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

Concept:

Binomial Distribution:

• A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times.
• The binomial is a type of distribution that has two possible outcomes (the prefix “bi” means two, or twice).
• For example, a coin toss has only two possible outcomes: heads or tails, and taking a test could have two possible outcomes: pass or fail.

Key PointsThe binomial distribution has the following properties:

• The mean of the distribution (μ) is equal to np.
• The variance (σ²) is npq.

where, n: The number of trials in the binomial experiment.

p: The probability of success on an individual trial.

q: The probability of failure on an individual trial.

(This is equal to q = 1 - p)

Calculation:

The variance of a binomial distribution is given by:

σ2 = npq

∴ Standard deviation, σ = 

∴ Standard deviation of a binomial distribution is given by, σ = 

The correct answer is Option 3

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)p^xq^{n – x}, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)p^x(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)p^xq^{n – x}, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)p^x(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] × λ^x[1 - λ/n]ⁿ

⇒ Lim x → ∞ b(x, n, p) = e^-λ × λ^x/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∑λ^x/x!

⇒ e^-λ× e^-λ = 1

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