Probability Distribution Function MCQ Quiz - Objective Question with Answer for Probability Distribution Function - Download Free PDF

Concept:

Let PX(x) = P(X = x) be the probability mass function of a random variable X then 

0 ≤ P_X(x) ≤ 1 and ∑ PX(x) = 1 

Explanation:

$\mathrm{P}(\mathrm{X}=\mathrm{x})=\{\begin{array}{ccc}k,& \mathrm{i}\mathrm{f}& x=0\\ 2k,& \mathrm{i}\mathrm{f}& x=1\\ 3k,& \mathrm{i}\mathrm{f}& x=2\\ 0,& & \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$ is the probability mass function of random variable X then

k + 2k + 3k = 1

⇒ 6k = 1 ⇒ k = 1/6

Now, P(X < 2) = P(X = 0) + P(X = 1) = k + 2k = 3k = 3 × $\frac{1}{6}$ = 1/2 

Option (4) is true.

Concept: 

To analyze the Random Variable ‘x’ two functions are used.

1) PDF (Probability Distribution Function)

2) pdf (Probability Density Function)

CDF and pdf are related as:

$CDF=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}PDFdx$  ----(1)

Properties of a valid PDF:

1) $f_X\left(x\right)\ge 0,\mathrm{\forall }xϵR$

∴ CDF will be bounded between 0 and 1

2) $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_X\left(x\right)dx=P_X\left(\mathrm{\infty }\right)=1$      ----(2)

Here PX is CDF

$P_X\left(\mathrm{\infty }\right)=\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ x\to \mathrm{\infty }\end{array}{P}_X\left(x\right)$

∴ CDF will always be a monotonically increasing function as the probability is always greater than or equal to 0. 

Calculation:

Given:

$\mathrm{P}\mathrm{D}\mathrm{F}=\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{cc}\frac{c}{\sqrt{x}},& 0<x<4\\ 0,& otherwise\end{array}$

Using Equation  (2):

$\underset{0}{\overset{4}{\int }}\frac{c}{\sqrt{x}}dx=1$

$[2c\sqrt{x}{]}_0^4=1$

4c = 1

$c=\frac{1}{4}$

Hence option (3) is the correct answer.

Formula

In poisson distribution

P(X = x) = e^-λλ^x/x! ; x = 0, 1, 2, 3, ------ λ > 0

In Binomial distribution

P(x) = ⁿc_xp^xq^{λ - x}

Calculation

P(xy = 0) = e^-2 × 2⁰/0! + ⁵c₀(0.4)⁵ × (0.6)⁰

⇒  e-2 × 20/0! + 5c0(0.4)⁵

⇒ e^-2 + (0.4)⁵ - e^-2 × (0.4)⁵

∴ The P(xy = 0) equals to e-2 + (0.4)5 - e-2 × (0.4)5

Given

m = 4

Formula

The probability distribution is given by 

p(x) = (e^-m × m^x)/x!

Calculation

According to the formula

⇒ p(x0 = (e^-4 × 4^x)/x!

Now the probability that in a given month there will be less than 4 acc idents

⇒ P(0) + P(1) + P(2) + P(3)

⇒ $\sum _{x=0}^{x=3}$e^-4 4^x/x! = e^-4[1 + 4 + 8 + 32/3]

⇒ 0.0183 × [13 + 32/3

⇒ 0.0183 × 71/3

∴ The probability that in a given month there will be less than 4 accidents is 0.4331

Formula

Variance X = σ²_x = E[(X – μ)²] = E(X²) – [E(X)]²

Calculation

E(X) = μ_x = μ

⇒ μ = ∑ x_if_i = ∑x_iP(X = x_j)

⇒ 1 × 1/6 + 2 × 1/6 + 3 × 1/6 + 4 × 1/6 + 5 × 1/6 + 6 × 1/6

⇒ (1/6)(1 + 2 + 3 + 4 + 5 + 6)

⇒ 7/2

E(X²) = ∑x_j²f(x_j) = ∑x_j²P(X = x_j)

⇒ 1² × 1/6 + 2² × 1/6 + ------6² × 1/6

⇒ 1/6(1² + 2² + 3² + 4² + 5² + 6²)

⇒ 1/6(1 +4 + 9 + 16 + 25 + 36)

⇒ 91/6

Variance X = E(X²) – [E(X)]²

⇒ 91/6 – (7/2)²

⇒ 91/6 – 49/4

∴ Variance X is 35/12

Given

In Poisson distribution

P(X = 0) = 0.6

Formula

Poisson distribution is given by

f(x) = e^-λλ^x/x!

Calculation

P(X = 0) = e^-λλ⁰/0!

⇒ 0.6 = e^-λ

⇒ 1/e^λ = 6/10 = 3/5

⇒ e^λ = 5/3

Taking log on both side

⇒ log_ee^λ = log_e(5/3)

∴ λ = Log_e(5/3)

Given

P(3) = P(4)

Concept

In poisson distribution = it is also a discrete distribution. Poisson distribution fulfil the limitation of nomal distribution

Poisson distribution is given by f(x) = (e^-λ × λ^x)/x!

Where x = 0, 1, 2, 3---- and λ = mean

Calculation

Here x = 3 and 4

⇒ (e^-λ × λ³)/3! = (e^-λ × λ⁴)/4!

⇒ λ  = 4!/3!

⇒ λ = 4

∴ mean of x in poisson distribution is 4

Given

X and Y are independent normal variables with

Mean (X) = 50

Mean (Y) = 80

SD (σ_x) = 4

SD(σ_y) = 3

Concept

Normal distribution is indicated by N(μ, √(σ²_x+ σ²₂))

Calculation

Since X and Y are independent then

⇒ Z = X + Y ~ N(X + Y, (σ²_x + σ²_y)

⇒ P(Z) = P(X + Y)  ~ N(130, (4² + 3²)

⇒ N(130, 25)

∴ N(μ, √(σ²_x+ σ²₂) is N(130, 5)

Given

Mean of poisson distribution = 9

Calculation

In poisson distribution mean is equal to variance

∴ Variance is 9

Formula

Variance X = σ²_x = E[(X – μ)²] = E(X²) – [E(X)]²

Calculation

E(X) = μ_x = μ

⇒ μ = ∑ x_if_i = ∑x_iP(X = x_j)

⇒ 1 × 1/6 + 2 × 1/6 + 3 × 1/6 + 4 × 1/6 + 5 × 1/6 + 6 × 1/6

⇒ (1/6)(1 + 2 + 3 + 4 + 5 + 6)

⇒ 7/2

E(X²) = ∑x_j²f(x_j) = ∑x_j²P(X = x_j)

⇒ 1² × 1/6 + 2² × 1/6 + ------6² × 1/6

⇒ 1/6(1² + 2² + 3² + 4² + 5² + 6²)

⇒ 1/6(1 +4 + 9 + 16 + 25 + 36)

⇒ 91/6

Variance X = E(X²) – [E(X)]²

⇒ 91/6 – (7/2)²

⇒ 91/6 – 49/4

∴ Variance X is 35/12

Formula

The probability distribution function o Binomial distribution is give asfollows

F(x) = ⁿc_xp^xq^{n – x}

The factorial moment = ⁿc_rp^r!r

X = 0, 1, 2 ---n

p = Probability of success event

q = 1 – p = Probability of failure of event.

Calculation

According to the formula of factorial moment of binomial distribution

⇒ E[(X)_r] =  ⁵c₃(0.4)³!3

⇒ [5!/(3!(5 – 3)!](0.4 × 0.4 × 0.4)!3

⇒ [5!/3! × 2!](0.064(3)

∴ The value of E[(X)_r] is 3.84

Given

n = 6

p = 1/4

Formula

S_k = (1 - 2p)/√(np(1 - p)

Calculation

Skewness of X = (1 - 2 × 1/4)/√(6 × 1/4(1 - 1/4)

⇒ (1 - 1/2(/√(3/2 × 3/4

⇒ 1/2/√(9/8)

∴ The value of skewness X is√2/3

Binomial distribution = A random variable X has a binomial distribution if its probability distribution is given by

b(x; n, p) = P(X = x) = (n x)p^xq^{n - x} x = 0, 1, 2, ----n

Normal distribution has the probability density function given by

f(x) = [1/(σ√2π)]e^{1/2(x - μ/σ)2}

Here, μ and σ are prameters of this distribution. The parameter μ is mean and σ is square root of variance of this distribution

-∞ < μ < ∞, σ > 0

The normal distribution is also called as the Gaussian distribution

Calculation

Normal distributiln = A normal distribution comes with a perfectly symmetrical shape that means distribution curve can divided in the middle to produce two equal halves.

The normal distribution function is denotted by

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}}{\left(\frac{x-\mu }{\sigma }\right)}^2$

f(x) = Probability density function

σ = Standard deviation

μ = Mean

In normal distribution the symmetry about the center and in it the mean = medan = mode = μ

∴ The mean = medan = mode = μ in normal distribution

Given that μ = 14, σ = 2.5

P(x = 16) = P(15.5 < × < 16.5)

$=P\left(\frac{15.5-u}{\sigma }<\frac{x-u}{\sigma }<\frac{16.5-u}{2.5}\right)$

$=P\left(\frac{15.5-u}{2.5}<z<\frac{16.5-u}{2.5}\right)$

= P(0.6 < z < 1)

= Area of OABCDE – Area of OADE

= 0.3413 – 0.2257 = 0.1156

% age of students = 100 × 0.1156

Explanation

The probability distribution function of Normal distribution us was given as

⇒ F(x) =[1/σ√2π ]e^{-(x – μ)2/2σ2}

If we put Z = (x – μ)/σ in Normal distribution then it is called a standard normal distribution

The probability distribution function of the standard normal distribution is as follows

⇒ f(x) = [1/σ√2π]e^{- z^2/2}

The value of E(X³) in a standard normal distribution is 0

The standard deviation of normal distribution  = σ = 1

The coefficient of skewness of Normal distribution is β₁ = 0