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The mean and variance of random variables help solve questions related to probability and statistics. Variance is known as the expected value of a squared deviation of a random variable from its sample mean. It can tell how far a set of numbers can spread out from their average value. A variance is an important tool in science, it is used to derive statistical data. The arithmetic mean of data is also known as arithmetic average, it is a central value of a finite set of numbers.
Read More: NCERT Solutions for Class 12 Mathematics Chapter 13 Probability
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Key Words: Mean, Variables, Probability, Statistics, variance, average value, arithmetic mean, arithmetic average, standard deviation
Definition of Mean
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A mean is a discrete random variable, denoted as X, it is the average of the possible values that the random variable can take. In other words, the sum of the values in the date divided by the number of values gives us the mean. The mean of a random variable gives each outcome Xi according to its probability Pi. The symbol for the mean is μ. The mean of a random variable gives the expected average outcomes of many observations.
μX = X1P1 + X2P2 + . . . + XkPk
=ΣXiPi
Definition of Variance
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A variance of a discrete random variable, say X measures the variability of the distribution. The standard deviation (σ) is the square root of variance. Variance measures the dispersion, which is how far the set data has spread out from the average value.
σx2 = Σ(xi -μi)2Pi
Read More: Inverse Trigonometric Formulas
Properties of Mean
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- The mean is generally defined by the density curve of the distribution for a continuous random variable.
- If X is a random variable which is adjusted by multiplying with the value b and adding the value a then the mean is given as μa+bX =a + bμx
- As there is an increase in the number of observations, the mean of these observations gets closer to the random variable’s true mean.
Read More: Difference Between Mean and Median
Properties of Variance
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- If x is a random number, which is adjusted by multiplying with b and adding a, then the variance is given as σ2 a+bx =b2σ2
- The variance of a constant is zero, V(a)=0, here a is any constant.
- If x is considered to be a random variable, and a, b are any constant, then V(aX+b) = a2V(X)
- There can be any pair wise independent random variable, be X1, X2,…, Xn, for any constant like a1, a2, … , an; V(a1X1 +a2 X2 +…+anXn) = a12 V(X1) + a22 V(X2) +…+ an2 V(Xn).
Also Read: Relation Between Mean Median and Mode
Things to Remember
- Variance is known as the expected value of a squared deviation of a random variable from its sample mean.
- The arithmetic mean of data is also known as arithmetic average, it is a central value of a finite set of numbers. The sum of the values in the date divided by the number of values gives us the mean.
- Mean expected is given as μX = X1P1 + X2P2 + . . . + XkPk = ΣXiPi
- The variance is σx2 = Σ(xi -μi )2Pi
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Sample Questions
Ques. A random variable x has distribution law as given below: (4 Marks)
What is the variance of distribution?
Ans.
| X | 1 | 2 | 3 |
| P(X=x) | 0.3 | 0.4 | 0.3 |
Variance: Var(X) = σ2= ΣPi(xi)2-( ΣPixi )2= ΣPi(xi)2-μ
We have X1= 1, X2= 2, X3= 3 and P1 = 0.3, P2 = 0.4 , P3 = 0.3.
Now, ΣPixi = (0.3 x 1)+(0.4 x 2)+(O.3 x 3)
= 0.3 + 0.8+0.9
=2
And ΣPi(xi)2 = (0.3 x 12)+(0.4 x 22)+(O.3 x 32)
=0.3+1.6+2.7
=4.6
Therefore, Var(X) = σ2= ΣPi(xi)2-( ΣPixi )2= 4.6-22=4.6-4 =0.6
The variance of distribution is Var(x) =0.6
Ques. In a week, the temperature of a certain place measured are 26oC, 24oC, 28oC, 31oC, 30oC, 26oC, 24oC. Find the mean temperature of the week. (3 Marks)
Ans. Mean temperature
= Sum of all temperature / Number of terms
= (26 + 24 + 28 + 31 + 30 + 26 + 24) / 7
= 189/7
= 27
Therefore the mean temperature is 27 oC.
Ques. 60 Kg is the mean weight of 4 members of a family. Three of them have the weight 56 kg, 68 kg and 72 kg respectively. What is the weight of the fourth member? (3 Marks)
Ans. Let ‘x’ be the weight of the fourth member.
Mean weight of 4 members = 60 kg
(56 + 68 + 72 + x)/4 = 60
196 + x = 60(4)
x = 240 - 196
x = 44 kg
Therefore the weight of the fourth member is 44 kg.
Ques. In a mathematics class test, 10 students got 75 marks, 12 students got 60 marks, 8 students got 40 marks and 3 students got 30 marks. What is the mean of their score? (3 Marks)
Ans. Number of students getting 75 marks = 10
Number of students getting 60 marks = 12
Number of students getting 40 marks = 8
Number of students getting 30 marks = 3
= [10(75) + 12(60) + 8(40) + 3(30)] / (10 + 12 + 8 + 3)
= (750 + 720 + 320 + 90) / 33
= 1880/33
= 56.96
Ques. For the set of data: 14, 16, 7, 9, 11, 13, 8, 10. Calculate the variance. (4 Marks)
Ans. First, determine the mean of the data set.
There are a total of n=data entries, hence the mean is given by¯x=(14+16+7+9+11+13+8+10)/8
=88/8
=11
Determine the distance of each entry from the mean by subtraction:
| Xi | ¯x−xi |
|---|---|
| 14 | -3 |
| 16 | -5 |
| 7 | 4 |
| 9 | 2 |
| 11 | 0 |
| 13 | -2 |
| 8 | 3 |
| 10 | 1 |
Square each of these distances:
| Xi | ¯x−xi | (¯x−xi)2 |
|---|---|---|
| 14 | -3 | 9 |
| 16 | -5 | 25 |
| 7 | 4 | 16 |
| 9 | 2 | 4 |
| 11 | 0 | 0 |
| 13 | -2 | 4 |
| 8 | 3 | 9 |
| 10 | 1 | 1 |
Find the mean of these squared distances. Which gives the value of the variance:
σ2=(9+25+16+4+0+4+9+1)/8
=68/8
=8.5
Ques. The Values of Mean and Mode are 30 and 15 respectively. What is the value of Median? (4 Marks)
(a) 25
(b) 28
(c) 26.5
(d) 23.5
Ans. The relation between mean, median, and mode is, Mean – Mode = 3 (Mean – Median).
Substituting the value, equation becomes 30 – 15 = 3 (30 – Median).
Therefore the median comes to 25.
Ques. What is the variance of the number 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. (4 Marks)
Ans. First, find the mean of the 10 values.
Mean = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) / 10
= (155) / 10
= 15.5
| Value N | N - N¯ | ( N -N¯)2 |
|---|---|---|
| 1 | -4.5 | 20.25 |
| 2 | -3.5 | 12.25 |
| 3 | -2. 5 | 6.25 |
| 4 | -1.5 | 2.25 |
| 5 | -0.5 | 0.25 |
| 6 | +0.5 | 0.25 |
| 7 | +1.5 | 2.25 |
| 8 | +2.5 | 6.25 |
| 9 | +3.5 | 12.25 |
| 19 | +4.5 | 20.25 |
| Total | 0 | 82.50 |
Now, we need to find the population variance
σ2 = 82.51082.510
= 8.25
Ques. There are three data sets A, B and C.
A = {9,10,11,7,13}
B = {10,10,10,10,10}
C = {1,1,10,19,19}
a) What is the mean of each data set?
b) What is the standard deviation of each data set?
c) Which set has the largest standard deviation?
d) Is it possible to answer question (c) without calculations of the standard deviation? (3 Marks)
Ans. a) Mean of Data set A = (9+10+11+7+13)/5 = 10
Mean of Data set B = (10+10+10+10+10)/5 = 10
Mean of Data set C = (1+1+10+19+19)/5 = 10
b) Standard Deviation Data set A = √[ ( (9-10)2+(10-10)2+(11-10)2+(7-10)2+(13-10)2 )/5 ]
= 2
Standard Deviation Data set B= √[ ( (10-10)2+(10-10)2+(10-10)2+(10-10)2+(10-10)2 )/5 ]
= 0
Standard Deviation Data set C= √[ ( (1-10)2+(1-10)2+(10-10)2+(19-10)2+(19-10)2 )/5 ]
= 8.05
c) Data set C has the largest standard deviation.
d)Yes, because data Set C has data values that are further away from the mean compared to sets A and E
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