Mean and Variance of Random variables MCQ Quiz in தமிழ் - Objective Question with Answer for Mean and Variance of Random variables - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

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

The variance is defined as:

f_x(X) is the probability function.

Property:

   ---(1)

Analysis:

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

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

= 

Hence option 2 is correct

Concept:

Calculation

⇒ C(1² + 2² + 3² + 4²) = 1

⇒  

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)

Variance:

V(X) = E(X2) – E(X)2

Calculation:

Given q = 0.4

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

Sum of n terms of AP is given by [2a + (n - 1) d] where a = first term and d = common difference

Calculation:

Given, a₁, a₂, a₃, a₄, a₅, a₆ are in A.P. and a₁ + a₃ = 10

Also, mean of a₁, a₂, a₃, a₄, a₅, a₆ = 

⇒ 

⇒ a₁ + a₂ + a₃ + a₄ + a₅ + a₆ = 57

Let common difference of AP be d.

⇒ S₆ =  = 57

⇒ 2a₁ + 5d = 19 ...(i) 

⇒ a₁ + a₁ + 2d = 10 [∵ a1 + a3 = 10]

⇒ 2a₁ + 2d = 10

⇒ a₁ + d = 5 ...(ii)

On solving equation (i) and equation (ii), we get

a₁ = 2 and d = 3

Now, variance = 

⇒ σ² = 

⇒ σ2 = 

⇒ σ2 =  

⇒ 8σ2 = 210

∴ The value of 8σ2 is 210.

The correct answer is Option 1.

Concept:

The expectation of a random variable E(X) is given by 

Calculation:

Given:

Expectation E(X) is calculated by 

⇒ E(X) = 30 ×  + 10 ×  +( - 10)×  

⇒ E(X) = 6 + 3 - 5

⇒ E(X) = 4

∴ E(X) is equal to 4.

The correct answer is option 2.

Concept:

If  is a probability mass function, then 

and the Expected value of X is given by 

Calculations:

Given:

Therefore the expected value of X is given by,

----(1)

It is arithmetic geometric series. Let us multiply it by the common ratio of geometric series. We get,

---(2)

Subtracting (2) from (1), We get

    sum of infinite of G.P. with first term a and common ratio r is given by 

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)}

E (x) = 7

⇒ E (x²) = ∑x₁² P(x_i)

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  and variance  of the random variable X is given as

Calculation:

As we have given 

So, mean of above is given as

​​

                   ( As  &  )

Now, the variance is given as 

​​

​​​

                    ( As  &  )

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