# Combining uniform (?) distributions

$C$ 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.

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.

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Hint: The variance of the sum of uncorrelated random variables is the sum of the variances. This does not hold true for correlated random variables; you need the values of the covariances as well in order to compute the variance of the sum of correlated random variables. Since you are not told whether the random variables are uncorrelated or not, you can't compute the variance (and hence the standard deviation) of their sum unless you boldly made a certain assumption.... – Dilip Sarwate Jan 31 '13 at 20:10

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.

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Variance of full box is $56^2 \times 1+48^2 \times 2.5^2$=17536 kg, so $\sigma=132.4kg$? – bbr4in Jan 31 '13 at 20:28
@user52187:I think you consider $\sum_{i=1}^{56} C_i = 56 C_1$. But this is not true. Your variance is $\sum_{i=1}^{56}Var(C_i)+\sum_{j=1}^{48}Var(D_i)=\sum_{i=1}^{56}1^2+\sum_{j=1}^{‌​48}2.5^2=56 \times 1+ 48 \times 2.5^2$. Next just square-root the variance and you get $\sigma$. – Argha Feb 1 '13 at 5:50