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GATE 2026 DA – Question 45

Probability and Statistics · Central limit theorem, confidence intervals and hypothesis tests · 2 marks · Multiple choice

Let

$$L = \lim_{n \to \infty}\sum_{k=0}^{n} e^{-n}\frac{n^k}{k!}$$

Which of the following is the value of $L$?

  1. 0.5
  2. 1.0
  3. 0
  4. $e^{-1}$

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Correct answer: (A) 0.5

Explanation

The sum is $P(N_n \le n)$, where $N_n$ has a Poisson distribution with mean $n$, which is also its variance. $N_n$ is the sum of $n$ independent Poisson(1) variables, so by the central limit theorem $\frac{N_n - n}{\sqrt{n}}$ tends to a standard normal. Then $P(N_n \le n) = P\left(\frac{N_n - n}{\sqrt{n}} \le 0\right) \to \Phi(0) = 0.5$.