Probability Sum of components > value

Question

I have a question that I cannot find the method in finding the solution.

Question: A device is made up of 5 subcomponents, denoted i=1,2,3,4,5. A subcomponents mean weight is 10i grams. All are normally distributed with a (population) variance of 4. What is the probability that the total weight of the device exceeds 160 grams?

Initially I did Z = 160-150/2 (as S^2 = 4, S = 2), but then thats Pr(Z > 5) which is basically 0.000001. But the answer doesn't seem right as the Probability of a single component being over by 2 grams is 0.1587, but then that would have to be 0.1587^5 which is essentially the same answer.

Basically my question is: Is there a proper way to do this kind of question? or am I doing it correctly and worrying about nothing?

Answer 1

Edit: Apparently I could not decipher that the means are $10i$. If I call the subcomponent weights $I_j$, and the total weight $W$, then $$E[W] = E[I_1]+\dotsb+E[I_5] = 10+20+30+40+50 = 150\text{ g}.$$

Assuming the weights are independent, the variance is $$\text{Var}(W) = \text{Var}(I_1)+\dotsb+\text{Var}(I_5) = 5(4) = 20\text{ g}^2.$$

Then \begin{align*} P(W> 160) &= 1-P(W\leq 160) \\ &= 1-P\left(Z\leq \frac{160-150}{\sqrt {20}}\right)\\ &= 1-\Phi(2.236067977) \\ &= 0.01267365936 \end{align*} which seems sensible.

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I forgot to include that the sum of independent normal random variables is again normal. If $X_i\sim N(\mu_i, \sigma^2_i)$ and independent, then $$X_1+\dotsb+X_n \sim N(\mu_1+\dotsb+\mu_n,\sigma^2_1+\dotsb+\sigma^2_n).$$ So above $$W\sim N(150, 20).$$

Answer 2

The mean and variance of a sum of normally distributed random variables are, respectively, the sums of the means and variances of those variables.

If $X_i\sim\mathcal N(10i, 2^2)$ and all are independently distributed, then : $(\sum_{i=1}^5 X_i)\sim\mathcal N(150, 5\cdot 2^2)$.

$$\mathsf P(Z> \tfrac{160-150}{2\surd 5}) \simeq 0.0126{\small 7}$$

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TL;DR The variance of a sum is going to be larger than any individual variance.