Which one of the following is the value of 'n' in a binomial distribution for which the mean is 4 and variance is 3. a) 4 b) 12 c) 16 d) 20 GROUP: B Short answer questions 8 × 5 = 40

For a binomial distribution, the mean $\mu$ and variance ${\sigma }^2$ are given by: $$\mu =np\text{and}{\sigma }^2=np(1-p)$$ Given: $$\mu =4,{\sigma }^2=3$$

Step 1: Express $p$ in terms of $n$ using the mean: $$p=\frac{\mu }{n}=\frac{4}{n}$$

Step 2: Substitute $p$ into the variance formula: $$3=n\left(\frac{4}{n}\right)\left(1-\frac{4}{n}\right)$$ $$3=4\left(1-\frac{4}{n}\right)$$

Step 3: Solve for $n$: $$3=4-\frac{16}{n}$$ $$\frac{16}{n}=4-3=1$$ $$n=16$$

Answer: c) 16