Mean and Variance of Binomial Distribution MCQ Quiz in मल्याळम - Objective Question with Answer for Mean and Variance of Binomial Distribution - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Mean and Variance of Binomial Distribution ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Mean and Variance of Binomial Distribution MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

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 .......

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