Mean and Variance of Binomial Distribution MCQ Quiz in தமிழ் - Objective Question with Answer for Mean and Variance of Binomial Distribution - இலவச PDF ஐப் பதிவிறக்கவும்

Last updated on Mar 22, 2025

பெறு Mean and Variance of Binomial Distribution பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Mean and Variance of Binomial Distribution MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Mean and Variance of Binomial Distribution MCQ Objective Questions

Top Mean and Variance of Binomial Distribution MCQ Objective Questions

Mean and Variance of Binomial Distribution Question 1:

If X has Binomial distribution with parameters n and p such that np =λ, then \(\mathop {\lim }\limits_{n \to \infty } b\left( {x,n,p} \right);x = 0,1,2,.....\) is equal to:

  1. \({\frac{{{e^{^{ - \lambda }}}\lambda }}{{x!}}^x}\), x = 0, 1, 2, … 
  2. Limit does not exist
  3. 0
  4. 1

Answer (Detailed Solution Below)

Option 1 : \({\frac{{{e^{^{ - \lambda }}}\lambda }}{{x!}}^x}\), x = 0, 1, 2, … 

Mean and Variance of Binomial Distribution Question 1 Detailed Solution

Explanation

Poisson distributionis a limiting case of binomial distribution if it follows conditions

n, the number trials is indefinitely large  that means n tends to infinite

p, the constant probability of success for each trial is indefinitely small p tends to 0

np = λ , is finite so λ/n = p, q = 1 – p

⇒ (1 – λ/n), λ is positive integer

The probability of x successes in a series of n independent trials is

⇒ b(x, n, p) = (n/x)pxqn – x, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)px(1 – p)n – x

∴ (n/x)(p/(1 – p)]x(1 – p)n - x

p , the constant probability of success for each trial is indefinitely small p tends to 0

np = λ , is finite so λ/n = p, q = 1 – p

⇒ (1 – λ/n), λ is positive integer

The probability of x successes in a series of n independent trials is

⇒ b(x, n, p) = (n/x)pxqn – x, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)px(1 – p)n – x

∴ (n/x)(p/(1 – p)]x(1 – p)n - x

 [n(n - 1)(n - 2)------(n - x + 1)/x!] × (λ/n)x/(1 - λ /n)x[1 - λ/n]n

⇒ [(1 - 1/n)(1 - 2/n)-----( 1 - (x - 1)/n/x!(1 - λ/n)x] × λx[1 - λ/n]n

⇒ Lim x → ∞ b(x, n, p) = e-λ × λx/x! ; x = 0, 1, 2, 3, 4 -------,n

Poisson distribution = A random variables X is said to follow poisson distribution if it assumes only non - negative values and its proportionality mass function i s given

by P)X = x) = e-λ × λx/x! where x = 0, 1, 2, 3 ------n and  λ > 0

⇒ p(x, λ) = ∑P(X - x)

⇒ e-λ∑λx/x!

⇒ e× e = 1

∴ The corresponding distribution function is F(x) = P(X = x) = ∑P(r) =  e ∑λ2/r!; x = 0, 1, 2 .......

Get Free Access Now
Hot Links: rummy teen patti teen patti master gold download teen patti gold download