在这道题中，你应该写出你所使用的任何分布的参数。 Crispy Cream 甜甜圈店销售 2 种甜甜圈：环形甜甜圈和夹心甜甜圈。环形甜甜圈和夹心甜甜圈的质量（单位：克）分别服从正态分布 $N(54,1.5^2)$ 和 $N(86,2^2)$。 12 个甜甜圈被包装在一个纸盒中。一个空纸盒的质量（单位：克）服从正态分布 $N(80,5^2)$。

1. Define the random variable representing the difference in mass and find its expected value.

Let $W = R_1 + R_2 + R_3 - 2F$, where $R_i \sim N(54, 1.5^2)$ and $F \sim N(86, 2^2)$.
$E(W) = 3E(R) - 2E(F)$
$E(W) = 3(54) - 2(86) = 162 - 172 = -10$

2. Calculate the variance of the linear combination.

$Var(W) = 3Var(R) + 2^2 Var(F)$
$Var(W) = 3(1.5^2) + 4(2^2)$
$Var(W) = 6.75 + 16 = 22.75$

3. State the parameters of the normal distribution for $W$ and compute the required probability.

Therefore, the distribution is $W \sim N(-10, 22.75)$.
We need to find $P(R_1 + R_2 + R_3 > 2F)$, which is equivalent to $P(W > 0)$.
Using a graphing calculator,
$P(W > 0) \approx 0.018015 \approx 0.0180 \text{ (to 3 s.f.)}$

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

Let $T$ be the total mass of the box and 12 donuts.
$T = R_1 + \dots + R_6 + F_1 + \dots + F_6 + B$
$E(T) = 6E(R) + 6E(F) + E(B)$
$E(T) = 6(54) + 6(86) + 80 = 920$

2. Calculate the variance of the total mass

$Var(T) = 6Var(R) + 6Var(F) + Var(B)$
$Var(T) = 6(1.5^2) + 6(2^2) + 5^2$
$Var(T) = 6(2.25) + 6(4) + 25 = 13.5 + 24 + 25 = 62.5$

3. State the distribution of $T$ and set up the probability equation

Since the individual masses are normally distributed and independent, their sum is also normally distributed. As required by the question preamble, we state the parameters of the distribution:
$T \sim N(920, 62.5)$
We are given:
$P(T > m) = 0.95$
$P(T \le m) = 0.05$

4. Solve for $m$ using the inverse normal distribution

Using the inverse normal function on a graphing calculator:
$m \approx 906.996$
$m = 907 \text{ (to 3 s.f.)}$

1. State the independence assumption required for the variance formulas used in parts (i) and (ii).

To calculate the variance of the combined masses (e.g. $R_1 + R_2 + R_3 - 2F$ or the total mass in a box), we add the individual variances. This requires the assumption that the mass of any given donut or box does not affect the mass of another.