Discrete Random Variable MCQ Quiz - Objective Question with Answer for Discrete Random Variable - Download Free PDF

Concept:

• The expected value of a random variable X follows linearity i.e.,
• E(a X + b Y) = a E(X) + b E(Y)
• The expectation and standard deviation formulas,

$E(X)=\sum x_{i}f(x_i)$

$Variance=\sum _{i=1}^4{p}_i(x_i-E(x){)}^2$

S.D = √Variance 

Mean,

$E(X)=\sum _{i=1}^4{p}_i{x}_i$ = 0.1 × 1 + 0.2  × 2 + 0.3 × 3 + 0.4  × 4 = 3

⇒ V = o.1(1 -3)2 + 0.2(2 -3)2+ 0.3(3 -3)2 + 0.4(4 -3)2

⇒ V = 0.4 + 0.2 + 0.4 = 1

⇒  S.D = √1 = 1

Given

PMF = (1 - p)^{k - 1}p

Concept used 

Mode is the value of x for which f(x) is the maximum. IIt can be obtained by equating the first derivative of f(x) to zero and checking that the second derivative of f(x)  is negative I;e

F’(x) = 0 and f’’(x) < o

Calculation

This PMF is in geometric expression and the mode of geometric distribution is 1

∴ Mode of X is 1

Explanation:

We know that for Fpr random variable X sum of Probaboility = 1

⇒ ∑P_i = 1

⇒ k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k = 1

⇒ 26k = 1

⇒ k = 1/26

∴ xf(x) = 103k = 103 × 1/26

∴ E(x) is 103/26

Concept:

For a continuous random variable, f(x) is called probability density function if it satisfies-

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

Relation between Probability density function f(x) and Probability distribution function is:

$f\left(x\right)=\frac{d}{dx}\left\{F\left(x\right)\right\}$

Calculation:

Here Probability distribution function F(x) is given 

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

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{3}{\int }}f\left(x\right)dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx=1$

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx+\underset{1}{\overset{3}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx=1$

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}\frac{d}{dx}\left\{0\right\}dx+\underset{1}{\overset{3}{\int }}\frac{d}{dx}\left\{k{\left(x-1\right)}^4\right\}dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}\frac{d}{dx}\left\{1\right\}dx=1$

$\Rightarrow 0+\underset{1}{\overset{3}{\int }}4k{\left(x-1\right)}^{3}dx+0=1$

$\Rightarrow 4\frac{k}{4}{\left[{\left(x-1\right)}^4\right]}_1^3=1$

$\Rightarrow k\left[{\left(3-1\right)}^4\right]=1$

$\Rightarrow k=\frac{1}{16}$

Derivative of g_x (z) at z = 1 gives the expectation E (X).

When g_y(z) is expanded, it results in a binomial distribution form and mean of a binomial distribution is in the form of N *p.

$g_x\left(z\right)=\sum _{j=0}^N{p}_j{z}^j$

$g_x^{\prime }\left(z\right)={\sum }_{j=1}^{N}j{p}_j{z}^{j-1}={\sum }_{j=1}^{N}j{p}_j=E\left(X\right)$ 

Similarly, take the derivate of g_y(z) at z= 1.

$E\left(Y\right)={{g^{\prime }}_y\left(z\right)|}_{z=1}={{\left({\left(1-\beta +\beta z\right)}^N\right)}^{\prime }|}_{z=1}=N\beta {{\left(1-\beta +\beta z\right)}^{N-1}|}_{z=1}=N\beta {\left(1-\beta +\beta z\right)}^{N-1}$

$E\left(Y\right)=N\beta$ 

Concept:

For a continuous random variable, f(x) is called probability density function if it satisfies-

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

Relation between Probability density function f(x) and Probability distribution function is:

$f\left(x\right)=\frac{d}{dx}\left\{F\left(x\right)\right\}$

Calculation:

Here Probability distribution function F(x) is given 

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

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{3}{\int }}f\left(x\right)dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx=1$

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx+\underset{1}{\overset{3}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx=1$

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}\frac{d}{dx}\left\{0\right\}dx+\underset{1}{\overset{3}{\int }}\frac{d}{dx}\left\{k{\left(x-1\right)}^4\right\}dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}\frac{d}{dx}\left\{1\right\}dx=1$

$\Rightarrow 0+\underset{1}{\overset{3}{\int }}4k{\left(x-1\right)}^{3}dx+0=1$

$\Rightarrow 4\frac{k}{4}{\left[{\left(x-1\right)}^4\right]}_1^3=1$

$\Rightarrow k\left[{\left(3-1\right)}^4\right]=1$

$\Rightarrow k=\frac{1}{16}$

Explanation:

We know that for Fpr random variable X sum of Probaboility = 1

⇒ ∑P_i = 1

⇒ k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k = 1

⇒ 26k = 1

⇒ k = 1/26

∴ xf(x) = 103k = 103 × 1/26

∴ E(x) is 103/26

Concept:

For a discrete probability density distribution mean value ‘or’ Average Value ‘or’ Expected Value is given as;

$\mathrm{E}\left[x\right]=\mu \left(Mean\right)=\sum _{i=1}^n{X}_{i}P\left(X_i\right)$

Calculation:

With, X = Discrete samples = Number of attempts that the passengers need to make to get a seat reserved.

The probability distribution is as shown:

(Here X = number of attempts)

Now, Average number of attempts that the passenger need to make to get a seat reserved is-

$E\left(X\right)=\mu =\sum _{i=1}^{\mathrm{\infty }}{X}_{i}P\left(X_i\right)$

= 1 × 0.4 + 2 × 0.6 × 0.4 + 3 × 0.6² × 0.4 + . . .

= 0.4 (1 + 2 × 0.6 + 3 × 0.6² + 4 × 0.6³ …)

= 0.4(1 – 0.6)^-2

= 0.4(0.4)^-2

$\Rightarrow 0.4\times \frac{1}{0.4}=\frac{1}{0.4}$

Given

PMF = (1 - p)^{k - 1}p

Concept used 

Mode is the value of x for which f(x) is the maximum. IIt can be obtained by equating the first derivative of f(x) to zero and checking that the second derivative of f(x)  is negative I;e

F’(x) = 0 and f’’(x) < o

Calculation

This PMF is in geometric expression and the mode of geometric distribution is 1

∴ Mode of X is 1

Concept:

Expectation of x is E(x) which is given by:

E(x) = ∑ x.p(x) (where p(x) is the probability of x)

Probability of ‘3’ as an outcome in the 1^st time (x=1) will be:

$P(x=1)=\frac{1}{6}$  

Probability of ‘3’ as an outcome in the 2^nd time (x = 2) will be:

$P(x=2)=\frac{5}{6}.\frac{1}{6}$ 

Similarly, we can write, the probability of ‘3’ as an outcome in the k^th time (x = k) as:

$P(x=k)={\left(\frac{5}{6}\right)}^{k-1}.\frac{1}{6}$ 

$Let,\begin{array}{c}\frac{5}{6}=x\\ \frac{1}{6}=y\end{array}\}$ 

$E[k]=?k.{(x)}^{(k-1)}.y=y{?}_{(k=1)}^k??k.{(x)}^{(k-1)}?$ 

E[k] = y (1 + 2x¹ + 3x² + 4x³ + …..)

E[k] = y (1-x)^-2 [Since (1-q)^-2 = 1+ 2q + 3q² + 4q³ + ….. ]

$E\left[k\right]=\left(\frac{1}{6}\right){(1-\frac{5}{6})}^{-2}[Sincex=\frac{5}{6}\mathrm{\&}y=\frac{1}{6}]$ 

$E\left[k\right]=\left(\frac{1}{6}\right){\left(\frac{1}{6}\right)}^{-2}=6$ 

Derivative of g_x (z) at z = 1 gives the expectation E (X).

When g_y(z) is expanded, it results in a binomial distribution form and mean of a binomial distribution is in the form of N *p.

$g_x\left(z\right)=\sum _{j=0}^N{p}_j{z}^j$

$g_x^{\prime }\left(z\right)={\sum }_{j=1}^{N}j{p}_j{z}^{j-1}={\sum }_{j=1}^{N}j{p}_j=E\left(X\right)$ 

Similarly, take the derivate of g_y(z) at z= 1.

$E\left(Y\right)={{g^{\prime }}_y\left(z\right)|}_{z=1}={{\left({\left(1-\beta +\beta z\right)}^N\right)}^{\prime }|}_{z=1}=N\beta {{\left(1-\beta +\beta z\right)}^{N-1}|}_{z=1}=N\beta {\left(1-\beta +\beta z\right)}^{N-1}$

$E\left(Y\right)=N\beta$ 

$E_1(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf(x)dx$

$\Rightarrow {\int }_{-a}^{0}xf(x)dx+{\int }_0^{a}af(x)dx$

$\Rightarrow {\int }_{-a}^{0}x(\frac{x}{a}+1)+{\int }_0^{a}x(-\frac{x}{a}+1)$

$\Rightarrow {\int }_{-a}^0\frac{x^2}{a}dx+{\int }_{-a}^{0}xds+{\int }_0^a\frac{-x^2}{a}dx+{\int }_0^{a}xdx$

⇒ 0

∴ $[{\int }_{-a}^0\frac{x^2}{a}=-{\int }_0^a-\frac{x^2}{a}]$

∴ $[{\int }_{-a}^{0}xdx=-{\int }_0^{a}xdx]$

$E_2(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xg(x)dx$

⇒ ${\int }_{-a}^{0}x(-\frac{x}{a})dx+{\int }_a^{0}x\left(\frac{x}{a}\right)dx$

⇒ ${\int }_{-a}^0-\frac{x^2}{a}dx+{\int }_a^0\frac{x^2}{a}dx$

⇒ 0

∴ $[{\int }_{-a}^0-\frac{x^2}{a}dx=-{\int }_a^0\frac{x^2}{a}dx]$

Variance

E(x²) - {E(x)}²

$E_1(x^2)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}f(x)$

⇒ ${\int }_{-a}^0{x}^2(\frac{x}{a}+1)dx+{\int }_0^a{x}^2(-\frac{x}{a}+1)dx$

$\Rightarrow {\int }_{-a}^0\frac{x^3}{a}dx+{\int }_{-a}^0{x}^{2}dx+{\int }_0^a-\frac{x^3}{a}dx+{\int }_0^a{x}^{2}dx$

$\Rightarrow -\frac{a^3}{4}+\frac{a^3}{3}-\frac{a^3}{4}+\frac{a^3}{3}=\frac{a^3}{6}$

$E_2(x^2)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}g(x)$

${\int }_{-a}^0{x}^2(-\frac{x}{a})dx+{\int }_0^a{x}^2\left(\frac{x}{a}\right)dx$

${\int }_{-a}^0\frac{-x^3}{a}dx+{\int }_0^a\frac{x^3}{a}dx$

${[-\frac{x^4}{4a}]}_{-a}^0+{\left[\frac{x^4}{4a}\right]}_0^a$

$\{0-\left[\frac{-(a{)}^4}{4a}\right]\}+\{\frac{a^4}{4a}-0\}$

$\frac{a^3}{4}+\frac{a^3}{4}=\frac{a^3}{2}$

Mean of f(x) is E(x):

$\begin{array}{l}=\underset{-a}{\overset{0}{\int }}X\left(\frac{X}{a}+1\right)dx+\underset{0}{\overset{a}{\int }}X\left(\frac{-X}{a}+1\right)dx\\ ={\left(\frac{X^3}{3a}+\frac{X^2}{2}\right)}_{-a}^0+{\left(\frac{-X^3}{3a}+\frac{X^3}{3}\right)}_0^a=0\end{array}$

$\begin{array}{l}=\underset{-a}{\overset{0}{\int }}{X}^2\left(\frac{X}{a}+1\right)dx+\underset{0}{\overset{a}{\int }}{X}^2\left(\frac{-X}{a}+1\right)dx\end{array}$

${\left(\frac{X^4}{4a}+\frac{X^3}{3}\right)}_{-a}^0+{\left(\frac{-X^4}{4a}+\frac{X^3}{3}\right)}_0^a=\frac{a^3}{6}$

⇒ Variance is $\frac{a^3}{6}$

Next, mean of g(x) is E(x)

$=\underset{-a}{\overset{0}{\int }}x\left(\frac{-x}{a}\right)dx+\underset{0}{\overset{a}{\int }}x\times \left(\frac{X}{a}\right)dx=0$

Variance of g(x) is E(x²) – {E(X)}², where

$E\left(X^2\right)=\underset{-a}{\overset{0}{\int }}{X}^2\left(\frac{-X}{a}\right)dX+\underset{0}{\overset{a}{\int }}{X}^2\left(\frac{X}{a}\right)dx=\frac{a^3}{2}$

⇒ Variance is $\frac{a^3}{2}$

∴ Mean of f(x) and g(x) are same but variance of f(x) and g(x) are different

Concept:

Binomial distribution

$P\left(X=r\right)=n_{c_r}{p}^r{q}^{n-r}$

Mean = np

Variance = npq

Standard deviation $=\sqrt{npq}$

Calculation:

Mean = np = 1

Variance = npq = 3/4

$\Rightarrow p=\frac{1}{4},q=\frac{3}{4},n=4$

$P\left(X=3\right)=4_{c_3}{\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^{4-3}$

$P(X=3)=4\times \frac{1}{64}\times \frac{3}{4}=\frac{3}{64}$

Concept:

For a continuous random variable, f(x) is called probability density function if it satisfies-

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

Relation between Probability density function f(x) and Probability distribution function is:

$f\left(x\right)=\frac{d}{dx}\left\{F\left(x\right)\right\}$

Calculation:

Here Probability distribution function F(x) is given 

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

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{3}{\int }}f\left(x\right)dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx=1$

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx+\underset{1}{\overset{3}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}\frac{d}{dx}\left\{F\left(x\right)\right\}dx=1$

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{1}{\int }}\frac{d}{dx}\left\{0\right\}dx+\underset{1}{\overset{3}{\int }}\frac{d}{dx}\left\{k{\left(x-1\right)}^4\right\}dx+\underset{3}{\overset{\mathrm{\infty }}{\int }}\frac{d}{dx}\left\{1\right\}dx=1$

$\Rightarrow 0+\underset{1}{\overset{3}{\int }}4k{\left(x-1\right)}^{3}dx+0=1$

$\Rightarrow 4\frac{k}{4}{\left[{\left(x-1\right)}^4\right]}_1^3=1$

$\Rightarrow k\left[{\left(3-1\right)}^4\right]=1$

$\Rightarrow k=\frac{1}{16}$

The variance cannot be negative, because it is an average of squared quantities.

$Var\left(X\right)\ge 0$

∴ -1 cannot be a variance of X.

Additional Information

 If X is a random variable with mean μ, then the variance of X, denoted by Var (X) is given by:

Var (X) = E[(X - μ)]2, where μ = E(X)

For a discrete random variable X, the variance of X is obtained as follows:

$Var(X)=\sum (x-\mu {)}^{2}pX(x)$