# Use CLT to find the probability

$X$ is a discrete random variable with the following density function: $$f_X(n) = \begin{cases} c e^{-2} \, \frac{2^n}{n!} & n \geq 0, \\ c 3^n & n < 0. \end{cases}$$ Define the variables $V$ as follows: $$V = \begin{cases} 0 & X < 0 \\ X & X \geq 0. \end{cases}$$ Let $V_1, V_2, V_3, \dotsc$ be a sequence of independent and identically distributed random variables each having the same distribution as $V$.

Use the central limit theorem to approximately find $P(20 < \sum_{i=1}^{20} V_i < 30)$.

(Original image here.)

Not a homework. The entire problem could be found at https://math.stackexchange.com/questions/2885578/find-the-expected-value-of-the-sum-of-random-variables. But it is not that relevant.

My work: In order to use CLT we assume that the variable is normal and then find the mean and std of the random variable.

$\mathbb E[V]=\sum_{n\geq 0}nce^{-2}2^n/n!=2c$

But how to find the variance for the Normal r.v.?

$\mathbb E[V^2]=\sum_{n\geq 0}n^2ce^{-2}2^n/n!=6c$ (not sure if it correct)

So $\operatorname{Var}[V]=6c-4c^2$

• No work at all? Then does not invite to help. Aug 17 '18 at 11:06
• @drhab Plz wait a sec for my work Aug 17 '18 at 11:07
• To avoid misunderstandings next time publish your question if it is completely ready to be published. Aug 17 '18 at 11:09
• To find variance apply the rule $\mathsf{Var}V=\mathbb EV^2-(\mathbb EV)^2$ Aug 17 '18 at 11:28

Based on your calculations (I think they are correct) you can find mean and variance of $\sum_{i=1}^{20}V_i$.
Denoting these by $\mu$ and $\sigma^2$ to be found is: $$P(20<\sum_{i=1}^{20}V_i<30)=P\left(\frac{20-\mu}{\sigma}<\frac{\sum_{i=1}^{20}V_i-\mu}{\sigma}<20<\frac{30-\mu}{\sigma}\right)\approx P\left(\frac{20-\mu}{\sigma}<U<\frac{30-\mu}{\sigma}\right)$$where $U$ has standard normal distribution.
• $\mu$ is just $20\times E[V]$ and $\sigma^2$ is just $20\times Var[V]$ because they are independent? Aug 17 '18 at 12:36