Let

L=\underset{n\to \infty }{\lim }\sum _{k=0}^n\frac{e^{-n}{n}^k}{k!}

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Value of $L$ is ______.

The correct option is 0.5.

Step-by-step Derivation and Explanation:

We are given the limit:

$L=\underset{n\to ��}{lim}\sum _{k=0}^n\frac{e^{-n}{n}^k}{k!}$

Let us analyze this sum using probability theory. Consider a Poisson random variable X with parameter λ = n.
The probability mass function of X is given by:

$P(X=k)=\frac{e^{-n}{n}^k}{k!}$

The sum inside the limit represents the cumulative distribution function (CDF) of this Poisson variable evaluated at X ≤ n:

$P(X\le n)=\sum _{k=0}^n\frac{e^{-n}{n}^k}{k!}$

For a Poisson distribution with parameter n, the mean (expected value) and the variance are both equal to n:
Mean, μ = n
Variance, σ² = n (which implies the standard deviation is σ = √n)

According to the Central Limit Theorem (CLT), as the parameter n approaches infinity, the standardized version of the Poisson random variable converges to the standard normal distribution Z ~ N(0, 1):

$Z=\frac{X-\mu }{\sigma }=\frac{X-n}{\sqrt{n}}$

Now, we rewrite the probability inequality in its standardized form:

$P(X\le n)=P\left(\frac{X-n}{\sqrt{n}}\le \frac{n-n}{\sqrt{n}}\right)=P\left(Z\le 0\right)$

Taking the limit as n approaches infinity, the probability converges to the probability of the standard normal distribution being less than or equal to 0:

$L=\underset{n\to \infty }{lim}P(X\le n)=P(Z\le 0)$

Since the standard normal distribution curve is perfectly symmetric about 0, the probability of being less than or equal to 0 is exactly half of the total area under the curve:

$P(Z\le 0)=0.5$

Thus, the value of the limit L is 0.5.