In this question, you should state clearly all the distributions that you use, together with the values of the appropriate parameters. A toy designer creates weighted wooden blocks used in balance-based educational toys. There are two types of blocks: cylindrical blocks and cuboidal blocks. The mass, in grams, of each cylindrical block and cuboidal block follow a normal distribution with the following parameters:

	Mean	Standard Deviation
cylindrical block	280	6
cuboidal block	220	5

To build a deluxe seesaw model, the designer uses:

• 2 cylindrical blocks,
• 3 cuboidal blocks,
• 7 screws,
• 1 wooden plank.

The blocks are drilled before assembly to fit screws:

• Each cylindrical block loses 5% of its mass after drilling.
• Each cuboidal block loses 3% of its mass after drilling.

The masses of the remaining components follow a normal distribution with the following parameters:

	Mean	Standard Deviation
screw	8	1.2
wooden plank	540	8

When assembling the deluxe seesaw model, all of its components are chosen at random.

1. Define the random variables and their initial distributions.

Let $X$ be the mass of a cylindrical block and $Y$ be the mass of a cuboidal block.
$X \sim \mathrm{N}(280, 6^2)$
$Y \sim \mathrm{N}(220, 5^2)$

2. Determine the expectation and variance of the difference $X - Y$.

$\mathrm{E}(X - Y) = \mathrm{E}(X) - \mathrm{E}(Y) = 280 - 220 = 60$
$\mathrm{Var}(X - Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) = 6^2 + 5^2 = 36 + 25 = 61$

3. State the distribution of $X - Y$ and calculate the required probability.

Since $X$ and $Y$ are independent normal variables,
$X - Y \sim \mathrm{N}(60, 61)$
$\mathrm{P}(X - Y > 65) \approx 0.26103 \approx 0.261$ (to 3 s.f.)

1. Formulate the random variable for the linear combination and calculate its expected value.

Let $X$ be the mass of a cylindrical block and $Y$ be the mass of a cuboidal block.
$X \sim \mathrm{N}(280, 6^2)$ and $Y \sim \mathrm{N}(220, 5^2)$.
Let $W$ be the total of twice the mass of a cylindrical block and thrice the mass of a cuboidal block. Then $W = 2X + 3Y$.
$\mathrm{E}(W) = 2\mathrm{E}(X) + 3\mathrm{E}(Y)$
$\mathrm{E}(W) = 2(280) + 3(220) = 1220$

2. Calculate the variance of the linear combination.

Since the masses of the blocks are randomly selected and thus independent, the variance of $W$ is:
$\mathrm{Var}(W) = 2^2\mathrm{Var}(X) + 3^2\mathrm{Var}(Y)$
$\mathrm{Var}(W) = 4(6^2) + 9(5^2)$
$\mathrm{Var}(W) = 144 + 225 = 369$

3. State the resulting normal distribution and use the inverse normal method to find $M$.

Since $X$ and $Y$ are normally distributed, their linear combination is also normally distributed:
$W \sim \mathrm{N}(1220, 369)$
We are given $\mathrm{P}(W < M) = 0.355$.
Using the inverse normal function on a graphing calculator:
$M = 1212.8569\dots$
Rounding to the nearest integer:
$M \approx 1213$

1. Define the random variable for the total mass of the assembled seesaw model.

Let $X$, $Y$, $S$, and $W$ be the masses of a cylindrical block, cuboidal block, screw, and wooden plank respectively.
Let $T$ be the total mass of the deluxe seesaw model.
$T = 0.95X_1 + 0.95X_2 + 0.97Y_1 + 0.97Y_2 + 0.97Y_3 + S_1 + \dots + S_7 + W$

2. Calculate the expected value of $T$.

$\text{E}(T) = 2(0.95 \times 280) + 3(0.97 \times 220) + 7(8) + 540$
$= 532 + 640.2 + 56 + 540$
$= 1768.2$

3. Calculate the variance of $T$ and state its distribution.

$\text{Var}(T) = 2(0.95^2 \times 6^2) + 3(0.97^2 \times 5^2) + 7(1.2^2) + 8^2$
$= 2(0.9025 \times 36) + 3(0.9409 \times 25) + 7(1.44) + 64$
$= 64.98 + 70.5675 + 10.08 + 64$
$= 209.6275$

Hence, $T \sim \text{N}(1768.2, 209.6275)$.

4. Calculate the probability that the total mass exceeds 1785 g.

$\text{P}(T > 1785) \approx 0.12297$
$\approx 0.123 \text{ (to 3 s.f.)}$