Combining uniform (?) distributions

Question

$C$ and $D$ make widgets. $C$ makes widgets with expected mass 10 kg and a standard deviation of 1kg and he makes 7 in an hour. $D$ makes widgets with expected mass 12 kg and a standard deviation of 2.5 kg and he makes 6 in an hour. They work 8 hours a day and at the end of the day all the widgets are put in a box of mass 20 kg.

a) What is the expected mass of the full box?

b) What is the expected value of the average mass of the widgets in the box?

c) Find the standard deviation of the mass of the full box. What assumption have you made?

d) Under the same assumption what is the standard deviation of the average mass of the widgets in the box?

If we assume that the masses of the widgets follow the normal distribution, and that they satisfy the assumption in c),

e) the probability that the total mass of the box is less than 1120 kg.

f) the probability that the average mass of the widgets in the box is greater than 11kg.

My answer:

a) $56 \times 10+48\times 12=1136,1136+20=1156$

b) $1136/(56+48)=10.9$

c) and d) What assumption?

I am sure I can do e) and f) after the above is explained.

Answer 1

Here $C$ makes $7\times 8=56$ widgets in one day. Let $C_i$ be the mass of the $i$th widget that $C$ makes $\forall i=1,2,\dots 56$. Similarly define $D_j$ for $D$ where $j=1.2.\dots 48.$ Lastly define the mass of the box as $B$.

Here total mass of the box is $=\sum_{i=1}^{56}C_i+\sum_{j=1}^{48}D_i+B$

So expected mass of the full box is $=E[\sum_{i=1}^{56}C_i+\sum_{j=1}^{48}D_i+B]$

So the expected value of the average mass of the widgets in the box is $=E\left[\dfrac{\sum_{i=1}^{56}C_i+\sum_{j=1}^{48}D_i}{56+48}\right]$

So the variance of the mass of the full box is $$\begin{align}&=Var\left[ \sum_{i=1}^{56}C_i+\sum_{j=1}^{48}D_i+B\right] \\ &= \sum_{i=1}^{56}Var(C_i)+\sum_{j=1}^{48}Var(D_i) \end{align}$$ Note that box mass is fixed and here we assume $C$ and $D$ are statistically independent.

I think from here you can proceed. Otherwise inform me.

Answer 2

For (c), (d), You'll want to make the assumption that each widget's mass is uncorrelated. You can then use the fact that the variance of the sum of uncorrelated random variables is just the sum of the variance of each random variable to then calculate the standard deviation.

$$\text{Var}\left( \sum_{i = 1}^{N} X_i \right) = \sum_{i = 1}^{N} \text{Var}(X_i)$$

Refer to the Wikipedia page for more information.