In a Binomial Distribution (BD) the mean is 15 and variance is 10, then parameter n (number of trials) is

  1. 28
  2. 16
  3. 45
  4. 25

Answer (Detailed Solution Below)

Option 3 : 45
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Detailed Solution

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Concept:

Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P( X = k) = nCk pk (1 - p)(n - k)

Where, n is number of observations, p is the probability of success & (1 - p) is probability of failure.

Properties:

  • Mean of the distribution (μX) = n × p
  • The variance (σ2x) = n × p × (1 - p)
  • Standard deviation (σx) = \(\sqrt{np(1-p)}\)

Calculation:

Given:

mean of BD = np = 15

And variance of BD = npq = 10

⇒ np(1 – p) = 10 (∵ p + q = 1)

\(\Rightarrow 1 - {\rm{p}}\frac{{10}}{{15}} = \frac{2}{3} \Rightarrow {\rm{P}} = 1 - \frac{2}{3}\)

\(\Rightarrow {\rm{p}} = \frac{1}{3}\)

\(\therefore {\rm{n}} = \frac{{15}}{{1/3}} = 15 \times 3 = 45\)
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