Question:medium

If $ A_{x:n} = \{k : |k - \tfrac{n}{2}| \le \tfrac{x\sqrt{n}}{2} \} $, then the value of

$ \lim_{n \to \infty} \sum_{k \in A_{1:n}} \binom{n}{k} 2^{-n} $

(where $ \Phi(\cdot) $ is the distribution function of the standard normal variate) is:

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The De Moivre-Laplace theorem is a special case of CLT. For $p=0.5$, the interval $[\mu - \sigma, \mu + \sigma]$ always contains approximately $68\%$ of the distribution in the limit.

Updated On: Jun 8, 2026

$\Phi(-1)$
$\Phi(1)$
$1-2\Phi(-1)$
$\frac{\Phi(1)}{2}$

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The Correct Option is C

Solution and Explanation

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If \( A_{x:n} = \{k : |k - \tfrac{n}{2}| \le \tfrac{x\sqrt{n}}{2} \} \), then the value of

\( \lim_{n \to \infty} \sum_{k \in A_{1:n}} \binom{n}{k} 2^{-n} \)

(where \( \Phi(\cdot) \) is the distribution function of the standard normal variate) is:

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Standard Distributions

Questions Asked in CUET (PG) exam

If \( A_{x:n} = \{k : |k - \tfrac{n}{2}| \le \tfrac{x\sqrt{n}}{2} \} \), then the value of \( \lim_{n \to \infty} \sum_{k \in A_{1:n}} \binom{n}{k} 2^{-n} \) (where \( \Phi(\cdot) \) is the distribution function of the standard normal variate) is:

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Standard Distributions

Questions Asked in CUET (PG) exam

If \( A_{x:n} = \{k : |k - \tfrac{n}{2}| \le \tfrac{x\sqrt{n}}{2} \} \), then the value of

\( \lim_{n \to \infty} \sum_{k \in A_{1:n}} \binom{n}{k} 2^{-n} \)

(where \( \Phi(\cdot) \) is the distribution function of the standard normal variate) is: