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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

Calculate the mean deviation from the median of the following frequency distribution :

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

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Solution:



Mean of any probability distribution is given by Mean = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:

xipixipixi2pi
20.20.40.8
30.31.54.5
40.51.24.8

∴ mean = 0.4 + 1.5 + 1.2 = 3.1

And variance = 0.8 + 4.5 + 4.8 – (3.1)2 = 0.49

∴ Standard deviation = √ 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:

xipixipixi2pi
10.40.40.4
30.10.30.9
40.20.83.2
50.31.57.5

∴ mean = 0.4 + 0.3+0.8+1.5= 3.0

And variance =0.4+0.9+3.2+7.5 – (3.0)2 = 3



∴ Standard deviation = √ 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:

xipixipixi2pi
-51/4-1.256.25
-41/8-0.52
11/20.50.5
21/80.250.5

∴ mean = -1.25-0.5+0.5+0.25 = -1

And variance = 6.25+2+0.5+0.5 – (-1)2 = 8.25

∴ Standard deviation = √8.25= 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:



xipixipixi2pi
-10.3-0.30.3
00.100
10.10.10.1
20.30.61.2
30.20.61.8

∴ mean = -0.3 + 0 + 0.1 + 0.6 + 0.6 = 1.0

And variance =0.3 + 0 + 0.1 + 1.2 + 1.8 – (1)2 = 2.4

∴ Standard deviation = √2.4 = 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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 :

xipixipixi2pi
10.40.40.4
20.30.61.2
30.20.61.8
40.10.41.6

∴ mean = 0.4+0.6+0.6+0.4 = 2.0

And variance = 0.4 +1.2 + 1.8 + 1.6– (2)2 = 1.0

∴ Standard deviation = √1 = 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2



∴ 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:

xipixipixi2pi
00.200
10.50.50.5
30.20.61.8
50.10.52.5

∴ mean = 0+0.5+0.6+0.5 = 1.6

And variance = 0 +0.5 + 1.8 + 2.5– (1.6)2 = 2.24

∴ Standard deviation = √2.24 = 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:

xipixipixi2pi
-20.1-0.20.4
-10.2-0.20.2
00.400
10.20.20.2
20.10.20.4

∴ mean = -0.2-0.2+0+0.2+0.2 = 0

And variance = 0 +0.4+0.2+0.2+0.4– (0)2 = 1.2

∴ Standard deviation = √1.2 = 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 = ∑xipi



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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:

xipi xipixi2pi
-30.05-0.150.45
-10.45-0.450.45
00.2000
10.250.250.25
30.050.150.45

∴ mean = -0.15-0.45+0+0.25+0.15 = -0.2

And variance = 0 +0.45+0.25+0.45+0.45– (-0.2)2 = 1.56

∴ Standard deviation = √1.56 = 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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 :

xipixipixi2pi
01/600
15/185/185/18
22/94/98/9
31/61/23/2
41/94/916/9
51/185/1825/18

∴ 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  

∴ standard deviation = √ (665/324) = √665/18

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.

∴ P(X=0.5) + P(X=1) + P(X=1.5) + P(X=2) = 1

∴ k + k2 + 2k2 + k = 1

⇒ 3k2 + 2k – 1 = 0

⇒ 3k2 + 3k – k – 1 = 0

⇒ 3k(k+1) – (k+1) = 0

⇒ (3k-1)(k+1) = 0

∴ k = 1/3 or k = -1

∵ k represents probability of an event. Hence 0≤P(X)≤1

∴ k = 1/3

Mean of any probability distribution is given by- Mean = ∑xipi

Now we have,

X: 0.5 1 1.5 2

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

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

∴ 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 = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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.

∴ p1x1 = ap and p2x2 = bq Similarly p1x12 = a2p and p2x22= b2q

∴ Mean = ap + bq

Variance = a2p + b2q – (ap + bq)2

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

=pq(a-b)2

∴ SD = √{pq(a-b)2 } = |a-b|√pq

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.

∵ 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

∴ 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(AՈB) = 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.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi2pi – (∑xipi)2

∴ 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: 

xipixipixi2pi
01/800
13/83/83/8
23/83/43/2
31/83/89/8

∴ 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.

∴ 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.

Let’s proceed to find mean and standard deviation.

Mean of any probability distribution is given by Mean = ∑xipi

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

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:

xipixipixi2pi
0188/22100
132/22132/22132/221
21/2212/2214/221

∴ mean = 0 + 32/221 + 2/221 = 34/221

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

∴ Standard deviation = √variance = √(400/2873) = 20/√2873

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.

∵ 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

∴ 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(AՈB) = 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.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi2pi – (∑xipi)2

Standard Deviation is given by SD = √Variance

∴ 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 :

xipixipixi2pi
1/800
13/83/83/8
23/83/43/2
31/83/89/8

∴ Mean = 0 + 3/8 + 3/4 + 3/8 = 3/2

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

Standard Deviation = √(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.

Let’s proceed to find mean

Mean of any probability distribution is given by Mean = ∑xipi

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

Following table gives the required products :

xipixipi
06/110
19/229/22
21/221/11

∴ 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

∴ 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.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi2pi – (∑xipi)2

Standard Deviation is given by SD = √Variance

∴ 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:

xipixipixi2pi
111/3611/3611/36
29/3618/361
37/3621/3663/36
45/3620/3680/36
53/3615/3675/36
61/366/361

∴ 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 = √variance = 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,THTH…………,TTTT}

A total of 24 = 16 possibilities.

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

∵ 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

∴ 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(AՈB) = 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.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi2pi – (∑xipi)2

∴ 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 :

xipi xipi xi2pi
01/1600
11/41/41/4
23/83/43/2
31/43/49/4
41/161/41

∴ 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.

∵ X denote twice the number appearing on die

∴ 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

∴ 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.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑xi2pi – (∑xipi)2

∴ 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:

xipixipixi2pi
21/62/64/6
41/64/616/6
61/6136/6
81/64/364/6
101/65/3100/6
121/62144/6

∴ 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.

∵ 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.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi2pi – (∑xipi)2

∴ 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:-

xipixipixi2pi
11/21/21/2
31/23/29/2

∴ 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,THTH…………,TTTT}

A total of 24 = 16 possibilities.

∵ 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

∴ 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(AՈB) = 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.

∴ 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)…]

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.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑xi2pi – (∑xipi)2

∴ 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 :

xipixipixi2pi
01/1600
17/167/167/16
25/1610/1620/16
32/166/1618/16
41/161/41

∴ 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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