Question

For a Binomial random variable \(X\), \(\mathrm{E}(X)\) and \(\mathrm{Var}(X)\) are the expectation and variance, respectively. Which one of the following statements CANNOT be true?

• A. \(\mathrm{E}(X)=20\) and \(\mathrm{Var}(X)=16\)
• B. \(\mathrm{E}(X)=6\) and \(\mathrm{Var}(X)=5.4\)
• C. \(\mathrm{E}(X)=10\) and \(\mathrm{Var}(X)=15\)
• D. \(\mathrm{E}(X)=64\) and \(\mathrm{Var}(X)=12.8\)

Hint:

For a binomial distribution, \(\mathrm{Var}(X)=\mathrm{E}(X)(1-p)\le \mathrm{E}(X)\). Any proposed pair with variance exceeding the mean is impossible.

Answer 1

The Correct Option is C

Step 1: For \(X\sim \text{Bin}(n,p)\), \(\mathrm{E}(X)=np\) and \(\mathrm{Var}(X)=np(1-p)=\mathrm{E}(X)\,(1-p)\). Hence \(\mathrm{Var}(X)\le \mathrm{E}(X)\).
Step 2: Option (C) violates this since \(15>10\). Thus it cannot be true.
Step 3: The others are feasible by taking \(p=0.2,0.1,0.8\) respectively with suitable integer \(n\).
Therefore, (C) is the only impossible pair.