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# Class 12 RD Sharma Solutions – Chapter 32 Mean and Variance of a Random Variable – Exercise 32.2 | Set 1

• Last Updated : 16 Oct, 2021

### Question 1(i): Find the mean and standard deviation of each of the following probability distributions:

xi: 2 3 4

pi: 0.3 0.5 0.3

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆšVariance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = 0.4 + 1.5 + 1.2 = 3.1

And variance = 0.8 + 4.5 + 4.8 â€“ (3.1)2 = 0.49

âˆ´ Standard deviation = âˆš 0.49 = 0.7

### Question 1(ii): Find the mean and standard deviation of each of the following probability distributions:

xi: 1 3 4 5

pi: 0.4 0.1 0.2 0.3

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = 0.4 + 0.3+0.8+1.5= 3.0

And variance =0.4+0.9+3.2+7.5 â€“ (3.0)2 = 3

âˆ´ Standard deviation = âˆš 3= 1.732

### Question 1(iii): Find the mean and standard deviation of each of the following probability distributions:

xi: -5 -4 1 2

pi: 1/4 1/8 1/2 1/8

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = -1.25-0.5+0.5+0.25 = -1

And variance = 6.25+2+0.5+0.5 â€“ (-1)2 = 8.25

âˆ´ Standard deviation = âˆš8.25= 2.9

### Question 1(iv): Find the mean and standard deviation of each of the following probability distributions:

xi: -1 0 1 2 3

pi: 0.3 0.1 0.1 0.3 0.2

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = -0.3 + 0 + 0.1 + 0.6 + 0.6 = 1.0

And variance =0.3 + 0 + 0.1 + 1.2 + 1.8 â€“ (1)2 = 2.4

âˆ´ Standard deviation = âˆš2.4 = 1.5

### Question 1(v): Find the mean and standard deviation of each of the following probability distributions:

xi: 1 2 3 4

pi: 0.4 0.3 0.2 0.1

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆšVariance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

âˆ´ mean = 0.4+0.6+0.6+0.4 = 2.0

And variance = 0.4 +1.2 + 1.8 + 1.6â€“ (2)2 = 1.0

âˆ´ Standard deviation = âˆš1 = 1

### Question 1(vi): Find the mean and standard deviation of each of the following probability distributions:

xi: 0 1 3 5

pi: 0.2 0.5 0.2 0.1

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = 0+0.5+0.6+0.5 = 1.6

And variance = 0 +0.5 + 1.8 + 2.5â€“ (1.6)2 = 2.24

âˆ´ Standard deviation = âˆš2.24 = 1.497

### Question 1(vii): Find the mean and standard deviation of each of the following probability distributions:

xi: -2 -1 0 1 2

pi: 0.1 0.2 0.4 0.2 0.1

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = -0.2-0.2+0+0.2+0.2 = 0

And variance = 0 +0.4+0.2+0.2+0.4â€“ (0)2 = 1.2

âˆ´ Standard deviation = âˆš1.2 = 1.095

### Question 1(viii): Find the mean and standard deviation of each of the following probability distributions:

xi: -3 -1 0 1 3

pi: 0.05 0.45 0.20 0.25 0.05

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = -0.15-0.45+0+0.25+0.15 = -0.2

And variance = 0 +0.45+0.25+0.45+0.45â€“ (-0.2)2 = 1.56

âˆ´ Standard deviation = âˆš1.56 = 1.248

### Question 1(ix): Find the mean and standard deviation of each of the following probability distributions:

xi: 0 1 2 3 4 5

pi: 1/6 5/18 2/9 1/6 1/9 1/18

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

âˆ´ Mean =  0+5/18+4/9+1/2+4/9+5/18 = 35/18

Variance =  0+5/18+8/9+3/2+16/9+25/18 -{35/18)2 = 665/324

âˆ´ standard deviation = âˆš (665/324) = âˆš665/18

### Question 2: A discrete random variable X has the probability distribution given below:

X: 0.5 1 1.5 2

P(X): k k2 2k2 k

(i) Find the value of k. (ii) Determine the mean of the distribution.

Solution:

To find the value of k we will be using the very basic idea of probability.

Note: We know that the sum of the probabilities of all random variables taken from a given sample space is equal to 1.

âˆ´ P(X=0.5) + P(X=1) + P(X=1.5) + P(X=2) = 1

âˆ´ k + k2 + 2k2 + k = 1

â‡’ 3k2 + 2k â€“ 1 = 0

â‡’ 3k2 + 3k – k â€“ 1 = 0

â‡’ 3k(k+1) â€“ (k+1) = 0

â‡’ (3k-1)(k+1) = 0

âˆ´ k = 1/3 or k = -1

âˆµ k represents probability of an event. Hence 0â‰¤P(X)â‰¤1

âˆ´ k = 1/3

Mean of any probability distribution is given by- Mean = âˆ‘xipi

Now we have,

X: 0.5 1 1.5 2

P(X): 1/3 1/9 2/9 1/3

âˆ´ first we need to find the product i.e. pixi and add them to get mean.

âˆ´ Mean = 0.5 x (1/3) + 1 x (1/9) + 1.5 x (2/9) +2 x (1/3) = 23/18.

### Question 3: Find the mean-variance and standard deviation of the following probability distribution

Xi: a b

Pi: p q

Where p+q=1.

Solution:

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

âˆ´ p1x1 = ap and p2x2 = bq Similarly p1x12 = a2p and p2x22= b2q

âˆ´ Mean = ap + bq

Variance = a2p + b2q â€“ (ap + bq)2

=a2pq + b2pq + 2abpq [p + q=1]

=pq(a-b)2

âˆ´ SD = âˆš{pq(a-b)2 } = |a-b|âˆšpq

### Question 4: Find the mean and variance of the number of tails in three tosses of a coin.

Solution:

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

âˆµ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

âˆ´ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

Note: P(AÕˆB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

Now we have pi and xi.

Letâ€™s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ Mean = 0 + 3/8 + 3/8 + 1/8 = 3/2

Variance = 0 + 3/8 + 3/4 + 3/8 – (3/2)2 = 3/4

### Question 5: Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.

Solution:

In a deck of 52 cards there are 4 kings each of one suit respectively.

Let X be the random variable denoting the number of kings for an event when two cards are drawn simultaneously.

âˆ´ X can take values 0 , 1 or 2.

P(X=0) = 48C2/52C2    = 48×47/52×51 = 188/221

[For selecting 0 kings, we removed all 4 kings from deck and selected out of 48]

P(X=1) = 4C1 x 48C1/52C2 = 48 x 4 x 2/52 x 51 = 32/221

[For selecting 1 king, we need to select and 1 out of 4 and not any other]

P(X=2) = 4C2/52C2 = 4 x 3/52 x 51 = 1/221

[For selecting 2 king, we need to select and 2 out of 4]

Now we have pi and xi.

Letâ€™s proceed to find mean and standard deviation.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Standard Deviation is given by SD = âˆš Variance where variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

âˆ´ mean = 0 + 32/221 + 2/221 = 34/221

Variance = 0 + 32/221 + 4/221 – (34/221) = 400/2873

âˆ´ Standard deviation = âˆšvariance = âˆš(400/2873) = 20/âˆš2873

### Question 6: Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.

Solution:

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

âˆµ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

âˆ´ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

Note: P(AÕˆB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

Now we have pi and xi.

Letâ€™s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

Standard Deviation is given by SD = âˆšVariance

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

âˆ´ Mean = 0 + 3/8 + 3/4 + 3/8 = 3/2

Variance = 0 + 3/8 + 3/2 + 9/8  – (3/2) = 3/4

Standard Deviation = âˆš(3/4) = 0.87

### Question 7: Two bad eggs are accidentally mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.

Solution:

As there are total of two bad eggs. Therefore while drawing 3 eggs we can draw 1 bad egg or 2 or 0 bad eggs.

Let X be the random variable denoting number of bad eggs that can be drawn in each draw.

Clearly X can take values 0,1 or 2

P(X=0) = P(all 3 are good eggs) = 2C0 x 10C3 /12C3 = 120/220 = 6/11

[Since there are 10 good eggs so for selecting all good we took all three from 10 and 0 eggs from 2 bad ones. Total sample points are no of ways of selecting 3 eggs from total of 12 eggs]

Similarly,

P(X=1) = P(1 bad and 2 good eggs) =  2C1 x 10C2/12C3  = 9/22

P(X=2) = P(2 Bad eggs and 1 good egg) =2C2 x 10C1 /12C3 = 1/22

Now we have pi and xi.

Letâ€™s proceed to find mean

Mean of any probability distribution is given by Mean = âˆ‘xipi

âˆ´ first we need to find the products i.e. pixi and add them to get mean.

Following table gives the required products :

âˆ´ mean = 0 + 9/22 + 1/11 = 1/2

### Question 8: A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

Solution:

When a pair of fair dice is thrown there are total 36 possible outcomes.

X denotes the minimum of two numbers which appear

âˆ´ X can take values 1,2,3,4,5 and 6

P(X=1) = 11/36

[Possible Pairs: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)]

P(X=2) = 9/36

[Possible Pairs: (2,2),(3,2),(4,2),(5,2),(6,2),(2,6),(2,5),(2,4),(2,3)]

P(X=3) = 7/36

[Possible Pairs: (3,3),(3,4),(4,3),(5,3),(3,5),(3,6),(6,3)]

P(X=4) = 5/36

[Possible Pairs: (4,4),(5,4),(4,5),(4,6),(6,4)]

P(X=5) = 3/36

[Possible Pairs (5,5),(5,6),(6,5)]

P(X=6) = 1/36

[Possible Pairs: (6,6)]

Now we have pi and xi.

Letâ€™s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

Standard Deviation is given by SD = âˆšVariance

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

Required Probability distribution table:

âˆ´ Mean = 11/36 + 18/36 + 21/ 36 + 20/36 + 15/36 + 6/36 = 91/36

Variance = 11/36 + 1 + 63/36 + 80/36 + 75/36 + 1 – (91/36)2 =2555/1296

Standard deviation = âˆšvariance = 1.403

### Question 9: A fair coin is tossed four times. Let X denote the number of heads occurring. Find the probability distribution, mean and variance of X.

Solution:

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTHâ€¦â€¦â€¦â€¦,TTTT}

A total of 24 = 16 possibilities.

Let X be a random variable representing number of heads occurring in 4 tosses of a coin.

âˆµ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

âˆ´ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4th toss) can be given by their individual products.

Note: P(AÕˆB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(TTTT) = P(T)P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 x 1/2 = 1/16

Selecting a coin out of 4 which will show head rest all showing tail

= 4C1 x P(HHHT) = 4C1 x(1/2) x (1/2) x (1/2) x (1/2) = 1/4

similarly ,

P(X=2) = 4C2 x(1/2)4 = 3/8

P(X=3) = 4C3 x (1/2)4 = 1/4

P(X=4) = P(HHHH) = 1/16

Now we have pi and xi.

Letâ€™s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

âˆ´ Mean = 0+ 1/4 + 3/4 + 3/4 + 1/4 = 2

Variance = 0 + 1/4 + 3/2 + 9/4 + 1 – (2) = 1

### Question 10: A fair die is tossed. Let X denote twice the number appearing. Find probability distribution, mean and variance of X.

Solution:

When a fair dice is thrown there are total 6 possible outcomes.

âˆµ X denote twice the number appearing on die

âˆ´ X can take values 2,4,6,8,10 and 12

As appearance of a number on a fair die is equally likely

i.e. P(appearing of 1) = P(appearing of 2) = P(appearing of 3) = P(appearing of 4) = P(appearing of 5) = P(appearing of 6) = 1/6

âˆ´ appearance of twice of the number is also equally likely with a probability of 1/6.

P(X=2)=P(X=4)=P(X=6)=P(X=8)=P(X=10)=P(X=12)=1/6

Now we have pi and xi.

Letâ€™s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Variance is given by:

Variance = âˆ‘xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

Required Probability distribution table:

âˆ´ Mean = 2/6 + 4/6 + 1 + 4/3 + 5/3 + 2 = 7

Variance = 4/6 + 16/6 + 36/6 + 64/6 + 100/6 + 144/6 – 72 = 70/6.

### Question 11: A fair die is tossed. Let X denote 1 or 3 according as an odd or an even number appears. Find the probability distribution, mean and variance of X.

Solution:

When a fair dice is thrown there are total 6 possible outcomes.

âˆµ X denote 1 or 3 according as an odd or an even number appears.

P(appearing of even number on a die) = 3/6 [favourable outcomes {2,4,6}]

P(appearing of an odd number on a die) = 3/6 [favourable outcomes {1,4,3}]

P(X=1) = 3/6 = 1/2

P(X=3) = 3/6 = 1/2

Now we have pi and xi.

Letâ€™s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Variance is given by:

Variance = âˆ‘ xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

Required Probability distribution table:-

âˆ´ Mean = 1/2 + 3/2 = 2

Variance = 1/2 + 9/2 – (2)2 = 1.

### Question 12: A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X.

Solution:

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTHâ€¦â€¦â€¦â€¦,TTTT}

A total of 24 = 16 possibilities.

âˆµ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

âˆ´ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4th toss) can be given by their individual products.

Note: P(AÕˆB) = P(A)P(B) where A and B are independent events.

As X is a random variable representing longest string of head occurring in 4 tosses.

âˆ´ X can take following values:

X = 0 [ all tails (TTTT) ]

X = 1 [Longest string contains only 1 head e.g. (HTTT),(TTTH),(HTHT)..]

X = 2 [ Longest string contain only 2 head e.g. (HHTT),(HHTH),(THHT)â€¦]

X = 3 [Longest string contain only 3 head e.g. ( HHHT) And (THHH)]

X = 4 [ Longest string contain 4 heads i.e. (HHHH) ]

Thus,

P(X=0) = 1/16

P(X=1) = 7/16 [by counting number of favourable outcomes as explained]

P(X=2) = 5/16

P(X=3) = 2/16

P(X=4) = 1/16

Now we have pi and xi.

Letâ€™s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = âˆ‘xipi

Variance is given by:

Variance = âˆ‘xi2pi â€“ (âˆ‘xipi)2

âˆ´ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

âˆ´ Mean = 0 + 7/16 + 10/16 + 6/16 + 1/4 = 1.7

Variance = 0 + 7/16 + 20/16 + 18/16 + 1 – (1.7)2 = 0.935

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