Use CLT to find the probability

Question

$X$ 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.

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

Answer 1

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.