Random Variables Basics MCQ Quiz in मल्याळम - Objective Question with Answer for Random Variables Basics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

$E\left[X^2\right]=\sum x_i^{2}P\left(x_i\right)$

Calculation:

Given: xi = i2, where i = 1, 2, 3, 4.

$E\left[X^2\right]=\sum x_i^{2}P\left(x_i\right)$

= (1)2 × 0.2 + (4)2 × 0.2 + (9)2 × 0.3 + (16)2 × 0.3

= 0.2 + 16 × 0.2 + 81 × 0.3 + (16)2 × 0.3

= 0.2 + 3.2 + 24.3 + 76.8

= 104.5

Concept:

$Var[aX+b]=a^{2}Var(X)$

$V(ax)=a^{2}V(x)$ where a is constant,

$V(a)=0$, i.e. variance of constant is zero ; where a is constant

Calculation:

Given:

$Var(x)=1$, $Var(3x+4)=?$

Let $y=3x+4$

$V(y)=V(3x+4)$

$V(y)=3^2[V(x)]$

$V(y)=9[1]$

$V(y)=V(3x+4)=9$.

For any density function,

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

$\Rightarrow \underset{0}{\overset{3}{\int }}kxdx=1$

$\Rightarrow k=\frac{2}{9}$

P(|X − 1.8| < 0.6) = P(1.2 < X < 2.4)

$=\underset{1.2}{\overset{2.4}{\int }}\frac{2}{9}xdx=\frac{1}{9}\left({2.4}^2-{1.2}^2\right)=0.48$

Concept:

A Gaussian random variable X with mean μ and variance σ² can be denoted as X N (μ, σ²) and its probability density function (pdf) is given by:

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

The expectation of any function of x is calculated as:

$E\left[g\left(x\right)\right]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g\left(x\right)\cdot f_x\left(x\right)dx$

If f(-x) = f(x) then f(x) is said to be even function of x.

And, an even function satisfies:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx=2\underset{0}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)d\tau$

Calculation:

Given:

X ∼ N (0, 1)

$f_X\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}$

$E\left[\left|X\right|\right]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left|x\right|f_X\left(x\right)dx$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left|x\right|\cdot \frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}dx$

$=\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left|x\right|\cdot e^{-x^2/2}dx$

Since:

$\left|x\right|\cdot e^{-x^2/2}$ is an even function of ‘x’, we can write:

$E\left[\left|x\right|\right]=\frac{1}{\sqrt{2\pi }}\times 2\underset{0}{\overset{\mathrm{\infty }}{\int }}x\cdot e^{-x^2/2}dx$

Using substitution of variables:

$Let\frac{x^2}{2}=y$

$xdx=dy$

$E\left[\left|x\right|\right]=\frac{1}{\sqrt{2\pi }}\times 2\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-y}dy$

$=\sqrt{\frac{2}{\pi }}\{-e^{-y}{|}_0^{\mathrm{\infty }}=\sqrt{\frac{2}{\pi }}\left\{1-0\right\}$

$E\left[\left|x\right|\right]=\sqrt{\frac{2}{\pi }}\simeq 0.798$

E[|x|] ≃ 0.8

$\left(\sum _{i=1}^n{x}_i{p}_i\right)=0.2+0.8+0.9+0.4=2.3$

$\left(\sum _{i=1}^n{x}_i^2{p}_i\right)=0.2+1.6+2.7+1.6=6.1$

$var\left(X\right)={\sigma }^2=\sum _{i=1}^n{x}_i^2{p}_i-{\left(\sum _{i=1}^n{x}_i{p}_i\right)}^2$

var(X) = σ² = 6.1 – 5.29

var(X) = σ² = 0.81

Given

PMF(x) = 1/2^x, x = 1, 2, 3 -----

Formula

P(X > 4) = 1 – P(X ≤ 4)

Calculation

P(X > 4) = 1 – P(X ≤ 4)

⇒ 1 – [P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]

⇒ 1 – [1/2¹ + ½²  + ½³ + ½⁴]

⇒ 1 – [1/2 + ¼ + 1/8 + 1/16]

⇒ 1 – [(8 + 4 + 2 + 1)/16]

⇒ 1 – [15/16]

⇒ (16 – 15)/16

∴ The value of P(X > 4) is 1/16

Probability mass function or  PMF is defined as the set of ordered pair [x, f(x)] if for each possible outcome x, f(x) satisfies the following condition

F(x) ≥ 0

∑ f(x) = 1

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)

Concept:

If f(x) is probability density function for random variable x then,

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)dx=1$

Calculation:

We are given a probability density function

$f\left(x\right)=K\cdot \frac{1}{1+x^2}forx\in \left(-\mathrm{\infty },\mathrm{\infty }\right)$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{K}{1+x^2}dx$

$=K{\left[{\tan }^{-1}x\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=1$

$=K\left[\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)\right]=1$

⇒ K(π) = 1