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

Concept:

The mean of a random variable also called the expected value is defined as:

\(E\left[ X \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } x.{f_x}\left( x \right)dx\)

The variance is defined as:

\(\sigma _x^2 = E\left[ {{{\left( {X - {\mu _x}} \right)}^2}} \right]\)

f_x(X) is the probability function.

\(E\left[ {{X^2}} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } {x^2}{f_x}\left( x \right)dx\)

Property:

\(Var\left[ {aX + b} \right] = E{\left[ {aX + b - \left( {a\mu + b} \right)} \right]^2}\)

\( = E{\left[ {aX - a\mu } \right]^2}\)

\( = E\left[ {{a^2}{{\left( {X - \mu } \right)}^2}} \right]\)

\( =a^2E\left[ {{}{{\left( {X - \mu } \right)}^2}} \right]\)

\(Var\left[ {aX + b} \right] = {a^2}\sigma _X^2\)   ---(1)

Analysis:

On comparing with Equation (1), we can write:

\(V\left[ {X + b} \right] = {\left( 1 \right)^2}.\sigma _x^2\)

\( = \sigma _x^2\)

Properties of mean:

1) E[K] = K, Where K is some constant

2) E[c X] = c. E[X], Where c is some constant

3) E[a X + b] = a E[X] + b, Where a and b are constants

4) E[X + Y] = E[X] + E[Y]

Properties of Variance:

1) V[K] = 0, Where K is some constant.

2) V[cX] = c² V[X]

3) V[aX + b] = a² V[X]

4) V[aX + bY] = a2 V[X] + b2 V[Y] + 2ab Cov(X,Y)

Cov.(X,Y) = E[XY] - E[X].E[Y]

Calculation 

HHH → 0

HHT → 0

HTH → 1

HTT → 0

THH → 1

THT → 1

TTH → 1

TTT → 0

Probability distribution

\(\begin{array}{c|c|c} \mathrm{x}_{\mathrm{i}} & 0 & 1 \\ \hline \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right) & 1 / 2 & 1 / 2 \end{array}\)

\(\mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{1}{2}\)

\(\sigma^{2}=\sum x_{i}^{2} p_{i}-\mu^{2}\)

= \(\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

\(64\left(\mu+\sigma^{2}\right)=64\left(\frac{1}{2}+\frac{1}{4}\right)=48\)

Hence option 2 is correct

Concept:

\(\sum\limits_{k = 0}^4 {p(X = K) = 1 } \)

Calculation

\(⇒ \sum\limits_{k = 0}^4 {C{K^2} = 1}\)

⇒ C(1² + 2² + 3² + 4²) = 1

⇒ \(C = \frac{1}{{30}}.\) 

Concept: 

Mean:

Let X is a discrete random variable having the possible values x1, x2, ……xn

If P(X = xi) = f(xi), where i = 1, 2……n, then

E(X) = mean = x1 f(x1) + x2 f(x2)+ …….xn f(xn)

\({\rm{i}}.{\rm{e\;E}}\left( {\rm{x}} \right) = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{X}}_{\rm{i}}}{\rm{\;f}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\)

\({\rm{E}}\left( {{{\rm{x}}^2}} \right) = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} X_i^2{\rm{\;f}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\)

Variance:

V(X) = E(X2) – E(X)2

Calculation:

Given q = 0.4

Required value = V(X) = E(X2) – [E(X)]2

\(\begin{array}{l} E\left( X \right) = \mathop \sum \limits_i {X_i}{p_i} \; \end{array}\)

\(= 0 \times 0.4 + 1 \times 0.6 = 0.6\)

\( E\left( {{X^2}} \right) = \mathop \sum \limits_i X_i^2{p_i} = {0^2} \times 0.4 + {1^2} \times 0.6 = 0.6\)

\( V\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} \)

⇒ V(x) = 0.6 - 0.36 = 0.24 

Explanation:

• For two independent events A and B :

P(A ∩ B) = P(A)P(B)   →    D - (II)

• A random variable X which takes values 0, 1, 2, ... ,n follow binomial distribution if its probability distribution function is given by,

P(X = x) = ⁿC_x qn-x px     →   C - (I)

• The variance of the random variable X,

Var(X) = E[(X - E[X])]²

⇒  Var(X) = E[X² - 2XE[X] + (E[X])²] 

⇒ Var(X) = E[X²] - 2E[X]E[X] +  (E[X])2

⇒ Var(X) = E[X2] - (E[X])2          →    A - (III)

• By definition,

Concept:

The expectation of a random variable E(X) is given by \(\Sigma {xP(x) }\)

Calculation:

Given:

Expectation E(X) is calculated by \(\Sigma {xP(x) }\)

⇒ E(X) = 30 × \(\frac{1}{5}\) + 10 × \(\frac{3}{10}\) +( - 10)× \(\frac{1}{2}\) 

⇒ E(X) = 6 + 3 - 5

⇒ E(X) = 4

∴ E(X) is equal to 4.

The correct answer is option 2.

Concept:

If \(P(X=x)=f(x)\) is a probability mass function, then 

\(\sum_{x\in Range(X)}f(x)=1\)

and the Expected value of X is given by \(E(X)=\sum_{x_{i}}x_{i}f(x_{i})\)

Calculations:

Given:

Therefore the expected value of X is given by,

\(E(X)=\sum_{x_{i}}x_{i}f(x_{i})\)

\(\Rightarrow E(X)=\sum_{x=1}^{\infty}x\frac{1}{2^x}\)

\(\Rightarrow E(X)=(1.\frac{1}{2})+(2.\frac{1}{2^2})+(3.\frac{1}{2^3})+...\)----(1)

It is arithmetic geometric series. Let us multiply it by the common ratio \(\frac{1}{2}\)of geometric series. We get,

\(\Rightarrow \frac{1}{2}E(X)=(\frac{1}{2}.\frac{1}{2})+(\frac{2}{2}.\frac{1}{2^2})+(\frac{3}{2}.\frac{1}{2^3})+...\)---(2)

Subtracting (2) from (1), We get

\(\Rightarrow \frac{1}{2}E(X)=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\)

\(\Rightarrow \frac{1}{2}E(X)=\frac{\frac{1}{2}}{1-\frac{1}{2}}\)    sum of infinite of G.P. with first term a and common ratio r is given by \(S_{\infty}=\frac{a}{1-r}\)

\(\Rightarrow \frac{1}{2}E(X)=\frac{\frac{1}{2}}{\frac{1}{2}}\)

\(\Rightarrow \frac{1}{2}E(X)=1\)

\(\therefore E(X)=2\)

Therefore option 1 is correct.

When two fair dice rolled, 6 × 6 = 36 observations are obtained.

P (x = 2) = P = (1, 1) 

P (x = 3) = P (1, 2) + P (2, 1) 

P (x = 4) = P (1, 3) + P (3, 1) + P (2, 2) 

P (x = 5) = P (1, 4) + P (4, 1) + P (2, 3) + P (3, 2) 

P (x = 6) = P (1, 5) + P (5, 1) + P (2, 4) + P (4, 2) + P (3, 3) 

P ( x = 7) = P (1, 6) + P (6, 1) + P (2, 5) + P (5, 2) + P (3, 4) + P (4, 3) 

P (x = 8) = P (2, 6) + P (6, 2) + P (3, 5) + P (5, 3) + P (4, 4) 

P (x = 9) = P (3, 6) + P (6, 3) + P (4, 5) + P (5, 4) 

P (x = 10) = P (5, 5) + P (6, 4) + P (4, 6) 

P (x = 11) = P (5, 6) + P (6, 5) 

P (x = 12) = P (6, 6) 

Therefore, the required probability distribution is as follows,

Then, E(x) = ∑x_i P (x_i)

= {x₁. P (x₁) + x₂ P (x₂) + x₃ P (x₃) + x₄ P (x₄) + x₅ P (x₅) + x₆ P (x₆) + x₇ P (x₇) + x₈ P (x₈) + x₉  P (x₉) + x₁₀ P (x₁₀) + x11 P(x11) + x₁₂ P(x₁₂)}

\( = 0 + \left( {2 \times \frac{1}{{36}}} \right) + \left( {3 \times \frac{1}{{18}}} \right) + \left( {4 \times \frac{1}{{12}}} \right) + \left( {5 \times \frac{1}{9}} \right) + \left( {6 \times \frac{5}{{36}}} \right) + \left( {7 \times \frac{1}{6}} \right)\)

\( + \left( {8 \times \frac{5}{{36}}} \right) + \left( {9 \times \frac{1}{9}} \right) + \left( {10 \times \frac{1}{6}} \right) + \left( {11 \times \frac{1}{{18}}} \right) + \left( {12 \times \frac{1}{{36}}} \right)\)

\( = \left\{ {\frac{1}{{18}} + \frac{1}{6} + \frac{1}{3} + \frac{5}{9} + \frac{5}{6} + \frac{7}{6} + \frac{{10}}{9} + \frac{1}{1} + \frac{5}{3} + \frac{{11}}{{18}} + \frac{1}{3}} \right\}\)

E (x) = 7

⇒ E (x²) = ∑x₁² P(x_i)

\( = \left( {4 \times \frac{1}{{36}}} \right) + \left( {9 \times \frac{1}{{18}}} \right) + \left( {16 \times \frac{1}{{12}}} \right) + \left( {25 \times \frac{1}{9}} \right) + \left( {36 \times \frac{5}{{36}}} \right) + \left( {49 \times \frac{1}{6}} \right)\)

\( + \left( {64 \times \frac{5}{{36}}} \right) + \left( {81 \times \frac{1}{9}} \right) + \left( {100 \times \frac{1}{6}} \right) + \left( {121 \times \frac{1}{{18}}} \right) + \left( {144 \times \frac{1}{{36}}} \right)\)

\( = \left( {\frac{1}{9} + \frac{1}{2} + \frac{4}{3} + \frac{{25}}{9} + \frac{5}{1} + \frac{{49}}{6} + \frac{{80}}{9} + 9 + \frac{{25}}{3} + \frac{{121}}{{18}} + 4} \right)\)

\( = \frac{{987}}{{18}} = \left( {\frac{{329}}{6}} \right) = 54.833\)

Variance x E(x²) -[E (x)]²

= 54.833 - 49

Variance = 5.833

Concept:

If X is the random variable with probability P(X = x) then mean \((\mu)\) and variance \((\sigma^2)\) of the random variable X is given as

\(\mu=E(X)= \sum_{i=0}^{\infty}x_iP(X=x_i)\)

\(\sigma^2=E(X^2)-\mu^2= \sum_{i=0}^{\infty}x^2_iP(X=x_i)-\mu^2\)

Calculation:

As we have given 

\( P(X = x) =\left\{ \begin{matrix} \dfrac {x+1}{2^{(x+2)}} & x = 0, 1, 2... \\\ 0 & \rm{otherwise} \end{matrix} \right.\)

So, mean of above is given as

\(\mu= \displaystyle\sum_{i=0}^{\infty}x\left(\dfrac {x+1}{2^{(x+2)}}\right)= \displaystyle\sum_{i=0}^{\infty}x\left(\dfrac {x+1}{2^x.2^2}\right)\)

​\(\mu= \displaystyle\frac{1}{4}\sum_{i=0}^{\infty}\left(\dfrac {x^2+x}{2^x}\right)\)​

\(\mu= \displaystyle\frac{1}{4}\left[\sum_{i=0}^{\infty}\left(\dfrac {x^2}{2^x}\right)+\sum_{i=0}^{\infty}\left(\dfrac {x}{2^x}\right)\right]\)

\(\mu=\frac{1}{4}\left[6+2\right]=\frac{8}{4}=2\)                   ( As \( \displaystyle \sum_{i=0}^{\infty}\left(\dfrac {x}{2^x}\right)=2\) &  \( \displaystyle \sum_{i=0}^{\infty}\left(\dfrac {x^2}{2^x}\right)=6\))

Now, the variance is given as 

\(\sigma^2== \displaystyle\sum_{i=0}^{\infty}x^2\left(\dfrac {x+1}{2^x.2^2}\right)-\mu^2\)​​

\(\sigma^2= \displaystyle\frac{1}{4}\left[\sum_{i=0}^{\infty}\left(\dfrac {x^3+x^2}{2^x}\right)\right]-(2)^2\)

​\(\sigma^2= \displaystyle\frac{1}{4}\left[\sum_{i=0}^{\infty}\left(\dfrac {x^3}{2^x}\right)+\sum_{i=0}^{\infty}\left(\dfrac {x^3}{2^x}\right)\right]-4\)​​

\(\sigma^2= \displaystyle\frac{1}{4}\left[26+6\right]-4\)                    ( As \( \displaystyle \sum_{i=0}^{\infty}\left(\dfrac {x^2}{2^x}\right)=6\) &  \( \displaystyle \sum_{i=0}^{\infty}\left(\dfrac {x^3}{2^x}\right)=26\))

\(\sigma^2=\frac{32}{4}-4=8-4=4\)

Hence, the mean and variance of the given probability function is 2 and 4 respectively.​

Concept:

Expected Value of a Discrete Random Variable:

• The expected value, denoted by E(X), of a discrete random variable is the sum of the products of each value of the variable with its corresponding probability.
• Formula: E(X) = Σ [X × P(X)]
• It represents the average outcome over a large number of repetitions of the experiment.

Calculation:

We have,

Σ X × P(X) = 1 × 1/2 + 2 × 2/5 + 4 × 12/25 + 2A × 2/10 + 3A × 3/25 + 5A × 5/25

⇒ Σ X × P(X) = (25 + 20 + 24 + 10A + 6A + 10A) ÷ 50

⇒ Σ X × P(X) = (69 + 26A) ÷ 50

Since E(X) = Σ X × P(X)

⇒ 2.94 = (69 + 26A) ÷ 50

⇒ 26A = 50 × 2.94 − 69

⇒ 26A = 147 − 69

⇒ A = 78 ÷ 26 = 3

∴ The value of A is 3.