Question

In a binomial distribution, \( n = 200 \) and \( p = 0.04 \). Taking Poisson distribution as an approximation to the binomial distribution:
Assertion (A): Mean of Poisson distribution = 8.
Reason (R): \( P(X = 4) = \frac{512}{3e^8} \).

Hint:

The Poisson distribution is a good approximation for a binomial distribution when \( n \) is large and \( p \) is small (\( np<10 \)). The mean of the Poisson distribution is given by \( \lambda = np \).

Answer 1

Step 1: For a binomial distribution, the mean is given by: \[ \mu = np = 200 \times 0.04 = 8. \] Since Poisson approximation is used, the mean of the Poisson distribution is also 8. Thus, Assertion (A) is true. 
Step 2: The probability mass function of a Poisson distribution is: \[ P(X = k) = \frac{e^{-\mu} \cdot \mu^k}{k!}. \] Substituting \( \mu = 8 \) and \( k = 4 \): \[ P(X = 4) = \frac{e^{-8} \cdot 8^4}{4!} = \frac{512}{3e^8}. \] Since the given expression matches this calculation, Reason (R) is also true. 
Step 3: However, Reason (R) does not directly explain why the mean of the Poisson distribution is 8. The mean of a Poisson distribution is derived from the binomial approximation (\( \lambda = np \)), not from the probability calculation of \( P(X = 4) \). Thus, Assertion (A) and Reason (R) are both true, but Reason (R) is not the correct explanation of Assertion (A).