GATE ME 2021 Set 1 — Question 15
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Engineering Mathematics → Statistics & Distributions → Binomial Distribution
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Question
Consider a binomial random variable . If are independent and identically distributed samples from the distribution of with sum , then the distribution of as can be approximated as
Correct answer
(D) Normal
Solution
To find the limiting distribution of the sum as , we apply the Central Limit Theorem (CLT).
Each follows a binomial distribution, , with a finite mean and a finite, non-zero variance :
$\mu$ = E[X_i] = mp $\sigma$^2 = \text{Var}(X_i) = mp(1-p)
E[Y] = E\left[\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} E[X_i] = n$\mu$ = nmp
Since the samples are independent, the variance of is:
\text{Var}(Y) = \text{Var}\left(\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} \text{Var}(X_i) = n$\sigma$^2 = nmp(1-p)
Z = \frac{Y - n$\mu$}{$\sigma$ \sqrt{n}} converges to the standard normal distribution . Thus, the distribution of can be approximated as a Normal distribution:
Y $\sim$ N\left(nmp, \, nmp(1-p)\right) \quad \text{as } n \to $\infty$Correct Option: D
Step-by-Step Derivation
1.Identify the Properties of the Samples:
We are given that are independent and identically distributed (i.i.d.) random variables. Each follows a binomial distribution, , with a finite mean and a finite, non-zero variance :
$\mu$ = E[X_i] = mp $\sigma$^2 = \text{Var}(X_i) = mp(1-p)
2.Apply the Central Limit Theorem (CLT):
The Central Limit Theorem states that if are i.i.d. random variables with a finite mean and finite variance , then the distribution of their sum approaches a normal distribution as .3.Determine the Parameters of the Limiting Distribution:
The mean of is:E[Y] = E\left[\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} E[X_i] = n$\mu$ = nmp
Since the samples are independent, the variance of is:
\text{Var}(Y) = \text{Var}\left(\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} \text{Var}(X_i) = n$\sigma$^2 = nmp(1-p)
4.Conclusion:
As , the cumulative distribution function of the standardized variable Z = \frac{Y - n$\mu$}{$\sigma$ \sqrt{n}} converges to the standard normal distribution . Thus, the distribution of can be approximated as a Normal distribution:
Y $\sim$ N\left(nmp, \, nmp(1-p)\right) \quad \text{as } n \to $\infty$Correct Option: D
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