MHT CET · Maths · Probability
A random variable X has p.m. f. \(\mathrm{P}(\mathrm{X}=x)=\frac{{ }^4 \mathrm{C}_x}{2^4}, x=0,1,2,3,4\) and \(\mu\) and \(\sigma^2\) are mean and variance respectively of random variable \(X\), then
- A \(\mu=2, \quad \sigma^2=4\)
- B \(\mu=2, \quad \sigma^2=1\)
- C \(\mu=3, \quad \sigma^2=4\)
- D \(\mu=2, \quad \sigma^2=5\)
Answer & Solution
Correct Answer
(B) \(\mu=2, \quad \sigma^2=1\)
Step-by-step Solution
Detailed explanation
The given p.m.f. is of a Binomial Distribution \(B(n,p)\). \(P(X=x)=\frac{{ }^4 C_x}{2^4} = { }^4 C_x \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{4-x}\)
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