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பெறு Mean and Variance of Random variables பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Mean and Variance of Random variables MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Mean and Variance of Random variables MCQ Objective Questions

Top Mean and Variance of Random variables MCQ Objective Questions

Mean and Variance of Random variables Question 1:

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let Y denote the random variable of number of Jacks obtained in the two drawn cards. Then P(Y = 1) + P(Y = 2) equals?

  1. 25169
  2. 24169
  3. 5169
  4. 1169

Answer (Detailed Solution Below)

Option 1 : 25169

Mean and Variance of Random variables Question 1 Detailed Solution

Calculation:

P(Y = 1) = P(jack and non jack) + P(non jack and jack)

452×4852+4852×452=24169

P(Y = 2) = P(jack and jack) = 452×452=1169

P(Y = 1) + P(Y = 2) = 25169

Mean and Variance of Random variables Question 2:

X is a random variable with variance σx2 . The variance of (X + a), where a is a constant

  1. (σx+a)1
  2. σx2 
  3. (σx2+a2)
  4. (σx2a2)

Answer (Detailed Solution Below)

Option 2 : σx2 

Mean and Variance of Random variables Question 2 Detailed Solution

Concept:

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

E[X]=+x.fx(x)dx

The variance is defined as:

σx2=E[(Xμx)2]

fx(X) is the probability function.

E[X2]=+x2fx(x)dx

Property:

Var[aX+b]=E[aX+b(aμ+b)]2

=E[aXaμ]2

=E[a2(Xμ)2]

=a2E[(Xμ)2]

Var[aX+b]=a2σX2   ---(1)

Analysis:

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

V[X+b]=(1)2.σx2

=σx2

26 June 1

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] = aV[X]

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

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

Mean and Variance of Random variables Question 3:

The value of C for which P (x = K) = CK2 can serve as the probability function of a random variable x that takes 0, 1, 2, 3, 4 is

  1. 130
  2. 110
  3. 13
  4. 115

Answer (Detailed Solution Below)

Option 1 : 130

Mean and Variance of Random variables Question 3 Detailed Solution

Concept:

k=04p(X=K)=1

Calculation

k=04CK2=1

⇒ C(12 + 22 + 32 + 42) = 1

⇒ C=130. 

Mean and Variance of Random variables Question 4:

Consider the following probability mass function (p.m.f.) of a random variable X:

p(x,q)={q1q0ifX=0ifX=1otherwise

If q = 0.4, the variance of X is___________.

  1. 0.24
  2. 0.36
  3. 0.60
  4. None

Answer (Detailed Solution Below)

Option 1 : 0.24

Mean and Variance of Random variables Question 4 Detailed Solution

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)

i.eE(x)=i=1nXif(Xi)

E(x2)=i=1nXi2f(Xi)

Variance:

V(X) = E(X2) – E(X)2

Calculation:

Given q = 0.4

        X      

       0        

       1        

p(X)

0.4

0.6

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

E(X)=iXipi

=0×0.4+1×0.6=0.6

E(X2)=iXi2pi=02×0.4+12×0.6=0.6

V(X)=E(X2)[E(X)]2

⇒ V(x) = 0.6 - 0.36 = 0.24 

Mean and Variance of Random variables Question 5:

The following is the probability distribution of a random variable.

X 3 2 1 0
P(X) 0.465  0.125  0.205 k

What is the value of k?

  1. 0.205
  2. 0
  3. 0.795
  4. 1

Answer (Detailed Solution Below)

Option 1 : 0.205

Mean and Variance of Random variables Question 5 Detailed Solution

Concept 

The sum of the probabilities in a probability distribution must equal 1

Calculation

Given:

X 3 2 1 0
P(X) 0.465  0.125  0.205 k
 

From above concept

P(3) + P(2) + P(1) + P(0) = 1

⇒ 0.465 + 0.125 + 0.205 + k = 1 

⇒ 0.795 + k = 1 

⇒ k = 1 - 0.795  = 0.205

Option 1 is correct

Mean and Variance of Random variables Question 6:

Let X be a discrete random variable. The probability distribution of X is given below:

X 30 10 –10
P(X) 15 310 12


Then E(X) is equal to

  1. 6
  2. 4
  3. 3
  4. –5

Answer (Detailed Solution Below)

Option 2 : 4

Mean and Variance of Random variables Question 6 Detailed Solution

Concept:

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

Calculation:

Given:

X 30 10 –10
P(X) 15 310 12

Expectation E(X) is calculated by ΣxP(x)

⇒ E(X) = 30 × 15 + 10 × 310 +( - 10)× 12 

⇒ E(X) = 6 + 3 - 5

E(X) = 4

∴ E(X) is equal to 4.

The correct answer is option 2.

Mean and Variance of Random variables Question 7:

Match the following:

  List I   List II
A. Var(X) I ncx qn-x px
B. E(X) II  P(E).P(F) = P(E ∩ F)
C. Binomial Distribution III E(X2) - [E(X)]2
D. E and F are independent IV i=1nxipi

Choose the  correct answer from the option given below:

  1. A - III, B - II, C - I, D - IV
  2. A - III, B - IV, C - II, D - I
  3. A - III, B - I, C - II, D - IV
  4. A - III, B - IV, C - I, D - II

Answer (Detailed Solution Below)

Option 4 : A - III, B - IV, C - I, D - II

Mean and Variance of Random variables Question 7 Detailed Solution

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) = nCx 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,

E(X) =  i=1nxipi        →        B - (IV)

Mean and Variance of Random variables Question 8:

Let X be a discrete random variable and f be a function given by P(X=x)=f(x)=12x for x=1,2,3,.... Then the expected value of X are 

  1. 2
  2. 3
  3. 1
  4. 4

Answer (Detailed Solution Below)

Option 1 : 2

Mean and Variance of Random variables Question 8 Detailed Solution

Concept:

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

xRange(X)f(x)=1

and the Expected value of X is given by E(X)=xixif(xi)

Calculations:

Given:

Therefore the expected value of is given by,

E(X)=xixif(xi)

E(X)=x=1x12x

E(X)=(1.12)+(2.122)+(3.123)+...----(1)

It is arithmetic geometric series. Let us multiply it by the common ratio 12of geometric series. We get,

12E(X)=(12.12)+(22.122)+(32.123)+...---(2)

Subtracting (2) from (1), We get

12E(X)=12+122+123+...

12E(X)=12112    sum of infinite of G.P. with first term a and common ratio r is given by S=a1r

12E(X)=1212

12E(X)=1

E(X)=2

Therefore option 1 is correct.

Mean and Variance of Random variables Question 9:

Let the six numbers a1, a2, a3, a4, a5, a6, be in A.P. and a1 + a3 = 10. If the mean of these six numbers is 192 and their variance is σ2, then 8σ2 is equal to: 

  1. 210
  2. 220
  3. 200
  4. 105

Answer (Detailed Solution Below)

Option 1 : 210

Mean and Variance of Random variables Question 9 Detailed Solution

Concept:

Sum of n terms of AP is given by n2[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, a6192

⇒ a1+a2+a3+a4+a5+a66=192

⇒ a1 + a2 + a3 + a4 + a5 + a6 = 57

Let common difference of AP be d.

⇒ S662(2a1+5d) = 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=xi2n(x¯)2

⇒ σ222+52+82+112+142+1726(192)2

⇒ σ2 = 69963614

⇒ σ2 = 1054 

⇒ 8σ2 = 210

∴ The value of 2 is 210.

The correct answer is Option 1.

Mean and Variance of Random variables Question 10:

The sum of the numbers obtained by throwing two identical dice is X. The variance of X will be:

  1. 7.0010
  2. 5.8330
  3. 54.833
  4. 27.417

Answer (Detailed Solution Below)

Option 2 : 5.8330

Mean and Variance of Random variables Question 10 Detailed Solution

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) = ∑xi P (xi)

= {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+(2×136)+(3×118)+(4×112)+(5×19)+(6×536)+(7×16)

+(8×536)+(9×19)+(10×16)+(11×118)+(12×136)

={118+16+13+59+56+76+109+11+53+1118+13}

E (x) = 7

⇒ E (x2) = ∑x12 P(xi)

=(4×136)+(9×118)+(16×112)+(25×19)+(36×536)+(49×16)

+(64×536)+(81×19)+(100×16)+(121×118)+(144×136)

=(19+12+43+259+51+496+809+9+253+12118+4)

=98718=(3296)=54.833

Variance x E(x2) -[E (x)]2

= 54.833 - 49

Variance = 5.833

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