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I have $6$ independent and identically distributed variables such that $C_i \sim N(1000,400)$.

1) Calculate the density functions, distribution function and characteristic function of $C = \sum_{k=1}^6 C_i$

2) Calculate the probability $P(C > 6100)$

1)

Each $C_i$ is a variable of a normal distribution, so $C$ is a linear combination of these variables therefore $C \sim N(6\cdot1000,6\cdot400) = N(6000,2400)$ So the density and distribution are the ones of a normal distribution. The characteristic function is $\varphi_C(w) = e^{iw6000 -\frac{1}{2} 2400 w^2}$

2)

\begin{align} P(C > 6100) & = 1-P(C \le 6100) = 1 - P\left( \frac{C - 6000}{\sqrt{2400}} \le \frac{6100-6000}{\sqrt{2400}}\right) \\[6pt] & = 1- \Phi\left(\frac{6100-6000}{\sqrt{2400}}\right) \end{align}

Where $\Phi$ is the standard normal distribution function.

I would like to know if it is solved in the right way.

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  • $\begingroup$ I changed $C_i$ ~ $N(1000,400)$ to $C_i\sim N(1000,400)$ and $6*1000$ to $6\cdot1000$ and $\sqrt(2400)$ to $\sqrt{2400}$ and otherwise brought some of the MathJax code into line with standard usages. $\endgroup$
    – Michael Hardy
    May 25, 2015 at 13:29

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The argument is OK, but where you base a conclusion on the fact that $C$ is a linear combination, I'd say specifically that it is their sum and they are independent. If they were not independent then there would be more work to do. Also, if this is homework, I'd write out the particular density function for the normal distribution with the particular mean and variance that you have.

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