A factory manufactures paperweights consisting of glass mounted on a wooden base. The volume of glass, in $\text{cm}^3$, in a randomly chosen paperweight has a normal distribution with mean 56.5 and standard deviation 2.9. The volume of wood, in $\text{cm}^3$, has an independent normal distribution with mean 38.4 and standard deviation $\sigma$. The probability that the total volume of a randomly chosen paperweight exceeds $100\text{ cm}^3$ is 0.05. Show that $\sigma$ is 1.10, correct to 3 significant figures.

1. State the distribution of the sample mean for the volume of glass.

Let $G$ be the volume of glass in a paperweight and $\bar{G}$ be the sample mean of 20 paperweights.
Since $G \sim \mathrm{N}(56.5, 2.9^2)$, the sample mean is exactly normally distributed:
$\bar{G} \sim \mathrm{N}\left(56.5, \frac{2.9^2}{20}\right)$

2. Calculate the probability that the sample mean is less than 57.1.

$\mathrm{P}(\bar{G} < 57.1) \approx 0.82259$

1. Define the random variable for the difference in volume of wood between two paperweights and find its distribution.

Let $W_1$ and $W_2$ be the volumes of wood in two randomly chosen paperweights.
$\mathrm{E}(W_1 - W_2) = 38.4 - 38.4 = 0$
$\mathrm{Var}(W_1 - W_2) = \sigma^2 + \sigma^2 = 2(1.10)^2 = 2.42$
$\therefore W_1 - W_2 \sim \mathrm{N}(0, 2.42)$
(Note: Using the more precise unrounded value $\sigma^2 \approx 1.2036$ from the previous part gives $\mathrm{Var}(W_1 - W_2) \approx 2.407$)

2. Formulate the required probability.

The probability that the volumes differ by at least $0.07\text{ cm}^3$ is given by:
$\mathrm{P}(|W_1 - W_2| \ge 0.07) = 1 - \mathrm{P}(-0.07 < W_1 - W_2 < 0.07)$

3. Calculate the probability using a graphing calculator.

Using a graphing calculator:
$\mathrm{P}(|W_1 - W_2| \ge 0.07) = 1 - 0.03590 \dots$
$= 0.96410 \dots$
$\approx 0.964 \text{ (to 3 s.f.)}$

1. Formulate the random variable for the total mass of the paperweight.

Let $G$ be the volume of glass and $W$ be the volume of wood in a paperweight.
Let the total mass of the paperweight be $M$.
Since mass is density multiplied by volume, we can express $M$ as a linear combination of $G$ and $W$:
$M = 3.1G + 0.8W$

2. Calculate the expected value of the total mass.

Since $G$ and $W$ are independent normally distributed variables, their linear combination $M$ is also normally distributed.
$\mathrm{E}(M) = 3.1\mathrm{E}(G) + 0.8\mathrm{E}(W)$
$\mathrm{E}(M) = 3.1(56.5) + 0.8(38.4) = 175.15 + 30.72 = 205.87$

3. Calculate the variance of the total mass.

Using the stated result $\sigma = 1.10$ from the problem stem:
$\mathrm{Var}(M) = 3.1^2\mathrm{Var}(G) + 0.8^2\mathrm{Var}(W)$
$\mathrm{Var}(M) = 3.1^2(2.9^2) + 0.8^2(1.10^2)$
$\mathrm{Var}(M) = 9.61(8.41) + 0.64(1.21)$
$\mathrm{Var}(M) = 80.8181 + 0.7744 = 81.5925$

4. Calculate the probability that the total mass is between 200 and 220 grams.

We have $M \sim \mathrm{N}(205.87, 81.5925)$.
We need to find $\mathrm{P}(200 < M < 220)$.
Using a graphing calculator:
$\mathrm{P}(200 < M < 220) \approx 0.68326$
$\mathrm{P}(200 < M < 220) = 0.683 \text{ (to 3 s.f.)}$