into each of the two length categories.

Let $Y$ be the length of a ball of 3-ply yarn from Machine $B$.
$Y \sim \mathrm{N}(220, 6^2)$
$\mathrm{P}(Y > 230) = 0.047790$
$\mathrm{P}(Y < 215) = 0.202328$

2. Calculate the combined probability for the 5 randomly chosen balls.

Out of 5 balls, we require exactly 2 to be $> 230$ m and the remaining 3 to be $< 215$ m. The number of such arrangements is $\binom{5}{2} = 10$.
$\text{Required probability} = \binom{5}{2} [\mathrm{P}(Y > 230)]^2 [\mathrm{P}(Y < 215)]^3$
$= 10 \times (0.047790)^2 \times (0.202328)^3$
$= 0.00018918 \approx 0.000189 \text{ (to 3 s.f.)}$

1. Define the random variable for the total length and find its expected value.

Let $X$ and $Y$ be the lengths of yarn from Machine $A$ and Machine $B$ respectively, and $T = X_1 + X_2 + Y_1 + Y_2 + Y_3$ be the total length of the 5 randomly chosen balls.
$\mathrm{E}(T) = 2\mathrm{E}(X) + 3\mathrm{E}(Y)$
$\mathrm{E}(T) = 2(300) + 3(220) = 1260$

2. Find the variance of the total length and state the distribution of $T$.

$\mathrm{Var}(T) = 2\mathrm{Var}(X) + 3\mathrm{Var}(Y)$
$\mathrm{Var}(T) = 2(20^2) + 3(6^2)$
$\mathrm{Var}(T) = 2(400) + 3(36) = 908$
Since the sum of independent normal random variables is normally distributed, we state:
$T \sim \mathrm{N}(1260, 908)$

3. Set up the probability equation and solve for $k$ using the inverse normal distribution.

We are given $\mathrm{P}(T > k) = 0.385$.
$\mathrm{P}(T \le k) = 1 - 0.385 = 0.615$
Using the inverse normal function on a graphing calculator with $\mu = 1260$ and $\sigma = \sqrt{908}$:
$k \approx 1268.8101$
$k = 1268.81 \text{ (to 2 d.p.)}$

1. Define the normal distributions for the new lengths of yarn from Machine A and Machine B.

Let $A'$ and $B'$ be the continuous random variables representing the new lengths of yarn from Machine A and B respectively.
$A' = 0.91A \implies \text{E}(A') = 0.91(300) = 273$ and $\text{Var}(A') = 0.91^2(20^2) = 331.24$
Thus, $A' \sim \text{N}(273, 331.24)$

$B' = 0.95B \implies \text{E}(B') = 0.95(220) = 209$ and $\text{Var}(B') = 0.95^2(6^2) = 32.49$
Thus, $B' \sim \text{N}(209, 32.49)$

2. Formulate the linear combination for the difference between the total lengths of 3 balls of $A'$ and 4 balls of $B'$, and find its distribution.

Let $W = (A'_1 + A'_2 + A'_3) - (B'_1 + B'_2 + B'_3 + B'_4)$.
$\text{E}(W) = 3\text{E}(A') - 4\text{E}(B') = 3(273) - 4(209) = -17$
Since the balls are chosen randomly and independently, the variances add up:
$\text{Var}(W) = 3\text{Var}(A') + 4\text{Var}(B') = 3(331.24) + 4(32.49) = 1123.68$
Therefore, $W \sim \text{N}(-17, 1123.68)$.

3. Calculate the probability that the difference in total lengths is within 50 metres.

We need to find the probability that the difference is within $50\text{ m}$.
$\text{P}(|W| \le 50) = \text{P}(-50 \le W \le 50)$
Using a graphing calculator,
$\text{P}(-50 \le W \le 50) \approx 0.81473$
$\text{P}(-50 \le W \le 50) = 0.815 \text{ (to 3 s.f.)}$