Question

Which one of the following probability mass function is impossible?

• A. A Poisson distribution with mean 16 and standard deviation 4
• B. A binomial distribution with mean 16 and standard deviation 4
• C. A binomial distribution with mean 18 and variance 6
• D. A Poisson distribution with mean 5 and variance 5

Hint:

For any non-degenerate Binomial distribution, the variance is always strictly less than the mean. If a statement claims they are equal, it is statistically impossible.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Every standard probability distribution has defined relationships between its parameters, such as mean, variance, and standard deviation.

Step 2: Detailed Explanation:
Let us analyze the validity of each option:
- Option (A): For a Poisson distribution, $\text{Mean} = \text{Variance} = \lambda$. Here, $\lambda = 16 \implies \text{SD} = \sqrt{16} = 4$. This is possible.
- Option (B): For a Binomial distribution, the parameters are $n$ (trials) and $p$ (probability of success, where $0 < p < 1$).
The mean is $\mu = np = 16$.
The standard deviation is $\sigma = \sqrt{np(1-p)} = 4 \implies \text{Variance} = np(1-p) = 16$.
Equating the mean and variance:
\[ np(1-p) = np \implies 16(1-p) = 16 \implies 1-p = 1 \implies p = 0 \]
But if $p = 0$, the mean $np$ must be 0, which contradicts the given mean of 16.
More generally, for any Binomial distribution with $p > 0$, the variance $np(1-p)$ must be strictly less than the mean $np$.
Therefore, a Binomial distribution with mean 16 and variance 16 is impossible.
- Option (C): $\text{Mean} = 18$, $\text{Variance} = 6$. Here, $\text{Variance} < \text{Mean}$, which is possible for a binomial distribution.
- Option (D): For a Poisson distribution, $\text{Mean} = \text{Variance} = 5$. This is possible.

Step 3: Final Answer
The correct option is (B).