Question

In the binomial distribution \(B(n,p)\), the variance \((\sigma^{2})\) and the mean \((\mu)\) are related by

• A. \(\mu=\sigma^{2}\)
• B. \(\mu=\sigma^{2}p^{2}\)
• C. \(\sigma^{2}=\dfrac{\mu}{p}\)
• D. \(\mu=\sigma^{2}+\mu p\)

Hint:

Always remember the standard formulas for a binomial distribution: \[ \boxed{ \mu=np,\qquad \sigma^{2}=npq=np(1-p). } \] From these, \[ \boxed{ \sigma^{2}=\mu(1-p) } \] and therefore, \[ \boxed{ \mu=\sigma^{2}+\mu p. } \] These identities are frequently asked in university examinations and competitive exams.

Answer 1

The Correct Option is D

Concept: A random variable \(X\) is said to follow a binomial distribution if \[ X\sim B(n,p), \] where
• \(n\) = number of independent trials,
• \(p\) = probability of success in each trial,
• \(q=1-p\) = probability of failure. The two most important characteristics of a binomial distribution are: \[ \boxed{\mu=np} \] and \[ \boxed{\sigma^{2}=npq=np(1-p).} \] Using these standard formulas, we can derive the required relation between the mean and the variance.

Step 1: Write the formulas for mean and variance.
For the binomial distribution, \[ \mu=np. \] Also, \[ \sigma^{2}=np(1-p). \] Since \[ np=\mu, \] the variance becomes \[ \sigma^{2} = \mu(1-p). \]

Step 2: Expand the expression for variance.
Expanding, \[ \sigma^{2} = \mu-\mu p. \] Rearranging the terms, \[ \mu = \sigma^{2} + \mu p. \]

Step 3: Compare with the given options.
The obtained relation is \[ \boxed{ \mu=\sigma^{2}+\mu p. } \] This matches option (D).

Step 4: Write the final answer.
Hence, \[ \boxed{\mu=\sigma^{2}+\mu p.} \] Therefore, the correct option is \[ \boxed{\text{(D)}}. \]