"Famous Amas" 制作两种曲奇饼干：杏仁曲奇和巧克力片曲奇。一块杏仁曲奇和一块巧克力片曲奇的质量（以克为单位）是相互独立的随机变量，分别服从分布 $N(32, 3.0^2)$ 和 $N(28, 1.8^2)$。 杏仁曲奇的价格为每 100 克 $6，巧克力片曲奇的价格为每 100 克 $5。约翰（John）买了 4 块杏仁曲奇和 4 块巧克力片曲奇。辛迪（Cindy）买了 10 块巧克力片曲奇。

1. Determine the distribution for the total mass of 4 randomly chosen chocolate chip cookies.

Let $C$ be the mass of a chocolate chip cookie, where $C \sim N(28, 1.8^2)$.
Let $T_C = C_1 + C_2 + C_3 + C_4$.
$\mathrm{E}(T_C) = 4 \times 28 = 112$
$\mathrm{Var}(T_C) = 4 \times 1.8^2 = 12.96$
Hence, $T_C \sim N(112, 12.96)$.

2. Evaluate the probability that the total mass is between 100 g and 120 g.

We require $\mathrm{P}(100 < T_C < 120)$.
Using a graphing calculator:
$\mathrm{P}(100 < T_C < 120) \approx 0.986427$
$\approx 0.986$ (to 3 s.f.)

1. Define the random variable for the difference and calculate its expected value.

Let $C_1, C_2, C_3, C_4$ be the masses of the 4 randomly chosen chocolate chip cookies and $A$ be the mass of the randomly chosen almond cookie.
Let $W = (C_1 + C_2 + C_3 + C_4) - 2A$.
$\text{E}(W) = 4\text{E}(C) - 2\text{E}(A)$
$\text{E}(W) = 4(28) - 2(32) = 48$

2. Calculate the variance of the combined random variable.

$\text{Var}(W) = 4\text{Var}(C) + (-2)^2\text{Var}(A)$
$\text{Var}(W) = 4(1.8^2) + 4(3.0^2) = 12.96 + 36 = 48.96$
Thus, $W \sim \text{N}(48, 48.96)$.

3. Formulate the probability expression and evaluate it.

The magnitude of the difference is less than 50 g, so we require $\text{P}(|W| < 50)$.
$\text{P}(-50 < W < 50) \approx 0.612487$
$\text{P}(-50 < W < 50) = 0.612 \text{ (to 3 s.f.)}$

1. Define the cost variables for John ($J$) and Cindy ($Y$) by applying the price rates to the respective total masses.

Let $W_A$ be the total mass of 4 almond cookies, $W_{C1}$ be the total mass of 4 chocolate chip cookies, and $W_{C2}$ be the total mass of 10 chocolate chip cookies.
From the given distributions:
$W_A \sim N(4 \times 32, 4 \times 3.0^2) \implies W_A \sim N(128, 36)$
$W_{C1} \sim N(4 \times 28, 4 \times 1.8^2) \implies W_{C1} \sim N(112, 12.96)$
$W_{C2} \sim N(10 \times 28, 10 \times 1.8^2) \implies W_{C2} \sim N(280, 32.4)$

The prices per gram are $\char36{}0.06$ for almond and $\char36{}0.05$ for chocolate chip.
Cost for John: $J = 0.06 W_A + 0.05 W_{C1}$
Cost for Cindy: $Y = 0.05 W_{C2}$

2. Calculate the expected value of the difference $J - Y$.

$\mathrm{E}(J) = 0.06(128) + 0.05(112) = 7.68 + 5.60 = 13.28$
$\mathrm{E}(Y) = 0.05(280) = 14$
$\mathrm{E}(J - Y) = 13.28 - 14 = -0.72$

3. Calculate the variance of the difference $J - Y$.

$\mathrm{Var}(J) = 0.06^2(36) + 0.05^2(12.96) = 0.1296 + 0.0324 = 0.162$
$\mathrm{Var}(Y) = 0.05^2(32.4) = 0.081$

Since $J$ and $Y$ are independent:
$\mathrm{Var}(J - Y) = \mathrm{Var}(J) + \mathrm{Var}(Y)$
$\mathrm{Var}(J - Y) = 0.162 + 0.081 = 0.243$

4. State the distribution of $J - Y$ and compute the probability $P(J > Y)$.

Since linear combinations of independent normal variables are also normally distributed:
$J - Y \sim N(-0.72, 0.243)$

We want to find $P(J > Y)$, which is equivalent to $P(J - Y > 0)$:
$P(J - Y > 0) = P\left(Z > \frac{0 - (-0.72)}{\sqrt{0.243}}\right)$
$P(J - Y > 0) = P(Z > 1.46059...)$
$P(J - Y > 0) \approx 0.072056$
$P(J > Y) = 0.0721 \text{ (to 3 s.f.)}$