𝑼𝑵𝑰𝑻 𝑰 𝑹𝒂𝒏𝒅𝒐𝒎 𝑽𝒂𝒓𝒊𝒂𝒃𝒍𝒆 2.

𝑃(𝑋 ≤ −∞) = 0
3. If 𝑥1 ≤ 𝑥2 then 𝑃(𝑋 = 𝑥1 ) ≤ 𝑃(𝑋 = 𝑥2 )
Introduction
4. 𝑃(𝑋 > 𝑥) = 1 − 𝑃(𝑋 ≤ 𝑥)
Consider an experiment of tossing a coin twice. The outcomes of the experiment
5. 𝑃(𝑋 ≤ 𝑥) = 1 − 𝑃(𝑋 > 𝑥)
are 𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇. Let 𝑋 denote the number of heads turning up. Then 𝑋 has 6.

the values 2,1,1,0. Here 𝑋 is a random variable which assigns a real number to Cumulative distribution function of 𝑿
every outcome of a random experiment. If X is a random variable, then the function 𝐹 ∶ 𝑅 → [0, 1] is defined by 𝐹(𝑥) =
Definition 𝑃(𝑋 ≤ 𝑥) is called the cumulative distribution function of X.
Let 𝑆 be a sample space associated with an experiment 𝐸. A function If X is a discrete random variable with values 𝑥1 < 𝑥2 < 𝑥3 … . . < 𝑥𝑖 …. Then
X : S  R which assigns to each element s  S , a unique real number 𝐹(𝑥) = ∑𝑥𝑖≤𝑥 𝑝𝑖

X s   R is called a random variable.

Illustrative Examples
Example 1 : A random variable X has the following probability function.
Types of Random variable
Discrete random variable Values of 𝑋 : 𝑥 0 1 2 3 4 5 6 7
A random variable whose set of possible values is either finite or countably
𝑃(𝑥) 0 𝑘 2𝑘 2𝑘 3𝑘 𝑘2 2𝑘 2 7𝑘 2 + 𝑘
infinite is called discrete random variable.
Example : The number of students who are absent for a particular period. (i) Find the value of 𝑘.
(ii) Evaluate 𝑃(𝑋 < 6), 𝑃(𝑋 ≥ 6) and 𝑃(0 < 𝑋 < 5)
Probability mass function (iii) Find the distribution function of X
Let 𝑋 be a discrete random variable which takes the values 𝑥1 , 𝑥2 , 𝑥3 , … … … Let 1
(iv) If 𝑃(𝑋 ≤ 𝑘) > find the minimum value of k and
2
𝑃(𝑋 = 𝑥𝑖 ) = 𝑝(𝑥𝑖 ) = 𝑝𝑖 then the function 𝑝(𝑥𝑖 ) is called the probability mass
(v) Find P(1.5 < X < 4.5 / X > 2)
function of X if satisfies the following conditions
Solution :
(𝑖) 𝑝(𝑥𝑖 ) ≥ 0, ∀𝑖 (𝑖𝑖) ∑∞
𝑖=1 𝑝(𝑥𝑖 ) = 1
(i) Since the total probability is equal to 1. i.e., ∑𝑖 𝑝(𝑥𝑖 ) = 1
The set of ordered pairs of numbers (𝑥𝑖 ,𝑝𝑖 ) is called the probability distribution
𝑖. 𝑒. , 0 + 𝑘 + 2𝑘 + 2𝑘 + 3𝑘 + 𝑘 2 + 2𝑘 2 + 7𝑘 2 + 𝑘 = 1
of the random variable 𝑋.
⟹ 10𝑘 2 + 9𝑘 = 1

Important Results ⟹ 10𝑘 2 + 9𝑘 − 1 = 0

1. 𝑃(𝑋 ≤ ∞) = 1 ⟹ 10𝑘 2 + 10𝑘 − 𝑘 − 1 = 0

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⟹ 10𝑘(𝑘 + 1) − (𝑘 + 1) = 0 (iv) The minimum value of 𝑘 may be determined by trial method.
⟹ (10𝑘 − 1)(𝑘 + 1) = 0 1
𝑃(𝑋 ≤ 0) = 0 <
1 2
⟹𝑘= , −1
10
1 1 1
𝑘 = – 1 is not possible since the probability cannot take negative value 𝑃(𝑋 ≤ 1) = 𝑝 (0) + 𝑝(1) = 0 + = <
10 10 2
1
∴𝑘=
10
3 1
𝑃(𝑋 ≤ 2) = 𝑝(0) + 𝑝(1) + 𝑝(2) = <
The probability distribution is 10 2
𝑥 0 1 2 3 4 5 6 7 5 1
𝑃(𝑋 ≤ 3) = 𝑝 (0) + 𝑝(1) + 𝑝(2) + 𝑝(3) = =
1 2 2 3 1 2 17 10 2
𝑝(𝑥) 0
10 10 10 10 100 100 100
8 1
𝑃(𝑋 ≤ 4) = >
10 2
(ii) 𝑃 (𝑋 < 6) = 𝑝 (0) + 𝑝(1) + 𝑝(2) + 𝑝(3) + 𝑝(4) + 𝑝(5)
1
1 2 2 3 1 The smallest value of 𝑘 for which 𝑃(𝑋 ≤ 𝑘) > is 4
= 0 + + + + + 2
10 10 10 10 100
(v) To find 𝑃(1.5 < 𝑋 < 4.5 / 𝑋 > 2)
10  20  20  30  1 81 We know that
= =
100 100
𝑃(𝐴 ∩ 𝐵)
𝑃(𝑋  6) = 1 – 𝑃(𝑋 < 6) 𝑃(𝐴 ∕ 𝐵) =
𝑃(𝐵)
81 19 𝑃(1.5 < 𝑋 < 4.5 ∩ 𝑋 > 2)
= 1– = 𝑃(1.5 < 𝑋 < 4.5⁄𝑋 > 2) =
100 100 𝑃(𝑋 > 2)

𝑃(0 < 𝑋 < 5) = 𝑝(1) + 𝑝(2) + 𝑝(3) + 𝑝(4) [𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) + 𝑃(𝑋 = 4)] ∩ [𝑃(𝑋 = 3) + ∙∙∙ +𝑃(𝑋 = 7)]
=
1 − 𝑃(𝑋 ≤ 2)
1 2 2 3 1 2  2  3 8
= + + + = = [𝑃(𝑋 = 3) + 𝑃(𝑋 = 4)]
10 10 10 10 10 10 =
1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)]
(iii) To find the cumulative distribution function of 𝑥. [𝑃(𝑋 = 3) + 𝑃(𝑋 = 4)]
=
𝑥 0 1 2 3 4 5 6 7 1− [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)]
2 3 5
+ 5
1 3 5 8 81 83 = 10 10
= 10
=
𝐹(𝑥) = 𝑃(𝑋  𝑥) 0 1 3 7
7
10 10 10 10 100 100 1− 10
10

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 Example 2 : If the random variable 𝑋 takes the values 1, 2, 3 and 4 such that 2 15 10 25
𝐹(2) = 𝑃(𝑋 ≤ 2) = 𝑝(1) + 𝑝(2) = + =
𝑃(𝑋 = 1) = 3 𝑃(𝑋 = 2) = 𝑃(𝑋 = 3) = 5𝑃(𝑋 = 4). Find the probability 61 61 61
15 10 30 55
distribution and find cumulative distribution function of X. 𝐹(3) = 𝑃(𝑋 ≤ 3) = 𝑝(1) + 𝑝(2) + 𝑝(3) = + + =
61 61 61 61
Solution 15 10 30 6
𝐹(4) = 𝑃(𝑋 ≤ 4) = 𝑝(1) + 𝑝(2) + 𝑝(3) + 𝑝(4) = + + + =1
Let 2𝑃(𝑋 = 1) = 3𝑃(𝑋 = 2) = 𝑃(𝑋 = 3) = 5𝑃(𝑋 = 4) = 𝑘 61 61 61 61
k k k
i.e., P(X = 1) = , P(X = 2) = , P(X = 3) = 𝑘, P(X = 4) =
2 3 5 Example 3: The probability mass function of a random variable is defined as
Since the total probability is equal to 1. i.e., ∑𝑖 𝑝(𝑥𝑖 ) = 1 𝑃(𝑋 = 0) = 3𝐶 2 , 𝑃(𝑋 = 1) = 4𝐶 − 10𝐶 2 and 𝑃(𝑋 = 2) = 5𝐶 − 1 where 𝐶 > 0

k k k 15k  10k  30k  6k and 𝑃(𝑋 = 𝑟) = 0 𝑖𝑓 𝑟 ≠ 0,1,2 (i) Find the value of 𝐶
i.e., + +𝑘+ =1  1
2 3 5 30 (ii) Find 𝑃(0 < 𝑋 < 2 ∕ 𝑋 > 0) (iii) The distribution function of 𝑋
1
61k 30 (iv) The largest value of 𝑋 for which 𝐹(𝑋) <
  1 k  2
30 61 Solution:
Next to find probability distribution of X: (i) To find 𝐶
𝑋= 𝑥 1 2 3 4
15 10 30 6
Since the total probability is equal to 1. i.e.,  p( x )  1
i
𝑃(𝑋 = 𝑥) i
61 61 61 61
⇒ 3𝐶 2 + 4𝐶 − 10𝐶 2 + 5𝐶 − 1 = 1
7𝐶 2 − 9𝐶 + 2 = 0
(i) The cumulative distribution 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥)
(7𝐶 − 2)(𝐶 − 1) = 0
x
(ii) F ( x)   p( x ) i ∴ 𝐶 = 1, 𝐶 =
2
xi   7
2
The value of 𝐶 =
7
𝑥 1 2 3 4
(∵ 𝐶 = 1 gives 𝑃(𝑋 = 1) < 0, which is not possible)
15 25 55 ∴ Probability distribution of 𝑋:
𝐹(𝑥) 1
61 61 61
X: 0 1 2

15
12 16 3
Since 𝐹(1) = 𝑃(𝑋 = 1) = p(x):
61 49 49 7

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 𝑃(0 < 𝑋 < 2) ∩ 𝑃(𝑋 > 0 Probability Density Function
(ii)𝑃(0 < 𝑋 < 2⁄𝑋 > 0) =
f  x  is called the
𝑃(𝑋 > 0)
Let X be a continuous random variable. The function
𝑃(𝑋 = 1) ∩ [𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)]
= probability density function (p.d.f) of the random variable X if it satisfies the
𝑃(𝑋 > 0)
𝑃(𝑋 = 1) following conditions:
=
𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) (i) 𝑓(𝑥) ≥ 0𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥, −∞ < 𝑥 < ∞
16/49 16 ∞
(ii) ∫−∞ 𝑓(𝑥)𝑑𝑥 = 1
= =
37/49 37
Cumulative distribution function of 𝑿
(iii) Distribution function
If X is a random variable, then the function 𝐹: 𝑅 → [0,1] is defined by 𝐹(𝑥) =
0 𝑖𝑓 𝑥 < 0
12 𝑃(𝑋 ≤ 𝑥) is called the cumulative distribution function of X.
𝑖𝑓 0 ≤ 𝑥 < 1
𝐹(𝑥) = 49 If X is a continuous random variable with p.d.f 𝑓(𝑥) defined for all 𝑥 ∈
28
𝑖𝑓 1 ≤ 𝑥 < 2 𝑥
49 (−∞, ∞) then F(𝑥) = ∫−∞ 𝑓(𝑥)𝑑𝑥
[ 1 𝑖𝑓 𝑥 ≥ 2
Properties of Distribution function
1
(iv) The largest value of 𝑥 for which 𝐹(𝑋) < is 𝑥 = 0 1. 𝐹(−∞) = 0
2

Problem for Practice: 2. 𝐹(∞) = 1

1. Suppose that the random variable X assumes three values 0, 1, 2 with 3. 0 ≤ 𝐹(𝑥) ≤ 1
1 1 1
probabilities , 𝑎𝑛𝑑 respectively. Obtain the cumulative distribution 4. 𝐹(𝑥1 ) ≤ 𝐹(𝑥2 ) 𝑖𝑓 𝑥1 < 𝑥2
3 6 2

function. 5. P(𝑥1 < 𝑋 ≤ 𝑥2 ) = 𝐹(𝑥2 ) − 𝐹(𝑥1 )

𝟏 𝟏
Ans: 𝑭(𝟎) = , 𝑭(𝟏) = , 𝑭(𝟐) = 𝟏
𝟑 𝟐
EXAMPLES UNDER CONTINUOUS RANDOM VARIABLES
Example 1:If the density function of continuous random variable X is given by
Continuous random variable
𝑎𝑥, 0≤𝑥≤1
A random variable X which takes all possible values in a given interval is called a 𝑎, 1≤𝑥≤2
𝑓(𝑥) = [
3𝑎 − 𝑎𝑥, 2 ≤ 𝑥 ≤ 3
Continuous random variable.
0, 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
Example:
(i) Find the value of ′𝑎′ (ii) Find the C.d.f of X (iii) Compute 𝑃[𝑋 ≤ 1.5]
If X denotes the lifetime of a transistor then X is a continuous random variable.
Solution:
(i) Since 𝑓(𝑥) is a p.d.f,

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 ∞
Example 2: The diameter of an electric cable 𝑋 is a continuous R.V with
∫ 𝑓(𝑥)𝑑𝑥 = 1
−∞ p.d.f 𝑓(𝑥) = 𝑘𝑥(1 − 𝑥), 0 ≤ 𝑥 ≤ 1. (i) Find the value of 𝑘 (ii) The c.d.f of 𝑋 (iii)
3
∫ 𝑓(𝑥)𝑑𝑥 = 1 a number 𝑏 such that 𝑃(𝑋 < 𝑏) = 𝑃(𝑋 > 𝑏)
0
Solution:
1 2 3
𝑖. 𝑒. , ∫ 𝑎𝑥 𝑑𝑥 + ∫ 𝑎 𝑑𝑥 + ∫ (3𝑎 − 𝑎𝑥)𝑑𝑥 = 1 (i) To find 𝑘
0 1 2 ∞

𝑎𝑥 2 1
𝑎𝑥 2 3 ∫ 𝑓(𝑥) 𝑑𝑥 = 1
⇒[ ] + (𝑎𝑥)12 + [3𝑎𝑥 − ] =1 −∞
2 0 2 2 1

𝑎 9𝑎 𝑘 ∫ (𝑥 − 𝑥 2 ) 𝑑𝑥 = 1
∴ +𝑎+ − 4𝑎 = 1 0
2 2 1 1
1 𝑘[ − ] = 1
⇒ 2𝑎 = 1 ⇒𝑎= 2 3
2 𝑘
(ii) 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) ⇒ =1 ⇒𝑘=6
6
𝐹(𝑥) = 0,when 𝑥 < 0 6𝑥(1 − 𝑥) 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 1
∴ 𝑓(𝑥) = [
𝑥𝑥 𝑥2 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
= ∫0 𝑑𝑥 = when 0 ≤ 𝑥 < 1
2 4 Hence, 𝑓(𝑥) is a PDF of random variable 𝑋.
1𝑥 𝑥1 𝑥 1
= ∫0 2 𝑑𝑥 + ∫1 2 𝑑𝑥 = − when 1 ≤ 𝑥 < 2 𝑥
(ii) C.d.f of 𝑋: 𝐹(𝑥) = ∫0 6𝑥(1 − 𝑥)𝑑𝑥
2 4
1𝑥 21 𝑥 3 𝑥 3𝑥 𝑥2 5
= ∫0 𝑑𝑥 + ∫1 𝑑𝑥 + ∫2 ( − ) 𝑑𝑥 = − − when 2 ≤ 𝑥 < 3 𝑥2 𝑥3
2 2 2 2 2 4 4 = 6[ − ]
2 3
= 3𝑥 2 − 2𝑥 3 , 𝑓𝑜𝑟 0 ≤ 𝑥 < 1
(iii) 𝑃[𝑋 ≤ 1.5] = 𝑃(0 ≤ 𝑋 ≤ 1) + 𝑃(1 ≤ 𝑋 ≤ 1.5)
1 1.5
0, 𝑓𝑜𝑟 𝑥 < 0
𝑥 1
=∫ 𝑑𝑥 + ∫ 𝑑𝑥 ∴ 𝐹(𝑥) = [3𝑥 2 − 2𝑥 3 , 𝑓𝑜𝑟 0 ≤ 𝑥 < 1
0 2 1 2 1, 𝑖𝑓 𝑥 ≥ 1
1
1 𝑥2 1 (iii) To find 𝑏 such that 𝑃(𝑋 < 𝑏) = 𝑃(𝑋 ≥ 𝑏)
= [ ] + [𝑥]1.5
2 2 0 2 1 𝑏 1
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥
1 1 1 0 𝑏
= + =
4 4 2 𝑏 1
6 ∫ (𝑥 − 𝑥 2 )𝑑𝑥 = 6 ∫ (𝑥 − 𝑥 2 )𝑑𝑥
0 𝑏

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 𝑏 1 3
𝑥2 𝑥3 𝑥2 𝑥3 (𝑥 − 1)4
[ − ] = [ − ] ⇒ 4𝑘 [ ] =1
2 3 0 2 3 𝑏 4 1

𝑏2 𝑏3 1 1 𝑏2 𝑏3 ⇒ 𝑘[(𝑥 − 1)4 ]13 = 1 ⇒ 𝑘[24 − 0] = 1

[ − ] = [ − ]−[ − ]
2 3 2 3 2 3 1
∴𝑘=
𝑏2 𝑏2 𝑏3 𝑏3 1 2𝑏 2 2𝑏 3 1 16
+ − − = ⇒ − = 0 𝑥≤1
2 2 3 3 6 2 3 6
1
1 Hence the CDF is 𝐹(𝑥) = [16 (𝑥 − 1)4 1 ≤ 𝑥 ≤ 3
⇒ 3𝑏 2 − 2𝑏 3 = ⇒ 6𝑏 2 − 4𝑏 3 − 1 = 0 0 𝑥>3
2
⇒ (2𝑏 − 1) = 0, (2𝑏 2 − 2𝑏 − 1) = 0 𝑃(𝑋 ≤ 2) = 𝑃(1 ≤ 𝑋 ≤ 2)

1 2 ± √4 + 8 2 ± 2√3 1 + √3 1 1 1 1
𝑏= ; 𝑏= = = = 𝐹(2) − 𝐹(1) = (2 − 1)4 − (1 − 1)4 = (1)4 − 0 =
2 4 4 2 16 16 16 16
1 1±√3 Example 4: The C.D.F of a continuous RV X is given by
Here 𝑏 = is the real valueand 𝑏 = lies outside the interval (0,1)
2 2 0, 𝑥<0
1 1 1
𝑖. 𝑒. , 𝑏 = lies in (0, 1) ∴ 𝑏 = 𝑥 2, 0≤𝑥≤
2 2
𝐹(𝑥) = 2
Example 3:A continuous random variable X has the distribution function 𝐹(𝑥) = 3 1
1− (3 − 𝑥)2 , ≤𝑥<3
0 𝑥≤1 25 2
[𝑘(𝑥 − 1)4 1 ≤ 𝑥 ≤ 3 find 𝑘, probability density function 𝑓(𝑥), 𝑃(𝑋 ≤ 2). { 1, 𝑥≥3
0 𝑥>3 Find the PDF of X and evaluate 𝑃(|𝑋| < 1) and 𝑃( 1/3 ≤ 𝑋 < 4) using both
the PDF and CDF.
Solution: First, we find the PDF 𝑓(𝑥) Solution:

We know that 𝑓(𝑥) =

d
dx
F (x) =
d
dx
 
k ( x  1) 4  4k ( x  1) 3 (i) We know that 𝑓(𝑥) =
d
𝐹(𝑥)
dx
0 𝑥≤1
0 𝑥<0
𝑓(𝑥) = [4𝑘(𝑥 − 1)3 1 ≤ 𝑥 ≤ 3 1
0 𝑥>3 2𝑥 0≤𝑥≤
∴ 𝑓(𝑥) = 2
To find 𝒌. 6 1
(3 − 𝑥) ≤𝑥<3
We know that
∞
∫−∞ 𝑓(𝑥)𝑑𝑥 =1 25 2
{ 0 𝑥≥3
3
𝑃(|𝑋| < 1) = 𝑃(−1 ≤ 𝑋 ≤ 1)
∫ 4𝑘(𝑥 − 1)3 = 1
1 1
1 0 1
2
= ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
1
−1 −1 0
2

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 1
6 1 𝑒 −𝑥 , 𝑥>0
2 𝑓(𝑥) = {
= 0 + 2 ∫ 𝑥 𝑑𝑥 + ∫ (3 − 𝑥) 𝑑𝑥 0, 𝑥≤0
0 25 1
2
Ans: (i) yes (ii) 𝑃(1 ≤ 𝑋 ≤ 2) = 0.233
1
2 6
2 2
= [𝑥 ]0 − [(3 − 𝑥)2 ]11
2 25(2) 2
MATHEMATICAL EXPECTATIONS
1 3 25 1 3 −9 13
= − [4 − ] = − [ ]= Definition:
4 25 4 4 25 4 25
Using the property of CDF Let 𝑋 be a random variable with p.d.f (p.m.f) 𝑓(𝑥). Then its mean or
𝑃[|𝑋| < 1] = 𝑃(−1 ≤ 𝑋 ≤ 1) mathematical expectation is denoted by 𝐸(𝑋) is given by
= 𝐹(1) − 𝐹(−1) 𝑋̅ = 𝐸(𝑋) = ∑𝑛𝑖=1 𝑥𝑖 𝑝𝑖 For discrete r.v
∞
3 13 𝑋̅ = 𝐸(𝑋) = ∫−∞ 𝑥𝑓(𝑥)𝑑𝑥 For continuous r.v
=1− (3 − 1)2 − 0 =
25 25
Properties of Expectation:
4
1
𝑃 ( ≤ 𝑋 ≤ 4) = ∫ 𝑓(𝑥) 𝑑𝑥 If 𝑋 and 𝑌 are random variables and 𝑎, 𝑏 are constants then
3 1
3
1. E(a) = a
1
3
2
= ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥 2. E(aX) = aE(X)
1 1
3 2 3. E(aX + b) = aE(X) + b
1
3
2 6 4. E(X + Y) = E(X) + E(Y)(Additive theorem)
= 2 ∫ 𝑥 𝑑𝑥 + ∫ (3 − 𝑥) 𝑑𝑥
1 25 1 5. E(XY) = E(X)E(Y) if X, Y are indepentdent random variables
3 2
1
6 6. E(X − X) = 0
2 2
= [𝑥 ] − 1 [(3 − 𝑥)2 ]31
3 25(2) 2
VARIANCE:
1 1 3 25 5 3 8
= − − [0 − ] = + = Let X be a random variable with mean E(X) then the variance of X is defined as
4 9 25 4 36 4 9
𝐸[𝑋 − E(X)]2 . It is denoted by Var(X) or σ2 𝑥 .
Using the property of CDF
1 1 1 8 If X is a discrete random variable taking values 𝑥1 , 𝑥2 , 𝑥3 , … … with probabilities
𝑃 ( ≤ 𝑋 ≤ 4) = 𝐹(4) − 𝐹 ( ) = 1 − =
3 3 9 9 𝑝(𝑥1 ), 𝑝(𝑥2 ), 𝑝(𝑥3 ), …. and E(X) = 𝜇 then 𝑉𝑎𝑟(𝑋) = ∑𝑖(𝑥𝑖 − 𝜇)2 𝑝(𝑥𝑖 )
Problem for Practice If X is a continuous random variable with pdf 𝑓(𝑥), 𝑥 ∈ (−∞, ∞) then 𝑉𝑎𝑟(𝑋) =
∞
1. (i) Is the function defined as follows a density function? (ii) If so determine ∫−∞(𝑥𝑖 − 𝜇)2 𝑓(𝑥)𝑑𝑥
the probability that the variate having this density will fall in the interval Properties of Variance
(1,2) If a & 𝑏 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 then

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 7

(𝑖) Var(aX) = a2 var (X)
∞
(ii) Var(aX + b) = a2 var (X) 3. Mean = μ1′ = ∑ x p (x) Mean = μ1′ = ∫ x f(x)dx
−∞
∞
MOMENTS: 4. μ′2 = ∑ x 2 p(x) μ′2 = ∫ x 2 f(x)dx
−∞
The expected value of an integral power of a random variable is called its
moment. 5.Variance = μ′2 − (μ1′ )2 Variance =μ′2 − (μ1′ )2
= E (X 2 ) − {E(X)}2 = E (X 2 ) − {E(X)}2
Definition: 1 ( 𝒓𝒕𝒉 moment about the origin of X)
The 𝑟 𝑡ℎ Moment about the origin of X is defined as E (X r ) and is denoted by
EXAMPLES BASED ON DISCRETE RANDOM VARIABLE
𝜇𝑟′
Example 1: Given the following probability distribution of X compute
∑ 𝑥𝑖𝑟 𝑝𝑖 = 𝜇𝑟′ , r ≥ 1 , for discrete r. v X (i) E(X) (ii) E(X2) (iii) E(2X  3) (iv) Var(2X  3)
𝑖
E (X r ) = ∞

∫ 𝑥 𝑟 𝑓(𝑥)𝑑𝑥 = μ′r , r ≥ 1 , 𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑟. 𝑣 𝑋

X -3 -2 -1 0 1 2 3
{−∞
Definition: 2( 𝒓𝒕𝒉 moment about the mean (or) central moments) 𝑝(𝑥) 0.05 0.10 0.30 0 0.30 0.15 0.10
The 𝑟 𝑡ℎ moment about the mean of a r.v X is defined as E [( X − X) r ] and is
Solution:
denoted by μr
(i) E(X) = ∑𝑥𝑝(𝑥) = (-3) (0.05) + (-2)(0.10) + (-1) (0.30) + 0 + (1)
For discrete R.V. X , E [( X − X) r ] = ∑i( X − X) r pi = μr
(0.30) + (2) (0.15) +(3) (0.10)
r ∞ r
For continuous R.V. X , E [( X − X) ] = ∫−∞( X − X) 𝑓(𝑥) 𝑑 𝑥 = μr = 0.25.
EXPECTATION TABLE (ii) E(X2) = 𝑥 2 𝑝( 𝑥)
= (-3)2 (0.15) + (-2)2 (0.10)+ (-1)2 (0.30) + 0 + (1)2 (0.30) +
Discrete random variables Continuous random variables (2)2 (0.15) + (3)2 (0.10)
∞ = 2.95
1. E(X) = ∑ x p (x) E (X) = ∫ x f(x)dx
−∞ (iii) E (2X  3) = 2 E(X) E(3) = 2(0.25)  3 = 0.5  3
∞ (iv) Var(2X  3) = 22Var(X)= 4 {E(X2)-[E(X)]2}
2. E(X r ) = μ′r = ∑ x r p(x) E(X r ) = μ′r = ∫ x r f(x)dx
−∞ = 4 {2.95 –(0.25)2}
= 4 [2.95 – 0.0625] = 11.55

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 ∞
Example 2: When a die is thrown ′𝑋′ denotes the number that turns up. Find
𝑘 ∫ 𝑒 −𝑥 𝑥 3−1 𝑑𝑥 = 1
𝐸(𝑋), 𝐸(𝑋)2 and Var(𝑋). 0
∞
Solution: ∴ 𝑘 Γ(3) = 1 ∵ Γ(𝑛) = ∫ 𝑒 −𝑥 𝑥 𝑛−1 𝑑𝑥 & Γ(3) = 2!
0
Given that ‘X’ denotes the number that turns up.
1
i.e., ‘X’ takes 1, 2, 3, 4, 5, 6 and with probability 1/6 for each ⇒ 2k = 1 ⇒ k =
2
∞
𝑋 1 2 3 4 5 6 Mean of X is defined as 𝐸(𝑋) = ∫−∞ 𝑥𝑓(𝑥)𝑑𝑥
p(x) 1/6 1/6 1/6 1/6 1/6 1/6 1 ∞ 3 −𝑥
𝐸(𝑋) = ∫ 𝑥 𝑒 𝑑𝑥
2 0

𝐸(𝑋) = 𝑥 𝑝(𝑥) 1 3!
= Γ(4) = = 3 ∵ Γ(4) = 3! = 6
2 2
1 1 1 1 1 1
= 1( ) + 2( ) + 3( ) + 4( )+ 5( ) + 6( ) Variance of X is defined as Var(𝑋) = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2
6 6 6 6 6 6
1 ∞
1  2  3  4  5  6 21 7 Where 𝐸(𝑋 2 ) = ∫0 𝑥 4 𝑒 −𝑥 𝑑𝑥
 
2
=
6 6 2 1 4!
= Γ(5) = = 12
𝐸(𝑋 2 ) = 𝑥 2 𝑝(𝑥) 2 2

1 1 1 1 1 1 Var(𝑋) = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2

= 1 ( ) + 4 ( ) + 9 ( ) + 16 ( ) + 25 ( ) + 36 ( )
6 6 6 6 6 6 = 12 − 9 = 3

1  4  9  16  25  36 91
= 
6 6 Example 2: The cumulative distribution function (CDF) of a random variable X is
Var(𝑋) = 𝐸(𝑋 2 ) – [𝐸(𝑋)]2 𝐹(𝑥) = 1 – (1 + 𝑥) 𝑒 −𝑥 , 𝑥 > 0. (i)Find the probability density function of X

91 49 364  294 70 35 Solution:

= – = = =
6 4 24 24 12 Given that 𝐹(𝑥) = 1 – (1 + 𝑥) 𝑒 −𝑥

d
We know that 𝑓(𝑥) = 𝐹(𝑥)
EXAMPLES UNDER CONTINUOUS RANDOM VARIABLES dx
Example 1: A continuous random variable X has a p.d.f 𝑓(𝑥) = 𝑘𝑥 2 𝑒 −𝑥 , 𝑥 ≥ 0. d
= [1 − (1 + 𝑥)𝑒 −𝑥 ]
Find k, mean and variance. dx
Solution: = (1 + 𝑥)𝑒 −𝑥 − 𝑒 −𝑥
∞
By property of p.d.f, ∫0 𝑘𝑥 2 𝑒 −𝑥 𝑑𝑥 = 1 = 𝑒 −𝑥 + 𝑥𝑒 −𝑥 – 𝑒 –𝑥 = 𝑥𝑒 −𝑥 , 𝑥 > 0

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 9

 ∞
Example 3: For the following density function 1
−|𝑥| = ∫ 𝑥 2 [ 𝑒 −|𝑥| ] 𝑑𝑥
𝑓(𝑥) = 𝑎𝑒 , −∞ < 𝑥 < ∞ Find (i) the value of ′𝑎′ (ii) Mean and Variance 2
−∞
Solution: ∞
1
−𝑥, 𝑥 < 0 = ∫ 𝑥 2 [𝑒 −|𝑥| ]𝑑𝑥
We know that |𝑥| = { 2
𝑥, 𝑥 > 0 −∞
−|𝑥| ∞
Given 𝑓(𝑥) = 𝑎𝑒 , −∞ < 𝑥 < ∞ 1
= 2 ∫ 𝑥 2 𝑒 −𝑥 𝑑𝑥
𝑎𝑒 𝑥 , 𝑥 < 0 2
= { −𝑥 0
𝑎𝑒 , 𝑥 > 0
∞
∞
(i) By the property of PDF ∫−∞ 𝑓(𝑥)𝑑𝑥 = 1 𝑒 −𝑥 𝑒 −𝑥
= [𝑥 2 [ ] − 2𝑥[𝑒 −𝑥 ] + 2 [ ]]
∞ 1 1
0
∫ 𝑎𝑒 −|𝑥| = 1 = (0 + 0 + 0) − (0 + 0 − 2) = 2
−∞
∞

⇒ 𝑎. 2 ∫ 𝑒 −𝑥 𝑑𝑥 = 1 Problem for Practice:

0 1. If a random variable X has the probability density function 𝑓(𝑥) =
−𝑥 ∞
𝑒 1 1
2𝑎 [ ] =1 ⇒ 2𝑎(0 + 1) = 1 ⇒ 𝑎 = (𝑥 + 1) , − 1 < 𝑥 < 1
−1 0 2 {2 Find the mean and variance.
0, otherwise
1
∴ 𝑓(𝑥) = 𝑒 −|𝑥| , −∞ < 𝑥 < ∞ 𝟏 𝟐
2 𝐀𝐧𝐬: 𝐌𝐞𝐚𝐧 = ; 𝐕𝐚𝐫𝐢𝐚𝐧𝐜𝐞 =
𝟑 𝟗
1 𝑥
𝑒 , 𝑥<0 MOMENT GENERATING FUNCTION (MGF):
= {2
1 −𝑥 The moment generating function (MGF) of a random variable 𝑋 about origin
𝑒 , 𝑥>0
2 ∞

(i) Mean ∑ etx p(x) For discrete probability distribution

∞ ∞ ∞ x=−∞
1 1 𝑀𝑋 (𝑡) = 𝐸(𝑒 𝑡𝑋 ) = ∞
= ∫ 𝑥𝑓(𝑥)𝑑𝑥 = ∫ 𝑥 𝑒 −|𝑥| 𝑑𝑥 = ∫ 𝑥𝑒 −|𝑥| 𝑑𝑥 = 0
2 2 ∫ etx f(x)dx For continuous probability distribution
−∞ −∞ −∞ {−∞
∵ 𝑥𝑒 −|𝑥| is an odd function Where ′𝑡′ is real parameter and the integration (or) summation being extended
∞
Variance = ∫−∞(𝑥 − 𝑚𝑒𝑎𝑛)2 𝑓(𝑥)𝑑𝑥 to the entire range of 𝑋.

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 ∞
Property of MGF: Prove that the 𝑟 𝑡ℎ moment of the R.V. X about origin is
∞ = ∑ 𝑝(𝑥)𝑒 𝑡𝑥
𝑡𝑟 𝑥=1
𝑀𝑋 (𝑡) = ∑ 𝜇𝑟 ′
𝑟! ∞
𝑟=0
= ∑ 𝑞 𝑥−1 𝑝 (𝑒 𝑡 )𝑥 ∵ 𝑝(𝑥) = 𝑞 𝑥−1 𝑝
Proof: We know that
𝑥=1
𝑀𝑋 (𝑡) = 𝐸(𝑒 𝑡𝑋 ) ∞
𝑝
𝑡𝑋 (𝑡𝑋)2 (𝑡𝑋)3 (𝑡𝑋)𝑟 = ∑ 𝑞 𝑥 (𝑒 𝑡 )𝑥
𝑞
= 𝐸 [1 + + + +∙∙∙∙∙∙∙ + +∙∙∙∙∙∙] 𝑥=1
1! 2! 3! 𝑟! ∞
𝑝
𝑡𝑋 𝑡 2𝑋 2 𝑡 3𝑋 3 𝑡𝑟𝑋𝑟 = ∑(𝑞𝑒 𝑡 )𝑥
= 𝐸(1) + 𝐸 ( ) + 𝐸 ( )+𝐸( ) +∙∙∙∙∙∙∙ +𝐸 ( ) +∙∙∙∙∙∙ 𝑞
1! 2! 3! 𝑟! 𝑥=1
∞
2 3 𝑟 𝑝
𝑡 𝑡 𝑡
= 1 + 𝑡𝐸(𝑋) + 𝐸(𝑋 2 ) + 𝐸(𝑋 3 ) +∙∙∙∙∙∙∙ + 𝐸(𝑋 𝑟 ) +∙∙∙∙∙∙∙∙∙ = (𝑞𝑒 𝑡 ) ∑(𝑞𝑒 𝑡 )𝑥−1
2! 3! 𝑟! 𝑞
𝑥=1
𝑡2 𝑡3 𝑡𝑟 𝑡 [1
= 1 + 𝑡𝜇1 ′ + 𝜇2 ′ + 𝜇3 ′ +∙∙∙∙∙∙∙ + 𝜇𝑟 ′ +∙∙∙∙∙∙∙ = 𝑝𝑒 + (𝑞𝑒 𝑡 ) + (𝑞𝑒 𝑡 )2 + (𝑞𝑒 𝑡 )3 +∙∙∙∙∙∙]
2! 3! 𝑟!
∞
𝑡𝑟 ′
𝑀𝑋 (𝑡) = ∑ 𝜇 = 𝑝𝑒 𝑡 [1 − (𝑞𝑒 𝑡 )]−1 ∵ (1 − 𝑥)−1 = 1 + 𝑥 + 𝑥 2 +∙∙∙∙∙∙∙∙
𝑟! 𝑟
𝑟=0
𝑝𝑒 𝑡
𝑡𝑟 𝑀𝑋 (𝑡) =
Thus 𝑟 𝑡ℎ moment = co efficient of 1 − 𝑞𝑒 𝑡
𝑟!

Note: To find mean and variance

The above result gives the MGF in terms of moments. Since 𝑀𝑋 (𝑡) generates 𝑑 𝑝𝑒 𝑡 (1 − 𝑞𝑒 𝑡 )(𝑝𝑒 𝑡 ) − 𝑝𝑒 𝑡 (−𝑞𝑒 𝑡 )
𝑀𝑋 ′ (𝑡) = [ ] =
𝑑𝑡 1 − 𝑞𝑒 𝑡 (1 − 𝑞𝑒 𝑡 )2
moments, it is called the moment generating function.
𝑝𝑒 𝑡
=[ ]
(1 − 𝑞𝑒 𝑡 )2
PROBLEMS BASED ON DISCRETE RANDOM VARIABLES
𝑑2 𝑝𝑒 𝑡 𝑑 𝑝𝑒 𝑡
Example 1: Find the MGF of the random variable with the probability law 𝑀𝑋 ′′ (𝑡) = [ ] = [ ]
𝑑𝑡 2 (1 − 𝑞𝑒 𝑡 ) 𝑑𝑡 (1 − 𝑞𝑒 𝑡 )2
𝑝(𝑥) = 𝑞 𝑥−1 𝑝 , 𝑥 = 1,2,3, …. Find the Mean and variance.
(1 − qet )2 (pet ) − pet [2(1 − qet )(−qet )]
Solution: =
(1 − qet )4
First we find MGF of X (1 − qet )[(1 − qet )pet + 2𝑝𝑞𝑒 2𝑡 ]
=
𝑀𝑋 (𝑡) = 𝐸(𝑒 𝑡𝑋 )by formula (1 − qet )4

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 11

 pet + pqe2t 𝑒𝑡 𝑒𝑡
−1
= = [1 − ]
(1 − qet )3 2 2
pet (1 + qet ) 𝑒𝑡 2 𝑒𝑡
= = [ ] =
(1 − qet )3 2 2 − 𝑒𝑡 2 − 𝑒𝑡
𝑝 𝑝 1 𝑒𝑡
Mean = 𝐸(𝑋) = 𝑀𝑋 ′ (0) = = = ∵𝑝+𝑞 =1 𝑀𝑋 (𝑡) =
(1 − 𝑞)2 𝑝2 𝑝 2 − 𝑒𝑡
𝑝(1 + 𝑞) 1 + 𝑞 𝑑 𝑒𝑡 2𝑒 𝑡
𝐸(𝑋 2 ) = 𝑀𝑋 ′′ (0) = = 𝑀𝑋 ′ (𝑡) = [ ] = [ ]
(1 − 𝑞)2 𝑝2 𝑑𝑡 2 − 𝑒 𝑡 (2 − 𝑒 𝑡 )2
1+𝑞 1 2 𝑞 𝑑2 𝑒𝑡 𝑑 2𝑒 𝑡 4𝑒 𝑡 + 2𝑒 2𝑡
Variance = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2 = 2
− ( ) = 2 𝑀𝑋 ′′ (𝑡) = [ ] = [ ] =
𝑝 𝑝 𝑝 𝑑𝑡 2 2 − 𝑒 𝑡 𝑑𝑡 (2 − 𝑒 𝑡 )2 (2 − 𝑒 𝑡 )3
Mean = 𝐸(𝑋) = 𝑀𝑋 ′ (0) = 2
Example 2:The probability function of an infinite discrete distribution is given by 𝐸(𝑋 2 ) = 𝑀𝑋 ′′ (0) = 6
1
𝑃(𝑋 = 𝑥) =
2𝑥
, 𝑥 = 1,2,3, … . ∞. Find the MGF, mean and variance of the Variance =𝐸(𝑋 2 ) − [𝐸(𝑋)]2 = 6 − 4 = 2

distribution. Also find P(X is even). To find P(X is even)

Solution: 𝑃(𝑋 = even) = 𝑃(𝑋 = 2) + 𝑃(𝑋 = 4) +∙∙∙∙∙∙∙

𝑀𝑋 (𝑡) = 𝐸(𝑒 𝑡𝑋 ) 1 1 1
= ( 2 ) + ( 4 ) + ( 6 ) +∙∙∙∙∙∙
∞ 2 2 2
= ∑ 𝑝(𝑥)𝑒 𝑡𝑥 1 2
1 2 1 4 1 6 ( ) 1
2
𝑥=1 = ( ) + ( ) + ( ) +∙∙∙∙∙∙∙ = =
2 2 2 1 2 3
∞ 1−( )
1 𝑡𝑥 2
=∑ 𝑒
2𝑥 𝑎 1 2 1 2
𝑥=1 ∵ The geometric series 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 +∙∙∙∙∙∙∙= ℎ𝑒𝑟𝑒 𝑎 = ( ) ; 𝑟 = ( )
∞ 𝑥 1−𝑟 2 2
𝑒𝑡
= ∑( ) EXAMPLES UNDER CONTINUOUS RANDOM VARIABLE
2
𝑥=1
Example 1:
𝑡 ∞ 𝑡 𝑥−1
𝑒 𝑒 Find the m.g.f of a. R.V X whose p.d.f is defined by 𝑓(𝑥) =
= ∑( )
2 2
𝑥=1 𝑥 , 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 1
𝑡 𝑡 𝑡 2 [2 − 𝑥 , 𝑓𝑜𝑟 1 < 𝑥 ≤ 2
𝑒 𝑒 𝑒
= [1 + ( ) + ( ) +∙∙∙∙∙∙∙∙∙] 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
2 2 2
Hence find mean and variance of X.

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 12

Solution: 𝑡 𝑡2 1 𝑡
𝑀𝑋′ (𝑡) = 2 (1 + + + ⋯ ) ( + + ⋯ )
∞ 2 6 2 3
𝑀𝑋 (𝑡) = ∫ 𝑒 𝑡𝑥 𝑓(𝑥)𝑑𝑥
1
−∞ ∴ 𝑀𝑋′ (0) = 2(1) ( ) = 1 ⇒ Mean= 1
2
1 2
= ∫ 𝑥𝑒 𝑡𝑥 𝑑𝑥 + ∫ (2 − 𝑥)𝑒 𝑡𝑥 𝑑𝑥 𝑡 𝑡2 1 1 𝑡 1 𝑡
0 1 𝑀𝑋′′ (𝑡) = 2 [1 + + + ⋯ ] [ + 0(𝑡)] + 2 [ + + ⋯ ] [ + + ⋯ ]
2 6 3 2 3 2 3
1 2
𝑒 𝑡𝑥 𝑒 𝑡𝑥 𝑒 𝑡𝑥 𝑒 𝑡𝑥 1 1 7 7
= [𝑥 ( ) − 2 ] + [(2 − 𝑥) + 2] ∴ 𝑀𝑋′′ (0) = 2 [ + ] = ⇒ 𝐸[𝑋 2 ] =
𝑡 𝑡 0 𝑡 𝑡 1 4 3 6 6
𝑒𝑡 𝑒𝑡 1 𝑒 2𝑡 𝑒𝑡 𝑒𝑡 7 1
= [( − 2 ) − (0 − 2 )] + [(0 + 2 ) − ( + 2 )] Variance 𝑜𝑓 𝑋 = 𝐸[𝑋 2 ] − {𝐸(𝑋)}2 = −1=
𝑡 𝑡 𝑡 𝑡 𝑡 𝑡 6 6
1
𝑒 𝑡 𝑒 𝑡 1 𝑒 𝑡 𝑒 2𝑡 𝑒 𝑡 Hence Mean = 1 and Variance =
= − + − + 2 − 2 6
𝑡 𝑡2 𝑡2 𝑡 𝑡 𝑡
2 𝜆𝑒 −𝜆𝑥 , 𝑥≥0
1 − 2𝑒 𝑡 + 𝑒 2𝑡 1 − 𝑒𝑡 Example 2:Find the m.g.f of a R.V with p.d.f 𝑓(𝑥) = {
= =( ) 0, otherwise
𝑡 2 𝑡
Hence find the first four moments about the origin.
(1 − 𝑒 𝑡 )2
∴ 𝑀𝑋 (𝑡) = Solution:
𝑡2 ∞
To find mean and variance 𝑀𝑋 (𝑡) = ∫ 𝑒 𝑡𝑥 𝜆𝑒 −𝜆𝑥 𝑑𝑥
0
2
(1 − 𝑒 𝑡 )2 1 𝑡 𝑡2 ∞
𝑀𝑋 (𝑡) = = [1 − (1 + + + ⋯ )] = 𝜆 ∫ 𝑒 −[𝜆−𝑡]𝑥 𝑑𝑥
𝑡2 𝑡2 1! 2!
0
2
1 t t2 𝑡 𝑡2 𝑒 −(𝜆−𝑡)𝑥
∞
= 2 [− − − ⋯ ] ∵ 𝑒𝑡 = 1 + + + ⋯ … … = 𝜆 [− ]
𝑡 1! 2! 1! 2! 𝜆−𝑡 0
𝜆
2 2 ∴ 𝑀𝑋 (𝑡) =
𝑡 𝑡 𝜆−𝑡
= (−1 − − −⋯)
2 6 𝑡 −1 𝑡 𝑡2 𝑡3
Also 𝑀𝑋 [𝑡] = [1 − ] = 1+ + + +⋯
𝜆 𝜆 𝜆2 𝜆3
2 2
𝑡 𝑡
= (1 + + + ⋯ ) The rth moment about the origin is given by
2 6
𝑡𝑟
𝑑 𝜇𝑟′ = 𝑐𝑜 − 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛 𝑀𝑋 [𝑡]
𝜇1′ = [ 𝑀𝑋 (𝑡)] = 𝑀𝑋′ (0) 𝑟!
𝑑𝑡 𝑡 =0 𝑟!
∴ 𝜇𝑟′ = , 𝑟 = 1,2 …
𝜆𝑟,

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 13

 1 2 6 24
⇒ 𝜇1′ = , 𝜇2′ = ; 𝜇3′ = and 𝜇4′ = Each trial has two possible outcomes, generally called success and failure. Such a
𝜆 𝜆2 𝜆3 𝜆4
trial is known as Bernoulli trial. The sample space for a Bernoulli trial is S = {s, f}
Example:
Problems for Practice
1. A toss of a single coin [head or tail]
1. A random variable X has density function given by
2. The throw of a die [even or odd numbers]
2𝑒 −2𝑥 , 𝑥 ≥ 0
𝑓(𝑥) = { . Obtain the m.g.f and the first four moments about the
0, 𝑥 < 0 DEFINITION: BINOMIAL DISTRIBUTION
𝟐 𝟏 𝟑 𝟑
origin. 𝐀𝐧𝐬: ; 𝝁′𝟏 = 𝝁′𝟐 = ; 𝝁′𝟑 = ; 𝝁′𝟒 = ; If the discrete R.V. X can take the values x = 0, 1, 2,…, n, such that 𝑃(𝑋 = 𝑥) =
𝟐−𝒕 𝟐 𝟒 𝟐
1 𝑥⁄ 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 where 𝑝 + 𝑞 = 1,
𝑒− 2, 𝑥>0
2. Find m.g.f corresponding to the distribution 𝑓(𝑥) = {2 and
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 then X is said to follow a Binomial distribution with parameters 𝑛 and 𝑝, denoted

hence find its mean and variance. by 𝐵(𝑛, 𝑝).

𝟏 Note:
𝐀𝐧𝐬: 𝑴𝒙 [𝒕] = ; 𝑬(𝑿) = 𝟐; 𝑬[𝑿]𝟐 = 𝟖; 𝒗𝒂𝒓[𝑿] = 𝟒
𝟏 − 𝟐𝒕 1. In the binomial distribution, p denote the probability of success in any trial
3. The density function of a random variable is given by and q denote the probability of failure.
𝑘𝑥(2 − 𝑥), 0 ≤ 𝑥 ≤ 2 2. ∑ 𝑃(𝑋 = 𝑥) = ∑𝑛𝑥=0 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 = (𝑞 + 𝑝)𝑛 = 1 ∵𝑞+𝑝=1
𝑓(𝑥) = {
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
Find the value of 𝑘, the mean , variance and 𝑟 𝑡ℎ moment.
𝟑 𝟔 𝟏 MOMENT GENERATING FUNCTION, MEAN AND VARIANCE OF BINOMIAL
Ans: 𝒌 = ; 𝑴𝒆𝒂𝒏 𝒐𝒇 𝑿 = 𝑬[𝑿] = 𝟏; 𝑬[𝑿𝟐 ] = ; 𝑽𝒂𝒓(𝑿) = ;
𝟒 𝟓 𝟓
DISTRIBUTION:
𝟑 𝟐𝒓+𝟑
𝝁′𝒓 = 𝑬[𝑿 𝒓]
= [(𝒓+𝟐)(𝒓+𝟑)] Find the moment generating function (MGF) of a Binomial distribution about
𝟒

origin and hence find mean and variance of a Binomial distribution.

DISCRETE DISTRIBUTIONS Proof:
The following distributions are discrete probability distributions: We know that the MGF of a random variable ‘X’ is 𝑀𝑋 [𝑡] =
1. Binomial Distribution ∑𝑛𝑥=0 𝑒 𝑡𝑥 𝑝(𝑥) [ ∵ 𝑝(𝑥) is pmf ]

2. Poisson Distribution Let X denote the random variable which follows binomial distribution, then MGF
3. Geometric Distribution about origin is

BINOMIAL DISTRIBUTION

BERNOULLI TRIAL :

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 14

 𝑛
Result:
𝑀𝑋 [𝑡] = 𝐸[𝑒 𝑡𝑥 ] = ∑ 𝑒 𝑡𝑥 𝑝(𝑥)
Find the moment generating function (MGF) of a binomial distribution about
𝑥=0
𝑛 mean (𝑛𝑝)
𝑡𝑥 𝑥 𝑛−𝑥 𝑥 𝑛−𝑥
= ∑ 𝑒 𝑛𝐶𝑥 𝑝 𝑞 ∵ 𝑝(𝑥) = 𝑛𝐶𝑥 𝑝 𝑞 Proof:
𝑥=0
𝑛 We know that the MGF of a random variable X about any point ‘𝑎’ is
= ∑(𝑝𝑒 𝑡 )𝑥 𝑛𝐶𝑥 𝑞 𝑛−𝑥 𝑀𝑋 (𝑡)( 𝑎𝑏𝑜𝑢𝑡 𝑥 = 𝑎) = 𝐸[𝑒 𝑡(𝑥−𝑎) ]
𝑥=0
𝑛
Here ‘a’ is mean of a binomial distribution
= ∑ 𝑛𝐶𝑥 𝑞 𝑛−𝑥 (𝑝𝑒 𝑡 )𝑥 𝑖. 𝑒. , 𝑀𝑋 (𝑡)( 𝑎𝑏𝑜𝑢𝑡 𝑥 = 𝑛𝑝) = 𝐸[𝑒 𝑡(𝑥−𝑛𝑝) ]
𝑥=0
= 𝐸[𝑒 𝑡𝑥 . 𝑒 −𝑡𝑛𝑝 ]
= 𝑞 𝑛 + 𝑛𝐶1 𝑞 𝑛−1 (𝑝𝑒 𝑡 )1 + 𝑛𝐶2 𝑞 𝑛−2 (𝑝𝑒 𝑡 )2 +∙∙∙∙∙∙ +(𝑝𝑒 𝑡 )𝑛
= 𝑒 −𝑡𝑛𝑝 . 𝐸[𝑒 𝑡𝑥 ] ∵ E(C) = C
𝑡 )𝑛
𝑀𝑋 [𝑡] = (𝑞 + 𝑝𝑒
= 𝑒 −𝑡𝑛𝑝 x 𝑀𝐺𝐹 𝑜𝑓 𝑋 𝑎𝑏𝑜𝑢𝑡 𝑜𝑟𝑖𝑔𝑖𝑛
is the MGF of a Binomial distribution
= 𝑒 −𝑡𝑛𝑝 . (𝑞 + 𝑝𝑒 𝑡 )𝑛
Mean and Variance of a Binomial distribution
∴ 𝑀𝑋 (𝑡) = (𝑒 −𝑡𝑝 )𝑛 . (𝑞 + 𝑝𝑒 𝑡 )𝑛
We know that 𝑀𝑋 [𝑡] = (𝑞 + 𝑝𝑒 𝑡 )𝑛
Note:
𝑀𝑋 ′ [𝑡] = 𝑛(𝑞 + 𝑝𝑒 𝑡 )𝑛−1 𝑝𝑒 𝑡 = (𝑛𝑝)(𝑞 + 𝑝𝑒 𝑡 )𝑛−1 . 𝑒 𝑡 … … … . (1)
The first four moments of Binomial distributions are
𝑀𝑋 ′′ [𝑡] = (𝑛𝑝)[(𝑞 + 𝑝𝑒 𝑡 )𝑛−1 . 𝑒 𝑡 + 𝑒 𝑡 (𝑛 − 1)(𝑞 + 𝑝𝑒 𝑡 )𝑛−2 . 𝑝𝑒 𝑡 ] … … … . (2)
(i) Mean = 𝐸(𝑋) = 𝑛𝑝
∵ 𝑑(𝑢𝑣) = 𝑢𝑣 ′ + 𝑣𝑢′
(ii) Variance(𝜇2 ) = 𝑛𝑝𝑞
Put 𝑡 = 0 in (1) we get
(iii) (𝜇3 ) = 𝑛𝑝𝑞(𝑞 − 𝑝)
Mean = 𝐸(𝑋) = 𝑀𝑋 ′ (0) = (𝑛𝑝)(𝑞 + 𝑝)𝑛−1 . 1 = 𝑛𝑝 ∵ 𝑞 + 𝑝 = 1
(iv) (𝜇4 ) = 𝑛𝑝𝑞[1 − 6𝑝𝑞 − 3𝑛𝑝𝑞]
Put 𝑡 = 0 in (2), we get
Example 1: The mean and S.D of a Binomial distribution are 5 and 2. Determine
𝐸(𝑋 2 ) = 𝑀𝑋 ′′ (0) = (𝑛𝑝)[(𝑞 + 𝑝)𝑛−1 + (𝑛 − 1)(𝑞 + 𝑝)𝑛−2 . 𝑝]
the distribution.
= 𝑛𝑝[1 + (𝑛 − 1)𝑝]
Solution: The Binomial distribution is
= 𝑛𝑝 + 𝑛2 𝑝2 − 𝑛𝑝2
𝑃(𝑋 = 𝑥) = 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 , 𝑥 = 0,1,2, … … . . (1)
2 2
= 𝑛 𝑝 + 𝑛𝑝(1 − 𝑝)
To find 𝑝, 𝑞and 𝑛.
2)
𝐸(𝑋 = 𝑛2 𝑝2 + 𝑛𝑝𝑞
Given Mean = 𝑛𝑝 = 5 … … . . (2)
Variance 𝑜𝑓 𝑋 = 𝐸[𝑋 2 ] − {𝐸(𝑋)}2 = 𝑛2 𝑝2 + 𝑛𝑝𝑞 − 𝑛2 𝑝2 = 𝑛𝑝𝑞
𝑆. 𝐷 = √𝑛𝑝𝑞 = 2

𝑖. 𝑒. , 𝑛𝑝𝑞 = 4 ∙∙∙∙∙∙∙∙∙ (3)

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 15

(3) 𝑛𝑝𝑞 4 4 6 1
⇒ = ⇒𝑞= 𝑝 = probability of getting a doublet = =
(2) 𝑛𝑝 5 5 36 6
4 1 1 5
∴𝑝 =1−𝑞 =1− 𝑖. 𝑒. , 𝑝 = ∙∙∙∙∙∙∙∙∙ (4) ∴ 𝑞 = 1−𝑝 =1− = ; here 𝑛 = 4
5 5 6 6
1 We know that 𝑃(𝑋 = 𝑥) = 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥
Substituting (4) in (2), we get 𝑛 ( ) = 5 ⇒ 𝑛 = 25
5
1 2 1 2 4x3 1 25 25
Hence, (1) becomes 𝑃(𝑋 = 2) = 4𝐶2 ( ) ( ) = x x =
6 6 1 x 2 36 36 216
1 𝑥 4 25−𝑥
𝑃(𝑋 = 𝑥) = 25𝐶𝑥 ( ) ( ) , 𝑥 = 0,1,2, … …is the required Binomial Note:
5 5

distribution. If we assume that n trials constitute a set and if we consider N sets, the

Example 2: Four coins are tossed simultaneous, what is the probability of getting frequency function of the binomial distribution is given by

(i) 2 heads (ii) at least 2 heads (iii) at most 2 heads. 𝑓(𝑥) = 𝑁𝑃(𝑥) = 𝑁 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥

Solution: We know that 𝑃(𝑋 = 𝑥) = 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 Example 4: Six dice are thrown 729 times. How many times do you expect at

1 1 least three dice to show a five or six?

Here 𝑝 = , 𝑞= 𝑎𝑛𝑑 𝑛 = 4
2 2 Solution:
1 2 1 2 4x3 1 1 3
(i) 𝑃(𝑋 = 2) = 4𝐶2 ( ) ( ) = ( )( ) = Let ‘X’ be the R.V. denoting the number of successes when 6 dice are thrown.
2 2 1x2 4 4 8
𝑝 = probability of getting 5 or 6 with one die.
(ii) 𝑃(𝑋 ≥ 2) = 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) + 𝑃(𝑋 = 4)
1 1 1
1 2 1 2 1 3 1 1 4 𝑝= + =
= 4𝐶2 ( ) ( ) + 4𝐶3 ( ) ( ) + 4𝐶4 ( ) . 1 6 6 3
2 2 2 2 2
1 2
3 24 1 1 6 + 4 + 1 11 𝑞 =1−𝑝 =1− = , 𝑛=6 ∵ 𝑁 = 729
= + ( )+ = = 3 3
8 6 16 16 16 16
𝑃 (at least three dice showing five or six) = 𝑃(𝑋 ≥ 3)
(iii) 𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)
= 𝑃(𝑋 = 3) + 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5) + 𝑃(𝑋 = 6)
1 0 1 4 1 1 3 1 2 1 2
= 4𝐶0 ( ) ( ) + 4𝐶1 ( ) ( ) + 4𝐶2 ( ) ( ) 1 3 2 6−3 1 4 2 6−4 1 5 2 6−5 1 6 2 6−6
2 2 2 2 2 2 = 6𝐶3 ( ) ( ) + 6𝐶4 ( ) ( ) + 6𝐶5 ( ) ( ) + 6𝐶6 ( ) ( )
3 3 3 3 3 3 3 3
1 1 4 3 1 1 3 11
= + 4( ) + = + + = 1 1 1 1 233
16 2 8 16 4 8 16 = 160 x 6 + 60 x 6 + 12 x 6 + 1 x 6 = 6
3 3 3 3 3
Example 3: A pair of dice is thrown 4 times. If getting a doublet is considered a
success, find the probability of 2 successes.

For 729 times, the expected number of times at least 3 dice showing five or six=
Solution: 233 233
𝑁x = 729 x = 233 times
In a throw of pair dice the doublet are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6). 36 36

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 16

Example 5: In a large consignment of electric bulbs, 10 percent are defective. A 2. Out of 800 families with 4 children each how many families would be
random sample of 20 is taken for inspection. Find the probability that (1) all expected to have (i) 2 boys and 2 girls (ii) atleast 1 boy (iii) atmost 2
are good bulbs (2) at most there are 3 defective bulbs (3) exactly there are 3 girls (iv) children of both sexes. Assume equal probability for boys and girls.
defective bulbs. 𝐀𝐧𝐬: (𝒊)𝟑𝟎𝟎 (𝒊𝒊) 𝟕𝟓𝟎 (𝒊𝒊𝒊)𝟓𝟓𝟎 (𝒊𝒗) 𝟕𝟎𝟎
Solution: Let X denote the number of defective bulbs in 20
1
Here 𝑛 = 20 , 𝑝 = 10% = POISSON DISTRIBUTION
10
1 9 DEFINITION:
∴𝑞 = 1−𝑝 =1− =
10 10 If the discrete R.V. X can take the values X = 0, 1, 2, 3…. Such that 𝑃(𝑋 = 𝑥) =
The random variable 𝑋 follows binomial distribution with parameters 𝑛 and 𝑝 𝑒 −𝜆 𝜆𝑥
𝑃(𝑥) = , 𝑥 = 0,1,2,3, … … ∞
𝑥!
∴ 𝑃(𝑋 = 𝑥) = 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 , 𝑥 = 0,1,2, … … 𝑛
Then X is said to follow a Poisson distribution with parameter λ.
1 𝑥 1 20−𝑥
= 20𝐶𝑥 ( ) ( ) , 𝑥 = 0,1,2, … , 20 POISSON DISTRIBUTION IS A LIMITING CASE OF BINOMIAL DISTRIBUTION
10 10
(1) P( all are good )= P( no defective ) Proof:

9 20 We know that the binomial distribution is 𝑃(𝑋 = 𝑥) = 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 , 𝑥 =

𝑛
= 𝑃(𝑋 = 0) = 𝑞 = ( ) = 0.1216
10 0,1,2, … … 𝑛
(2) 𝑃(𝑋 ≤ 3) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) To consider it under limiting case when
9 20 1 9 19 1 2 9 18 (i) n is indefinitely large i.e., 𝑛 → ∞
= ( ) + 20𝐶1 ( ) ( ) + 20𝐶2 ( ) ( )
10 10 10 10 10 (ii) p is very small such that 𝑝 → 0
1 3 9 17 𝜆 𝜆
+ 20𝐶3 ( ) ( ) 𝑛𝑝 = 𝜆( 𝑎 𝑓𝑖𝑛𝑖𝑡𝑒 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦) ⇒ 𝑝 = 𝑎𝑛𝑑 𝑞 = 1 − 𝑝 = 1 −
10 10 𝑛 𝑛
9 20 919 918 917 𝑛!
= ( ) + 20 x 20 + 190 x 20 + 1140 x 20 𝑃(𝑋 = 𝑥) = 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 ∵ 𝑛𝐶𝑥 =
10 10 10 10 (𝑛 − 𝑥)! 𝑥!
= 0.1216 + 0.2702 + 0.2852 + 0.1901 = 0.8671 𝑛 (𝑛 − 1)(𝑛 − 2) ∙∙∙∙∙∙∙ (𝑛 − 𝑥) ∙∙∙∙∙∙∙∙∙ 2.1 𝜆 𝑥 𝜆 𝑛−𝑥
= ( ) (1 − )
1 3 9 7 [1.2 … … … (𝑛 − 𝑥)]𝑥! 𝑛 𝑛
(3) 𝑃(𝑋 = 3) = 20𝐶3 ( ) ( ) = 0.1901
10 10

𝑛 (𝑛 − 1)(𝑛 − 2) ∙∙∙∙∙∙∙ (𝑛 − 𝑥 + 1) 𝜆 𝑥 𝜆 𝑛−𝑥

Problems for Practice = ( ) (1 − )
𝑥! 𝑛 𝑛
1. The mean and variance of a binomial variate are 8 and 6. Find 𝑃(𝑋 ≥ 2) 1 2 (𝑥−1
𝑛 𝑥 (1 − ) (1 − ) … … . (1 − ) λ𝑥 𝜆 𝑛−𝑥
𝑛 𝑛 𝑛
𝟏 𝟑
𝐀𝐧𝐬: 𝒑 = , 𝒒 = , 𝑷(𝑿 ≥ 𝟐) = 𝟎. 𝟗𝟗𝟔𝟒 = (1 − ) ∙∙∙∙∙∙∙∙∙∙∙∙ (1)
𝟒 𝟒 𝑥! 𝑛𝑥 𝑛
Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 17
 λ 𝑛−𝑥 𝑡 −1)
We know that lim (1 − ) = 𝑒 −λ and ∴ 𝑀𝑋 (𝑡) = 𝑒 λ(𝑒
𝑛→∞ 𝑛
𝑡 −1)
1 2 𝑥−1 𝑀𝑋 ′ (𝑡) = 𝑒 λ(𝑒 λ𝑒 𝑡
lim (1 − ) = lim (1 − ) = ∙∙∙∙∙∙∙∙ lim {1 − ( )} = 1 𝑡 −1) 𝑡 −1)
𝑛→∞ 𝑛 𝑛→∞ 𝑛 𝑛→∞ 𝑛 𝑀𝑋 ′′ (𝑡) = 𝑒 λ(𝑒 λ𝑒 𝑡 + λ𝑒 𝑡 . 𝑒 λ(𝑒 . λ𝑒 𝑡
Hence (1) becomes ∴ 𝑀𝑋 ′ (0) = 𝐸(𝑋) = 𝑚𝑒𝑎𝑛 = λ
𝑛 𝑥 λ𝑥 −λ
𝑃(𝑋 = 𝑥) = .𝑒 𝑤ℎ𝑒𝑛 𝑛 → ∞ 𝑀𝑋 ′′ (0) = 𝐸(𝑋 2 ) = λ[1 + λ] = λ + λ2
𝑥! 𝑛 𝑥
∴ 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝐸(𝑋 2 ) − 𝐸[𝑋]2 = λ + λ2 − λ2 = λ
𝑒 −λ λ𝑥
∴ 𝑃(𝑋 = 𝑥) = 𝑥 = 0,1,2, … … . ∞
𝑥!
is the probability mass function of the Poisson random variable ‘X’ RESULT : State and prove the additive property of independent Poisson variate.
(OR) Prove that Sum of the independent Poisson variates is also a Poisson
MGF, MEAN AND VARIANCE OF POISSON DISTRIBUTION variate.
Find the moment generating function of the Poisson distribution and hence find Statement: If 𝑥1 , 𝑥2 , … . . , 𝑥𝑛 are ‘𝑛’ independent Poisson variates with
mean and variance. parameters 𝜆1 , 𝜆2 , … . . , 𝜆𝑛 then 𝑋 = ∑𝑛𝑖=1 𝑋𝑖 is also a Poisson variate with
Solution: parameter is 𝜆 = 𝜆1 + 𝜆2 + , … . . , + 𝜆𝑛
We know that MGF of X is Solution:
∞ We know that the MGF of the Poisson variate𝑋𝑖 is given by
∴ 𝑀𝑋 (𝑡) = ∑ 𝑒 𝑡𝑥 𝑝(𝑥) 𝑡 −1)
∴ 𝑀𝑋𝑖 (𝑡) = 𝑒 λi (𝑒 , 𝑖 = 1,2, … . . 𝑛
𝑥=0
∞ 𝑀𝑋1+𝑋2+𝑋3…….𝑋𝑛 (𝑡) = 𝑀𝑋1 (𝑡). 𝑀𝑋2 (𝑡) … … . . 𝑀𝑋𝑛 (𝑡)
𝑡𝑥
𝑒 −λ λ𝑥 𝑒 −λ λ𝑥
= ∑𝑒 ( ) ∵ 𝑝(𝑥) = 𝑡 −1) 𝑡 −1) 𝑡 −1)
𝑥! 𝑥! = 𝑒 λ1(𝑒 .𝑒 λ2(𝑒 … … … 𝑒 λn (𝑒
𝑥=0
∞ 𝑡 −1)
(λ𝑒 𝑡 )𝑥 . 𝑒 −λ = 𝑒 (λ1 +λ2 +⋯….λ𝑛 )(𝑒
=∑ 𝑡 −1)
𝑥! ∴ 𝑀𝑋 (𝑡) = 𝑒 λ(𝑒
𝑥=0
∞
(λ𝑒 𝑡 )𝑥 Which gives the MGF of the Poisson variate with parameter
−λ
=𝑒 ∑
𝑥! 𝜆 = 𝜆1 + 𝜆2 + , … . . , + 𝜆𝑛
𝑥=0
REMARK:

(λ𝑒 𝑡 ) (λ𝑒 𝑡 )2 (λ𝑒 𝑡 )3 When an event occurs rarely, the number of occurrences of such an event may
= 𝑒 −λ [1 + + + +∙∙∙∙∙∙∙∙∙]
1! 2! 3! be assumed to follow a Poisson distribution. The following are some of the

= 𝑒 −λ . 𝑒 λ𝑒 𝑡
=𝑒 λ(𝑒 𝑡 −1) examples, which may be analyzed using Poisson distribution.

To Find Mean and Variance (i) The number of telephone calls received at a telephone exchange

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 18

 in a given time interval. Solution:
(ii) The number of defective articles in a packet of 100 Let 𝑋 be the R.V. denoting the number of defective bulbs.
3 1 3
Example 1: If X is a Poisson variate such that 𝑃(𝑋 = 1) = and 𝑃(𝑋 = 2) = , Given 𝑝 = 𝑃 (a bulb is defective) = = 0.03 and 𝑛 = 100
10 5 100

find 𝑃(𝑋 = 0) and 𝑃 (𝑋 = 3). Mean = 𝜆 = 𝑛𝑝 = 100 x 0.03 = 3

𝑒 −λ λ𝑥 𝑒 −λ λ𝑥
Solution: We know that 𝑃(𝑋 = 𝑥) = Given that We know that 𝑃(𝑋 = 𝑥) =
𝑥! 𝑥!

3 𝑒 −λ λ1 3 3 𝑒 −3 (3)5
𝑃(𝑋 = 1) = ⇒ = ⇒ 𝑒 −λ 𝜆 = ∙∙∙∙∙∙∙∙ (1) ∴ 𝑃(exactly 5 bulbs are defective) = 𝑃(𝑋 = 5) =
10 1! 10 10 5!
1 𝑒 −λ λ2 1 2 0.0498 x 243
𝑃(𝑋 = 2) = ⇒ = ⇒ 𝑒 −λ 𝜆2 = ∙∙∙∙∙∙∙∙ (2) = = 0.1008
5 2! 5 5 120
(2) 𝑒 −λ 𝜆2 4 4 Example 4: If 𝑋 is a Poisson R.V such that (𝑋 = 2) = 9𝑃(𝑋 = 4) + 90𝑃(𝑋 = 6) ,
∴ ⇒ = ⇒ 𝜆=
(1) 𝑒 −λ λ1 3 3 then find the variance, mean , 𝐸(𝑋 2 ) and 𝑃(𝑋 ≥ 2).
4
4 0 𝑒 −λ λ𝑥
𝑒 −3 ( ) 4 Solution: Poisson distribution is 𝑃(𝑋 = 𝑥) =
𝑥!
, 𝑥 = 0,1,2, …
3 −
𝑃(𝑋 = 0) = =𝑒 3 = 0.2635
0! Given 𝑃(𝑋 = 2) = 9𝑃(𝑋 = 4) + 90𝑃(𝑋 = 6)
−
4
4 3
𝑒 3 ( )
3
𝑒 −λ 𝜆2 𝑒 −λ 𝜆4 𝑒 −λ 𝜆6
𝑃(𝑋 = 3) = = 0.1041 ⇒ = 9. + 90
3! 2! 4! 6!
Example 2: The number of monthly breakdown of a computer is R.V having 1 3 1
⇒ = 𝜆2 + 𝜆4 ( 𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 𝑒 −λ 𝜆2 )
2 8 8
Poisson distribution with mean equal to 1.8. Find the probability that this
⇒ 𝜆4 + 3𝜆2 − 4 = 0 ⇒ (𝜆2 + 4)(𝜆2 − 1) = 0 ⇒ 𝜆 = 1 [∵ 𝜆 > 0]
computer will function for a month with only one breakdown.
We know that 𝑉𝑎𝑟 (𝑋) = 𝜆 = 1
Solution:
𝑀𝑒𝑎𝑛 = 𝜆 = 1 ; 𝐸(𝑋 2 ) = 𝜆2 + 𝜆 = 2
Let X be the R.V. denoting the monthly breakdown of a computer.
𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 < 2)
Here, Mean = λ = 1.8
= 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1)] = 1 − [𝑒 −1 + 𝑒 −1 ] = 1 − 2𝑒 −1
𝑒 −λ λ𝑥 𝑒 −1.8 (1.8)𝑥
We know that 𝑃(𝑋 = 𝑥) = =
𝑥! 𝑥! Example 5: A manufacturer of cotter pins knows that 5% of his product is
𝑒 −1.8 (1.8)
P(Computer will function for 1 month) = 𝑃(𝑋 = 1)= = 0.2975 defective. If he sells cotter pins in boxes of 100 and guarantees that not more
1!

Example 3 : If 3% of the electric bulbs manufactured by a company are defective, than 10 pins will be defective. What is the approximate probability that a box

find the probability that in a sample of 100 bulbs exactly 5 bulbs are defective. will fail to meet the guaranteed quality?

Given 𝑒 −3 = 0.0498 Solution:

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Let X be the R.V denoting the no. of defective cotter pins. A discrete random variable ‘X’ is said to follow Geometric distribution, if it
5 assumes only non-negative values and its probability mass function is given by
𝐺𝑖𝑣𝑒𝑛 𝑛 = 100 , 𝑝 = 5% = = 0.05
100 𝑃(𝑋 = 𝑥) = 𝑞 𝑥−1 𝑝 , 𝑥 = 1,2, … … ∞ , 0 < 𝑝 ≤ 1 , where 𝑞 = 1 − 𝑝
∴ 𝑀𝑒𝑎𝑛 = 𝜆 = 𝑛𝑝 = 100 x 0.05 = 5
NOTE:
𝑒 −λ λ𝑥 𝑒 −5 5𝑥
The Poisson distribution is 𝑃(𝑋 = 𝑥) = = Suppose that we have a series of independent trails. If the first success is to come
𝑥! 𝑥!

𝑃(a box will fail to meet the guaranteed quality) = 𝑃(𝑋 > 10) on the 𝑘 𝑡ℎ trial, it has to be proceeded by ‘𝑘 − 1’ failures. If we take the
= 1 − 𝑃(𝑋 ≤ 10) probability of a success is 𝑝, then the probability of ‘𝑘 − 1’ failures in ‘𝑘 − 1’
= 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) +∙∙∙∙∙∙∙∙∙∙ 𝑃(𝑋 = 10)] trials is (1 − 𝑝)𝑘−1 𝑖. 𝑒. , 𝑞 𝑘−1
𝑒 −5 50 𝑒 −5 51 𝑒 −5 52 𝑒 −5 53 𝑒 −5 54 𝑒 −5 55 𝑒 −5 56 𝑒 −5 57 Therefore, the probability of getting the first success on the 𝑘 𝑡ℎ trial is given by
=1−[ + + + + + + +
0! 1! 2! 3! 4! 5! 6! 7! 𝑞 𝑘−1 𝑝, 𝑘 = 1,2, … . . 0 < 𝑝 ≤ 1
𝑃(𝑋 = 𝑘) = {
𝑒 −5 8
5 𝑒 5−5 9
𝑒 5 −5 10 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
+ + + ]
8! 9! 10!
= 1 − 𝑒 −5 [1 + 5 + 125 + 20.8333 + 26.0417 + 26.0417 + 21.7014 + 15.5010 MGF, MEAN AND VARIANCE OF A GEOMETRIC DISTRIBUTION
+ 9.6881 + 5.3823 + 2.6911] Find the moment generating function of geometric distribution and hence find
= 1 − 0.9863 = 0.014 mean and variance of a geometric distribution.
Problem for Practice: Solution:
1. A manufacturer knows that the condensers he makes contain on the The MGF of X is MX [t] = E[etx ]
∞
average 1% of defectives. He packs them in boxes of 100. What is the
= ∑ etx P(X = x)
probability that a box picked at random will contain 3 or more faulty x=1
condensers? Ans: 𝝀 = 𝟏, 𝑷(𝑿 ≥ 𝟑) = 𝟎. 𝟎𝟖𝟎𝟑 ∞

= ∑ etx qx−1 p
2. The probability that an individual suffers from a bad reaction from a certain
x=1
injection is 0.001, determine the probability that out of 2000 individuals. (a) ∞
𝑝
exactly 3 (b) more than 2 individuals will suffer from a bad reaction. = ∑(𝑞𝑒 𝑡 )𝑥
𝑞
x=1
𝑨𝒏𝒔: 𝛌 = 𝟐, 𝐏(𝐗 = 𝟑) = 𝟎. 𝟏𝟖𝟎𝟒, 𝐏(𝐗 > 𝟐) = 𝟎. 𝟑𝟐𝟑𝟓 𝑝
= [𝑞𝑒 𝑡 + (qet )2 + (qet )3 +∙∙∙∙∙∙∙∙∙]
𝑞
GEOMETRIC DISTRIBUTION 𝑝
= 𝑞𝑒 𝑡 [1 + 𝑞𝑒 𝑡 + (qet )2 + (qet )3 +∙∙∙∙∙∙∙∙∙]
𝑞
DEFINITION:
= 𝑝𝑒 𝑡 (1 − 𝑞et )−1 ∵ (1 − 𝑥)−1 = 1 + 𝑥 + 𝑥 2 +∙∙∙∙∙∙
Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 20
 pet MEMORY LESS PROBABILITY OF GEOMETRIC DISTRIBUTION
MX [t] =
1 − qet State and prove memory less property of Geometric distribution.
Next to find E(X) and Variance (OR) State and prove forgetfulness property of Geometric distribution.
We know that STATEMENT:
t
pe If X is a discrete random variable following a Geometric distribution then
MX [t] =
1 − qet
𝑃(𝑋 > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) = 𝑃(𝑋 > 𝑛), where m and n are two positive integers.
(1 − qet )pet − pet (−qet )
MX ′ [t] = PROOF:
(1 − qet )2
Given that X is a discrete random variable following a Geometric distribution.
pet − pqe2t + 𝑝𝑞𝑒 2𝑡
= Now,
(1 − qet )2
pet P(X > 𝑚 + 𝑛 𝑎𝑛𝑑 𝑋 > 𝑚)
MX ′ [t] = ∙∙∙∙∙∙∙∙∙∙∙∙ (1) 𝑃(𝑋 > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) =
(1 − qet )2 𝑃(𝑋 > 𝑚)

(1 − qet )2 pet − pet [2(1 − qet )](−qet ) P(X > 𝑚 + 𝑛 )

MX ′′ [t] = = ∙∙∙∙∙∙∙∙ (1)
(1 − qet )4 P(X > 𝑚)
∞
(1 − qet )[(1 − qet )pet + 2𝑝𝑞𝑒 2𝑡 ]
= P(X > 𝑚 + 𝑛 ) = ∑ 𝑞 𝑥−1 𝑝
(1 − qet )4 𝑥=𝑚+𝑛+1
pet − 𝑝𝑞𝑒 2𝑡 + 2𝑝𝑞𝑒 2𝑡 ∞
= = ∑ 𝑞 𝑥 𝑞 −1 𝑝
1 − qet
𝑥=𝑚+𝑛+1
pet + pqe2t
MX ′′ [t] = ∙∙∙∙∙∙∙∙ (2) 𝑝
∞
(1 − qet )3 = ∑ qx
𝑝 𝑝 1 𝑞
Put t=0 in (1) we get, MX ′ [0] = (1−𝑞)2 = = x=m+n+1
𝑝2 𝑝
𝑝
1 = [𝑞 𝑚+𝑛+1 + 𝑞 𝑚+𝑛+2 + ⋯ ]
∴ 𝑀𝑒𝑎𝑛 = 𝐸(𝑋) = 𝑞
𝑝
Put t=0 in (2), we get 𝑝 𝑚+𝑛+1
= 𝑞 [1 + 𝑞 + 𝑞 2 + ⋯ ]
(𝑝 + 𝑝𝑞) 𝑝(1 + 𝑞) (1 + 𝑞) 1 𝑞 𝑞
MX ′′ [0] = 3
= 3
= 2
= 2+ 2
(1 − 𝑞) 𝑝 𝑝 𝑝 p 𝑝
= 𝑞 𝑚+𝑛+1 (1 − 𝑞)−1
1 𝑞 1 2 𝑞 𝑞
∴ 𝑉𝑎𝑟(𝑋) = 𝐸[𝑋 2 ] − {𝐸(𝑋)}2 = + − ( ) = 2
𝑝 2 p 2 𝑝 p 𝑝 𝑞 𝑚+𝑛
= = 𝑞 𝑚+𝑛 ∵ 𝑝+𝑞 =1 ⇒𝑝 = 1−𝑞
1−𝑞

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 21

∴ P(X > 𝑚 + 𝑛 ) = 𝑞 𝑚+𝑛 = 𝑝[1 + (𝑞 2 )1 + (𝑞 2 )2 + (𝑞 2 )3 + ⋯ ]
𝑃(𝑋 > 𝑚) = 𝑞 𝑚 and 𝑃(𝑋 > 𝑛) = 𝑞 𝑛 = 𝑝[1 − 𝑞 2 ]−1
Hence (1) becomes, 𝑝
= ∵ 𝑝+𝑞 =1 ⇒𝑝 =1−𝑞
1 − 𝑞2
𝑞 𝑚+𝑛 𝑞𝑚 𝑞𝑛
𝑃(𝑋 > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) = = = 𝑞 𝑛 = 𝑃(𝑋 > 𝑛) (1 − 𝑞) 1
𝑞𝑚 𝑞𝑚 = =
(1 − 𝑞)(1 + 𝑞) 1 + 𝑞
Hence proved.
Example 3:
Example 1:
If the probability that a target is destroyed on any one shot is 0.5, what is the
A die is cast until 6 appear. What is the probability that it must be cast more than
probability that it would be destroyed on 6th attempt?
five times?
Solution:
Solution:
Let X be the number of attempts are required to destroy the target
Let 𝑋 be the R.V. denoting the number of times 6 appears.
1 1 5 5
Probability of destroying a target 𝑝 = 0.5
Here, 𝑝 = , 𝑞 =1−𝑝 = 1− = ,∴ 𝑞 =
6 6 6 6 ∵ 𝑞 = 1 − 𝑝 = 1 − 0.5 = 0.5
We know that the geometric distribution is
We know that by geometric distribution
𝑃(𝑋 = 𝑋) = 𝑞 𝑥−1 . 𝑝, 𝑥 = 1,2,3 … …
𝑃(𝑋 = 𝑋) = 𝑞 𝑥−1 . 𝑝, 𝑥 = 1,2, …
∴ 𝑃(𝑋 > 5) = 1 − 𝑃(𝑋 ≤ 5)
∴ 𝑃(𝑋 = 6) = (0.5)6−1 . (0.5)
= 1 − [𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + ⋯ + 𝑃(𝑋 = 5)]
= (0.5)6
0 1 2 3 4
5 1 5 1 5 1 5 1 5 1 = 0.0156
= 1 − [( ) ( ) + ( ) ( ) + ( ) ( ) + ( ) ( ) + ( ) ( )]
6 6 6 6 6 6 6 6 6 6
Example 4:
1 5 1 5 2 5 3 5 4
= 1 − ( ) [1 + ( ) + ( ) + ( ) + ( ) ] = 0.4018 If the probability that an applicant for a driver’s license will pass road test on
6 6 6 6 6
any given trial is 0.8,what is the probability that he will finally pass the test?
Example 2:
a) On the fourth trial and
If 𝑋 is a geometric variate taking values 1,2, … .. find 𝑃(𝑋 𝑖𝑠 𝑜𝑑𝑑)
b) In fewer than four trials
Solution:
Solution:
We know the geometric distribution is 𝑃(𝑋 = 𝑋) = 𝑞 𝑥−1 . 𝑝, 𝑥 = 1,2, …
Let X denote the number trials are required to achieve the first success.
Now, 𝑃(𝑋 = 𝑜𝑑𝑑) = 𝑞 𝑥−1 . 𝑝, 𝑥 = 1,3,5 …
By Geometric distribution, 𝑃(𝑋 = 𝑋) = 𝑞 𝑥−1 . 𝑝, 𝑥 = 1,2, …
1−1 3−1 5−1
=𝑞 .𝑝 + 𝑞 .𝑝 +𝑞 .𝑝 + ⋯
Here, 𝑝 = 0.08, 𝑞 = 1 − 𝑝 = 1 − 0.8 = 0.2
2 4
= 𝑝 + 𝑞 .𝑝 +𝑞 .𝑝 + ⋯
a) 𝑃(𝑋 = 4) = (0.2)4−1 (0.8)

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 22

 = (0.2)3 (0.8) = 0.0064 1 𝑒 𝑡𝑥
𝑏
= [ ]
𝑏) 𝑃(𝑋 < 4) = 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) 𝑏−𝑎 𝑡 𝑎
= (0.2)1−1 (0.8) + (0.2)2−1 (0.8) + (0.2)3−1 (0.8) 𝑒 𝑏𝑡 −𝑒 𝑎𝑡
∴ The MGF of the uniform distribution is𝑀𝑥 (𝑡) = (𝑏−𝑎)𝑡
= (0.8) + (0.2)1 (0.8) + (0.2)2 (0.8) = 0.992
∞
Mean 𝐸(𝑋) = ∫−∞ 𝑥𝑓(𝑥)𝑑𝑥
𝑏
CONTINUOUS DISTRIBUTION 1
=∫ 𝑥 𝑑𝑥
𝑎 𝑏−𝑎
If X is a continuous R.V., then we have the following distributions
1
1. Uniform(OR)Rectangular distribution = [𝑥 2 ]𝑏𝑎
2(𝑏 − 𝑎)
2. Exponential distribution
𝑏 2 − 𝑎2
3. Normal distribution =
2(𝑏 − 𝑎)
(𝑏 − 𝑎)(𝑏 + 𝑎)
UNIFORM (OR) RECTANGULAR DISTRIBUTION =
2(𝑏 − 𝑎)
If the PDF of a continuous R.V. X is 𝑓(𝑥) =
1
𝑏−𝑎
, a  x  b then 𝑋 follows a Mean 𝐸(𝑋) =
𝑏+𝑎
2
uniform distribution. It is denoted by 𝑈(𝑎, 𝑏), where ′𝑎′ and ′𝑏′ are two ∞
𝐸(𝑋 2 ) = ∫ 𝑥 2 𝑓(𝑥)𝑑𝑥
parameters. −∞

MGF, MEAN AND VARIANCE OF A UNIFORM DISTRIBUTION 𝑏

1
= ∫ 𝑥2 𝑑𝑥
Find the moment generating function of Uniform distribution and find Mean and 𝑎 𝑏−𝑎
Variance of a Uniform distribution. 1
= [𝑥 3 ]𝑏𝑎
3(𝑏 − 𝑎)
SOLUTION:
𝑏 3 − 𝑎3
We know that MGF of X is =
∞
3(𝑏 − 𝑎)
𝑀𝑋 (𝑡) = ∫−∞ 𝑒 𝑡𝑥 𝑓(𝑥)𝑑𝑥
(𝑏 − 𝑎)(𝑏 2 + 𝑎𝑏 + 𝑎2 )
𝑏 =
1 3(𝑏 − 𝑎)
= ∫ 𝑒 𝑡𝑥 𝑑𝑥
𝑎 𝑏−𝑎 𝑆𝑖𝑛𝑐𝑒(𝑎3 − 𝑏 3 ) = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ))
Since 𝑓(𝑥) =
1
𝑏−𝑎
, a xb (𝑏 2 + 𝑎𝑏 + 𝑎2 )
=
1 𝑏 3
= ∫ 𝑒 𝑡𝑥 𝑑𝑥 (𝑏 2 + 𝑎𝑏 + 𝑎2 )
𝑏−𝑎 𝑎 𝐸(𝑋 2 ) =
3

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 23

𝑉𝑎𝑥(𝑋) = 𝐸(𝑋 2 ) − (𝐸(𝑋))2 ⇒ 3 ∝ −3 − 2 ∝= 0 ⇒∝= 3
(𝑏 2 + 𝑎𝑏 + 𝑎2 ) 𝑏+𝑎 2 𝑃(|𝑋| < 1) = 𝑃(|𝑋| > 1).
= −( )
3 2 = 1 − 𝑃(|𝑋| < 1)
(𝑏 2 + 𝑎𝑏 + 𝑎2 ) (𝑏 2 + 2𝑎𝑏 + 𝑎2 ) 𝑃(|𝑋| < 1) + 𝑃(|𝑋| < 1) = 1
= −
3 4
2𝑃(|𝑋| < 1) = 1
4(𝑏 2 + 𝑎𝑏 + 𝑎2 ) − 3(𝑏 2 + 2𝑎𝑏 + 𝑎2 )
= 2𝑃(−1 < 𝑋 < 1) = 1
12
2 2 1
(𝑏 − 2𝑎𝑏 + 𝑎 )
𝑉𝑎𝑥(𝑋) = 2 ∫ 𝑓(𝑥)𝑑𝑥 = 1
12 −1

Remark: The PDF of a Uniform variable 𝑋 in (−𝑎, 𝑎) is given by 𝑓(𝑥) = 1

1
2∫ 𝑑𝑥 = 1
−1 ∝
2
1
, −𝑎 ≤ 𝑥 ≤ 𝑎
2𝑎
1 1 2
⇒ (𝑥)1−1 = 1 ⇒ (1 + 1) = 1 ⇒ = 1 ⇒ ∝= 2
∝ ∝ ∝
Example 1: If X is uniformly distributed over (−∝, ∝) then its PDF is 𝑓(𝑋) =
1
, −∝≤ 𝑥 ≤∝ Example : 2 Subway trains on a certain line run every half hour between mid-
{2∝ Find ∝
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 night and six in the morning. What is the probability that a man entering the
1
Show that (i)𝑃(𝑋 > 1) = (ii) 𝑃(|𝑋| < 1) = 𝑃(|𝑋| > 1). station at a random time during this period will have to wait at least twenty
3

Solution : minutes?
1 Solution: Let ‘𝑋’ be the R.V. which denotes the waiting time when a man arrives
, −∝≤ 𝑥 ≤∝
If X is uniformly distribution (−∝, ∝) then its PDF is 𝑓(𝑋) = {2∝
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 at the station.
1
(i)𝑃(𝑋 > 1) = Then X is uniformly distribution over (0, 30) with PDF
3
∝ 1 1
i.e∫1 𝑓(𝑥)𝑑𝑥 =
3 𝑓(𝑋) = {30 − 0 , 0 ≤ 𝑥 ≤ 30
∝
1 1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
∫ 𝑑𝑥 =
1 2∝ 3 P( the man has to wait at least 30 minutes.)= 𝑃(𝑋 ≥ 20)
30
1 ∝ 1 1 1
∫ 𝑑𝑥 = ⇒ [𝑋]1∝ = = ∫ 𝑓(𝑥)𝑑𝑥
2∝ 1 3 2∝ 3 20

1 1 30
1
⇒ [∝ −1] = =∫ 𝑑𝑥
2∝ 3 20 30
⇒ 3[∝ −1] = 2 ∝

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 24

 30 3 18
1 1 1
=∫ 𝑑𝑥 =∫ 𝑑𝑥 + ∫ 𝑑𝑥
20 30 0 30 15 30

1 30 1 1
= ∫ 𝑑𝑥 = (𝑥)30 + (𝑥)18
15
30 20 30 30
1 1 10 1 3 3 6 1
= (𝑥)30 (30 − 20) = = + = =
20 = = 30 30 30 5
30 30 30 3
Example : 3 4 buses arrive at a specified stop at 15 min. intervals starting at 7 Example : 4 The number of personal computer (PC) sold daily at a computer

A.M, that is, they arrive at 7, 7.15, 7.30, 7.45 and so on. If a passenger arrives world is uniformly distributed with a minimum of 2000 PCs and a maximum of

at the stop at a random time that is uniformly distributed between 7 and 7:30 5000 PCs. Find

A.M. find the probability that he waits (1)The probability that daily sales will fall between 2500 and 3000 PCs.

(a) Less than 5 min for a bus and (2)What is the probability that the computer world will sell at least 4000 PCs?

(b) At least 12min for a bus. (3) What is the probability that the computer world will sell exactly 2500 PCs?

Solution: Let‘𝑋’ denotes the time that a passenger arrives between 7 and 7.30 am Solution : Let ‘X’ be the R.V denoting the number of computer sold daily at a

Then X is uniformly distributed over (0, 30) computer world.

1 The PDF of the uniform distribution is
, 0 ≤ 𝑥 ≤ 30
i.e 𝑓(𝑋) = {30−0 1
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝑓(𝑋) = {5000 − 2000 , 2000 ≤ 𝑥 ≤ 5000
(a) Passenger waits less than 5 minutes, that is he arrives between 7.10-7.15 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
or 7.25-7.30 1
, 2000 ≤ 𝑥 ≤ 5000
i.e 𝑓(𝑋) = {3000
i.e. , 𝑃(waiting time less than 5 minutes) 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
= 𝑃(10 < 𝑋 < 15) + 𝑃(25 < 𝑋 < 30) 3000 1
(1) 𝑃(2500 < 𝑋 < 3000) = ∫2500 𝑑𝑥
3000
15 30
1 1 1 1 1 3000 − 2500 500 1
=∫ 𝑑𝑥 + ∫ 𝑑𝑥 = (𝑥)15
10 + (𝑥)30
25 = (𝑥)3000 = = =
10 30 25 30 30 30 3000 2500
3000 3000 6
1 1 5 5 10 1 5000 1
= (15 − 10) + (30 − 25) = + = = (2) 𝑃(𝑋 ≥ 4000) = ∫4000 𝑑𝑥
3000
30 30 30 30 30 3
5000
(b) Passenger waits at least 12 min. i.e., he arrives between 7 - 7.03 or 1
= ∫ 𝑑𝑥
3000 4000
7.15 – 7.18
1 1000 1
𝑃(𝑤𝑎𝑖𝑡𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 12 𝑚𝑖𝑛. )  P(0  X  3)  P (15  X  18) = [𝑥]5000
4000 = =
3000 3000 3
(2) 𝑃(𝑋 = 2500) = 0 ∵ 𝑃(𝑋 = 𝐶) = 0

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 25

EXPONENTIAL DISTRIBUTION tr
Now μ′r E(X r ) = the coeffi. of in MX (t)
r!
DEFINITION : A continuous R.V. ‘X’ is said to follow an exponential distribution r!
μ′r = , 𝑟 = 1,2, … … . (1)
λe−λx , x>0 λr,
with parameter λ > 0 its PDF is given by 𝑓(𝑥) = {
0, otherwise 1
Put 𝑟 = 1 in(1) , we get 𝑀𝑒𝑎𝑛 = 𝐸(𝑋) =
λ
MGF, MEAN AND VARIANCE OF THE EXPONENTIAL DISTRIBUTION
2
Find the moment generating function of the exponential distribution, hence find Put 𝑥 = 2 in (1), we get E[X 2 ] =
λ2

Mean and Variance. 2 1 1

Variance of X = E[X 2 ] − {E(X)}2 = 2
− 2= 2
𝜆 λ λ
Solution:
1 1
∞ Hence 𝑀𝑒𝑎𝑛 = , 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 =
The MGF of X is MX (t) = ∫0 etx f(x)dx λ λ2

∞
= ∫ etx λe−λx dx Example 1: Supposse the duration X in minutes of long distance calls from your
0
∞ home follows exponential law with PDF
= λ ∫ e−(λ−t)x dx
0
1 −𝑥
𝑓(𝑥) = { 5 𝑒 , 𝑥>0
5
λ ∞
= [𝑒 −(𝜆−𝑡)𝑥 ]0 0 , 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
−(λ − t)
Find (i) 𝑃(𝑋 > 5) (ii) 𝑃(3 ≤ 𝑋 ≤ 6) (iii) Mean of X and Variance of X (iv) MGF
𝜆
= [0 − 1] of X
−(𝜆 − 𝑡)
𝜆 Solution:
MX (t) = ,λ > 𝑡
(𝜆 − 𝑡) By Exponential distribution, 𝑓(𝑥) = λe−λx , for x > 0
Next, to find Mean and Variance Here λ =
1
5
𝜆 λ 𝑥
MX (t) = = ∞
(i) 𝑃(𝑋 > 5) = ∫5 𝑓(𝑥)𝑑𝑥 = ∫5
∞1
𝑒 −5 𝑑𝑥
(𝜆 − 𝑡) λ (1 − t ) 5
λ
x ∞
𝑡 −1 1 e− 5 x ∞
= (1 − ) = [ 1 ] = −1 [e−5 ] = −1[0 − 𝑒 −1 ] = 0.3678
𝜆 5 − 5
5 5
𝑡 t 2 𝑡 3 t 𝑟
= 1 + ( ) + ( ) + ( ) +∙∙∙∙∙∙∙∙ + ( ) +∙∙∙∙∙∙∙∙ 6 61 𝑥
𝜆 λ 𝜆 λ (ii) 𝑃(3 ≤ 𝑋 ≤ 6) = ∫3 𝑓(𝑥)𝑑𝑥 = ∫3 𝑒 −5 𝑑𝑥
5
∞
t 𝑟 x 6
MX (t) = ∑ ( ) 1 e−5 x 6 6 3
λ = [ 1 ] = −1 [e−5 ] = − [𝑒 −5 − 𝑒 −5 ] = 0.2476
r=0
5 − 3
5 3

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 26

(iii) In the exponential distribution RESULT : State and prove memory less property of the exponential distribution.
1 1 STATEMENT : If ‘X’ is exponentially distributed with parameter λ, then
𝑀𝑒𝑎𝑛 = =5 ∴λ=
𝜆 5 P(X > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) = P(X > n) for any 𝑚, 𝑛 > 0
1
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 2 = 25 Proof :
λ
λ 1⁄ 1 The PDF of X if
5
(iv) MGF is MX (t) = (λ−t) = 1 =
1−5𝑡
5
−𝑡 λe−λx , x>0
f(x) = {
Example 2 : The mileage which car owners get with a certain kind of radial tyre 0, otherwise
P(X > 𝑚 + 𝑛 𝑎𝑛𝑑 𝑋 > 𝑚)
is a R.V having an exponential distribution with mean 40,000 km. Find the P(X > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) =
P(X > 𝑚)
probability that one of these tyres will last (i) at least 29,000 km (ii) at most
P(X > 𝑚 + 𝑛 )
30,000 km. = ∙∙∙∙∙∙∙∙ (1)
P(X > 𝑚)
Solution: ∞
P(X > 𝑚) = ∫ 𝑓(𝑥) dx
λe−λx , x>0 m
The exponential distribution is f(x) = {
0, otherwise ∞
∞
e−λx
1
Given that Mean = = 40,000 ∴λ=
1 = ∫ λe−λx dx = λ [ ] = −1[0 − e−λm ] = 𝑒 −𝜆𝑚 ∙∙∙∙∙∙∙∙∙ (2)
λ 40000 m −λ m
Let X denotes the mileage obtained with the tire P(X > 𝑚 + 𝑛) = e−λ(m+n) … … … (3)
∞
(𝑖)𝑃(𝑋 ≥ 20000) = ∫ 𝜆𝑒 −𝜆𝑥 𝑑𝑥 Hence (1) becomes
20000
e−λ(m+n) e−λm e−λn
∞
1 x P(X > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) = = = e−λn = P(X > 𝑛)
=∫ e−40000 dx e−λm e−λm
20000 40000 ∴ P(X > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) = P(X > n)
𝑥 ∞
− PROBLEMS USING MEMORY LESS PROPERTY OF THE EXPONENTIAL
1 𝑒 40000
= [ 1 ]
40000 − DISTRIBUTION
40000 20000
20000 Example : 3
= −1 [0 − e−40000] = −[0 − e−0.5 ] = 0.6065
The time (in hours) required to repair a machine is exponentially distributed
𝑥
30000 1
(ii) 𝑃(𝑋 ≤ 30000) = ∫0 𝑒 −40000 𝑑𝑥 with parameter λ =
1
(a) What is the probability that the repair time exceed
40000 2
30000
𝑥 2h? (b) What is the conditional probability that a repair takes at 11h given that
1 𝑒 −40000 30000
= [ ] = −1 [e−40000 − 1] = [1 − e−0.75 ] = 0.5270 its duration exceeds 8h?
40000 − 1
40000 0
Solution :

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 27

Let X be the R.V which represent the time to repair the machine. Then the PDF is The probability that of 2 days selected at random, the stock is insufficient for
𝜆𝑒 −𝜆𝑥 , 𝑥>0 both days = 𝑒 −5 . 𝑒 −5 = 𝑒 −10 =0.0000453
𝑓(𝑥) = {
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
NORMAL DISTRIBUTION:
1 −x
1 e 2, x>0 A continuous R.V. X is said to follow a normal distribution with parameters
Given λ = 𝑖. 𝑒. , f(x) = {2
2
0, otherwise
𝜇 𝑎𝑛𝑑 𝜎, if its probability density function is given by
1
− (2) −1
(a) P(X > 2) = e 2 =e = 0.3678 1 2 ⁄2𝜎 2
𝑓(𝑥) = 𝑒 (𝑥−𝜇) − ∞ < 𝑥 < ∞, −∞ < 𝜇 < ∞, 𝜎 > 0
−λk
1 𝜎√2𝜋
(∵ by memory less property , 𝑃(𝑋 > 𝑘) = e here 𝑘 = 2, λ = )
2 Symbolically, Normal distribution is denoted by 𝑁(𝜇, 𝜎)
(b) P(X ≥ 11⁄X > 8) = 𝑃(𝑋 ≥ 8 + 3⁄𝑋 > 8) = P(X > 3)
(∵ by memory less property) STANDARD NORMAL DISTRIBUTION:
P(X > 𝑚 + 𝑛 ∕ 𝑋 > 𝑚) = P(X > 𝑛)) The normal distribution 𝑁(0, 1) is called the standard normal distribution,
1 3
− (3) − whose density function is given by
=𝑒 2 =𝑒 2 = 𝑒 −1.5 = 0.2231
1 2 ⁄2 𝑋−𝜇
Example : 4 The daily consumption of milk in excess of 20,000 gallons is ∅(𝑧) = 𝑒 −𝑧 , −∞ < 𝑧 < ∞ . Here 𝑍 = is the standard normal
√2𝜋 𝜎
approximately exponentially distributed with mean 3000. The city has a daily variate.
stock of 35,000 gallons. What is the probability that of 2 days selected at Note:
random, the stock is insufficient for both days. The total area under the normal curve is equal to 1. The density function is bell
Solution: shaped curve that is symmetric about . The value  and 2 represent the
Let X denote the R.V which denotes the excess amount of milk consumed in a average value (mean) and Var(X).
day, and Y denotes the daily consumption of milk, then X = Y – 20000
λe−λx , x>0
By exponential distribution is f(x) = {
0, otherwise
x
1 1 1
Given that Mean = = 3000 ⇒λ= ∴ 𝑓(𝑥) = e−3000
λ 3000 3000

P( the stock is insufficient for one day ) = P(Y > 35000)

= P(X + 20000 > 35000)
= P(X > 15000)
1
= 𝑒 −3000𝑥(15000) {∵ P(X > 𝑘) = e−λk }
= e−5

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 28

 ∞ −z2
Properties of Normal Distribution N(μ, σ): 1
∴ Area = ∫ e 2 σdz
1. The values of f(x) approaches zero as 𝑥 – ∞ and 𝑥 ∞. (i.e.) 𝑥 axis is −∞ σ√2π
∞
asymptote to the normal curve. 1 −z2
= ∫e 2 dz
2. The PDF is symmetric about . √2π
−∞

3. The maximum of the PDF occurs at 𝑥 = . 2

∞
−z2 −z2
= ∫e 2 dz [∵ e 2 is an even function]
4. Mean, median and mode coincide. √2π
0
5. Skewness is zero and the curve is mesokurtic (bell shaped). z
Put = u , dz = √2du
√2
6. The points of inflection of the curve occur at 𝑥 =  ± .
7. Area Property: 2 ∞ 2
∴ 𝐀𝐫𝐞𝐚 = √ ∫ e−u √2du
X−μ π 0
(i) 𝑃(𝑎 ≤ 𝑋 ≤ 𝑏) = 𝑃(𝑧1 ≤ 𝑍 ≤ 𝑧2 ), where Z = is the standard
σ
∞
normal variate 𝟐 2
= ∫ e−u du
√𝛑
(ii) 𝑃(– ∞ ≤ 𝑋 ≤ ) = 𝑃(– ∞ ≤ 𝑍 ≤ 0) = 0.5, 0

𝑃( ≤ 𝑋 ≤ ∞) = 𝑃(0 ≤ 𝑍 ≤ ∞) = 0.5 =

2 √π ∞
= 1 [∫0 e−u du =
2 √π
]
√π 2 2
(iii) 𝑃(– 𝑧1 ≤ 𝑍 ≤ 0) = 𝑃(0 ≤ 𝑍 ≤ 𝑧1 ) ∵ Area bounded by the above normal curve is 1.
(iv) 𝑃(– ∞ ≤ 𝑍 ≤ 𝑧1 ) = 0.5 – 𝑃(0 ≤ 𝑍 ≤ 𝑧1 ), 𝑃(𝑧1 ≤ 𝑍 ≤ ∞) = RESULT : Obtain the MGF of normal distribution.
0.5 – 𝑃(0 ≤ 𝑍 ≤ 𝑧1 ) PROOF :
∞

MX (t) = ∫ etx f(x)dx

RESULT: Show that the total area bounded by the above normal curve is 1.
−∞
PROOF: −(x−μ)2
∞ 1
∞ ∞ 1
−(x−μ)2 =∫−∞ etx e 2σ2 dx
σ√2π
Area= ∫−∞ f(x)dx = ∫−∞ e 2σ2 dx
σ√2π −(x−μ)2
1 ∞
Put = ∫ etx e
σ√2π −∞
2σ2 dx

𝑥−𝜇 Put
=𝑧 𝑥 = ∞ ,𝑧 = ∞
𝜎
𝑥−𝜇
=𝑧 𝑥 = ∞ ,𝑧 = ∞
𝜎𝑑𝑧 = 𝑑𝑥 𝑥 = −∞, 𝑧 = − ∞ 𝜎

𝜎𝑑𝑧 = 𝑑𝑥 𝑥 = −∞, 𝑧 = − ∞

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 29

 −z2 2 3 n
=
1 ∞
∫ et(σz+μ) e 2 σdz 1 t 2 σ2 1 (t 2 σ2 ) 1 (t 2 σ2 ) 1 (t 2 σ2 )
σ√2π −∞ =1+ ( )+ ( )+ ( )+⋯+ ( n )
1! 2 2! 4 3! 8 n! 2
−(z2 −2tσz)
eμt ∞
= ∫−∞ e 2 dz 1 t 2 σ2 1 t 4 σ4 1 t 6 σ6 1 t 2n σ2n
√2π
=1+ ( )+ ( )+ ( ) + ⋯ + ( n ) − − − (1)
−(z2 −2tσz) 1 (z−σt)2 σ2 t2 1! 2 2! 4 3! 8 n! 2
[∵ = − [(z − σt)2 − σ2 t 2 ] = − + ]
2 2 2 2
1 σ2 2 1 σ4 4 1 σ6 6
1 ∞ −1(z−σt)2 +σ t
2 2 =1+ ( )t + ( )t + ( )t + ⋯
=eμt ∫ e 2 2 dz 1! 2 2! 4 3! 8
√2π −∞
t2 σ2 1 Using (1), we get
1 ∞ − (z−σt)2
=eμt+ 2 ∫−∞ e 2 dz tn
√2π
μn = coefficient of
t2 σ2 1 ∞ −u2 n!
=eμt+ 2 ∫−∞ e 2 du
√2π t1
μ1 = coefficient of =0
Put u = z − σt , z = ∞ , u = ∞ 1!
du = dz , z = −∞, u = −∞ t2
μ2 = coefficient of = σ2
−u2
2!
1 ∞
= √2π [∵ ∫−∞ e 2 du =√2π] t3
√2π
μ3 = coefficient of =0
𝑡2 𝜎2 3!
MX (t) = 𝑒 𝜇𝑡+ 2
t4
μ4 = coefficient of = 3σ4
4!
RESULT: The first four moments are given by
Find the nth central moments of normal distribution and hence find the mean ∴ Mean = μ1 = 0
and variance. Variance=μ2 = σ2
PROOF : The MGF (about mean) is given by μ3 = 0
E[et(X−μ) ] = e−μt E(etX ) μ4 = 3σ4
= e−μt MX (t) Thus Mean = μ1 = 0
t2 σ2 Variance=μ2 = σ2 .
But MX (t)= eμt+ 2
NOTE:
∴ E[et(X−μ) ] = e−μt E(eXt )
1. Mean, median and mode of the normal distribution coincide.
𝐭 𝟐 𝛔𝟐
𝛍𝐭+
= e−μt MX (t) = 𝐞−𝛍𝐭 𝐞 𝟐 2. Q.D : MD : SD = 10:12:15.
t2 σ2
=e 2

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 30

ADDITIVE PROPERTY OF NORMAL DISTRIBUTION:
If Xi (i = 1,2, … , n) be n independent normal RV’s with mean μi and variance σi 2 ,
then ∑ni=1 ai Xi is also a normal RV with mean∑ni=1 a i μi and variance ∑ni=1 a i 2 σi 2
Proof :
M(∑ni=1 ai Xi ) (t) = M𝑎1𝑋1 (t)M𝑎2𝑋2 (t) … … … ManX𝑛 (t) by independence
2 𝜎 2 𝑡 2⁄2 2 𝜎 2 𝑡 2 ⁄2 2 𝜎 2 𝑡 2⁄2
= 𝑒 𝑎1 𝜇1𝑡+𝑎1 1 𝑋 𝑒 𝑎2𝜇2𝑡+𝑎2 2 𝑋 … … … 𝑋 𝑒 𝑎𝑛 𝜇𝑛𝑡+𝑎𝑛 𝑛

2 σ 2 t2 )⁄2
= e(∑ 𝑎𝑖 𝜇𝑖)t+∑(ai 𝑖

Which is the MGF of a normal R.V with mean ∑ 𝑎𝑖 𝜇𝑖 and variance∑ a i 2 σ𝑖 2

NOTE:

(i) If 𝑋1 is 𝑁(𝜇1 , 𝜎1 ) and 𝑋2 is 𝑁(𝜇2 , 𝜎2 ) , then 𝑋1 + 𝑋2 is 𝑁 (𝜇1 +

𝜇2 , √𝜎12 + 𝜎22 ) similarly 𝑋1 − 𝑋2 is 𝑁 (𝜇1 − 𝜇2 , √𝜎12 + 𝜎22 )

(ii) If 𝑋𝑖 (𝑖 = 1,2, … , 𝑛) are independent and identically distributed normal

variables with mean 𝜇 and standard deviation 𝜎 then their mean 𝑋̅ is
𝑁(𝜇, 𝜎⁄√𝑛)
Calculation of area under the Normal Curve

Example: 1 If X is a normal variate with mean 30 and S. D. = 5. Find (i)P(26 ≤

X ≤ 40),(ii)P(X ≥ 45).
Solution:
Given  = 30 𝑎𝑛𝑑  = 5
𝑥−𝜇
Let 𝑍 = be the standard normal variate
𝜎
x − 30
𝑖. 𝑒. , Z =
5
26 − 30 40 − 30
(𝑖) 𝑃(26 ≤ 𝑋 ≤ 40) = 𝑃 ( ≤𝑍≤ )
5 5

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 31

 = 𝑃(−0.8 ≤ 𝑍 ≤ 2) Example: 3The independent RV’s X and Y have distributions 𝑁(45 , 2) and
= 𝑃(−0.8 ≤ 𝑍 ≤ ∞ − 𝑃(2 ≤ 𝑍 ≤ ∞) 𝑁(44 , 1.5) respectively. What is the probability that randomly chosen values of
= 0.78814 − 0.02275 = 0.76539 X and Y differ by 1.5 or more?
45 − 30 Solution :
(𝑖𝑖)𝑃(𝑋 ≥ 45) = 𝑃 (𝑍 ≥ ) = 𝑃(𝑍 ≥ 3) = 0.0013
5
X is 𝑁(45 , 2) and Y is 𝑁(44 , 1.5) Therefore by the additive property, 𝑈 = 𝑋 – 𝑌
follows the distribution 𝑁(1 , √4 + 2.25) = 𝑁(1 , 2.5)
Example: 2 The marks obtained by a number of students in a certain student are
Now P( X and Y differ by 1.5 or more)
approximately distributed with mean 65 and standard deviation 5. If 3 students
=𝑃(|𝑋 − 𝑌| > 1.5)
are selected at random from this group, what is the probability that at least one
=𝑃(|𝑈| > 1.5)
of them would have scored above 75?
=1 − 𝑃(|𝑈| ≤ 1.5)
Solution :
= 1 − 𝑃(−1.5 ≤ 𝑈 ≤ 1.5)
If X represents the marks obtained by students, X follows the
−1.5 − 1 𝑈 − 1 1.5 − 1
distribution𝑁(65 , 5). = 1−𝑃{ ≤ ≤ }
2.5 2.5 2.5
P (A student scores above 75) = 𝑃 (𝑋 > 75 ) = 1 − 𝑃[−1 ≤ 𝑍 ≤ 0.2]
75 − 65 𝑋 − 65
= 𝑃( < < ∞) = 1 − {𝑃[−1 ≤ 𝑍 ≤ ∞] − 𝑃[0.2 ≤ 𝑍 ≤ ∞]}
5 5
= 1 − (0.8413 − 0.4207) = 0.5794
= 𝑃 (2 < 𝑍 < ∞)
= 0.0228
Example: 4 If X and Y are independent RV’s following N(8 , 2) and N(12 , 4√3)
Let p = P(A student scores above 75) = 0.0228 then q = 0.9772 and n = 3
Since p is the same for all the students, the number Y, of (successes) students respectively, find the value of  such that

scoring above 75, follows a binomial distribution. 𝑃(2𝑋 − 𝑌 ≤ 2𝜆) = 𝑃(𝑋 + 2𝑌 ≥ 𝜆)

P (at least 1 student scores above 75) = P(at least one success ) Solution:
= 𝑃(𝑌 ≥ 1) = 1 − 𝑃(𝑌 = 0) By the additive property of normal distribution
= 1 − 𝑛𝐶0 × 𝑝0 𝑞 𝑛 𝑈 = 2𝑋 − 𝑌 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑁(2 × 8 − 12, √4 × 4 + 1 × 48) = 𝑁(4 , 8) and 𝑉=𝑋+
= 1 − 3𝐶0 (0.9772)3 2𝑌 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑁(8 + 2 × 12, √4 + 4 × 48) = 𝑁(32 , 14)
= 1 − 0.9333 = 0.0667 Now 𝑃(2𝑋 − 𝑌 ≤ 2𝜆) = 𝑃(𝑋 + 2𝑌 ≥ 𝜆)
= 𝑃(𝑈 ≤ 2𝜆) = 𝑃(𝑉 ≥ 𝜆)

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 32

 𝑈−4 2𝜆 − 4 𝑉 − 32 𝜆 − 32
= 𝑃( ≤ ) = 𝑃( ≥ ) UNIVERSITY QUESTIONS
8 8 14 14
2𝜆 − 4 𝜆 − 32 APRIL/MAY 2015
= 𝑃 (𝑍 ≤ ) = 𝑃 (𝑍 ≥ )
8 14 PART A
2𝜆 − 4 𝜆 − 32 {|𝑥|; −1 ≤ 𝑥 ≤ 1
∴ = −( ) Since by the symmetry 1. Test whether 𝑓(𝑥) = { can be the probability density
8 14 0; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
⇒ 28 𝜆 − 56 = 256 − 8𝜆 function of a continuous random variable.
26 2. What are the limitations of Poisson distribution?
∴ 𝜆 =
3
PART B
1. (i) The CDF of the random variable of X is given by
Example : 5 Given that X is normally distributed,
0; 𝑥 < 0
If P(X < 45) = 0.31 and P(X > 64) = 0.08, find the mean and standard 𝑥 +
1
; 0≤𝑥≤
1
𝐹𝑋 (𝑥) = { 2 2
deviation of the distribution. 1
1; 𝑥 >
2
Solution:
1 1 1
Draw the graph of the CDF. Compute 𝑃 (𝑋 > ) , 𝑃 ( < 𝑋 ≤ )
Let X follows the distribution 𝑁(𝜇 , 𝜎)Given𝑃(𝑋 < 45) = 0.3 𝑎𝑛𝑑 4 3 2

𝑃(𝑋 > 64) = 0.08 (ii) Find the moment generating function of a geometrically distributed
45− 𝜇 64− 𝜇 random variable and hence find the mean and variance.
That is 𝑃 (−∞ < 𝑍 < ) = 0.31 and 𝑃 ( < 𝑍 < ∞) = 0.08
𝜎 𝜎
2. (i) Messages arrive at a switch board in a Poisson manner at an average
−(𝜇 − 45)
𝑃 (−∞ < 𝑍 < ) = 0.31 and rate of six per hour. Find the probability for each of the following
𝜎
64− 𝜇
events:
𝑃( < 𝑍 < ∞) = 0.08
𝜎 1. Exactly two messages arrive within one hour
(𝜇 − 45) 64 − 𝜇 2. No message arrives within one hour
𝑃 (0 < 𝑍 < ) = 0.19 and 𝑃 (0 < 𝑍 < ) = 0.42
𝜎 𝜎
3. At least three messages arrive within one hour.
𝜇−45 64− 𝜇
From the table of areas we get = 0.48 𝑎𝑛𝑑 = 1.42 (ii) The peak temperature T, as measured in degrees Fahrenheit, on a
𝜎 𝜎

𝜇 − 0.48 𝜎 = 45 ------ (1) and𝜇 + 1.42 𝜎 = 64 ------ (2) particular day is the Gaussian (85,10) random variable. What is
Solving (2) – (1), 𝑃(𝑇 > 100), 𝑃(𝑇 < 60) and 𝑃(70 ≤ 𝑇 ≤ 100)?
1.9 𝜎 = 19 ⟹ 𝜎 = 10
substitute in(1) ⟹ 𝜇 = 49.8

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 33

UNIVERSITY QUESTIONS
(2) What is the probability that it takes him less than 4 shots?
DEC 2014:
(3) What is the probability that it takes him an even number of shots?
PART A
B. Determine the moment generating function of an exponential random
{|𝑥|; −1 ≤ 𝑥 ≤ 1
1. Test whether 𝑓(𝑥) = { can be the probability density variable and hence find its mean and variance.
0; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
function of a continuous random variable. 2. Determine the mean, variance and moment generating function of a random
2. What do you mean by MGF? Why it is called so? variable X following Poisson distribution with parameter λ.
PART B B. Trains arrive at a station at 15 minutes intervals starting at 4.a.m. If a
𝑥2 passenger arrives at a station at a time that is uniformly distributed between
− ;
1. (i) If 𝑓(𝑥) = {𝑥𝑒 2 𝑥≥0 then show that 𝑓(𝑥) is a pdf and find 𝐹(𝑥).
0 ; 𝑥≤0 9.00 and 9.30, find the probability that he has to wait for the train for (1) less
(ii) Find the MGF of a Poisson random variable and hence find its mean than 6 minutes (2) more than 10 minutes.
variance.
2. (i) A random variable X takes the values -2, -1, 0 and 1 with probabilities DEC 2013:
1 1 1 1 PART-A
, , and respectively. Find and draw the probability function.
8 8 4 2
1. A coin is tossed 2 times, if x denotes the number of heads; find the probability
(ii) In a normal distribution, 31% of the items are under 45 and 8% are over
distribution of x.
64. Find the mean and variance of the distribution.
2. If the probability that a target is destroyed on any one shot is 0.5, find the
probability that it would be destroyed on sixth attempt.
MAY 2014:
PART: B
PART A
1. (a) Find the Moment generating function of the random variable X having the
1. A continuous random variable X has the probability density function given
𝑥, 0≤𝑥≤1
𝑎(1 + 𝑥 2 ), 2 ≤ 𝑥 ≤ 5
by 𝑓(𝑥) = { Find a and p(X<4). PDF 𝑓(𝑥) = { 2 − 𝑥, 1 ≤ 𝑥 ≤ 2
0, otherwise 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
2. For a binomial distribution with mean 6 and standard deviation√2 find the b) A manufacturer of pins knows that 2% of his products are defective. If he sells
first two terms of the distribution pins in boxes of 100 and guarantees that not more than 4 pins will be defective.
PART B What is the probability that a box fail to meet the guaranteed quality?
1. Suppose that a trainee soldier shoots a target in an independent fashion. If the 2. (a) 6 dice are thrown 729 times. How many times do you expect at least three
probability that the target is shot on any one shot is 0.7, dice to show a five or six?
(1) What is the probability that the target would be hit on tenth attempt?

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 34

b) If a continuous random variable X follows uniform distribution in the interval 2. Identify the random variable and name the distribution it follows
(0,2) and a continuous random variable Y follows exponential distribution with From the following statement: “ A realtor claims that only 30% of the houses in
parameter 𝜆. Find 𝜆 such that P (X <1)= P(Y<1). a certain neighbourhood, are appraised at less than Rs 20 lakhs. A random
sample of 10 houses from that neighbourhood is selected and appraised to check
MAY 2013: the realtor’s claim is acceptable are not”
PART-A PART:B
1. If X and Y are two independent random variables with variances 2 and 3 1. (a) If the random variable X takes the values 1,2,3, and 4 such that
find the variance of 3 X+4Y 2P(X=1)= 3P(X=2) =P(X=3) =5 P (X=4). Then find the probability
2. State memory less property of exponential distribution. Distribution and cumulative distribution function of X.
PART:B (b) Find the MGF of the Binomial distribution and find its mean.
1. a) A continuous random variable has the PDF f(x) = k x4-1 , -1<x<0. Find the 2. (a) If the probability that an applicant for a driver’s license will pass the road
1 1 test on any given trial is 0.8. What is the probability that he will finally pass the
value of K and also 𝑃 {𝑋 > (− ) /𝑋 < (− )}
2 4
test (1) on the 4th trial 2) in fewer than 4 trials?
b) Find the moment generating function of uniform distribution hence find its
(b) The number of monthly breakdowns of computer is a random variable
mean and variance
having a Poisson distribution with mean equal to 1.8.Find the probability that
2. a) Find the moment generating function and rth moment for the distribution
this computer will function for a month (1) without a breakdown (2) with only
whose PDF if f(x) = K e-x , 0 <x<∞. Hence find its mean and variance
one breakdown.
b) In a large consignment of electric bulbs, 10 percent are defective. A random
MAY 2012
sample of 20 is taken for inspection. Find the probability that (1) all are good
PART-A
bulbs (2) at most there are 3 defective bulbs (3) exactly there are 3 defective
1 Check whether the following is a probability density function or
bulbs.
𝜆𝑒 −𝜆𝑥 , 𝑥 > 0, 𝜆 > 0
DEC 2012 not f(x) ={
0, otherwise
PART-A
2. If a random variable has the moment generating given by
1. A continuous random variable X that can assume any value between x=2 and
2
x=5 has a density function given by f(x)= k(1+x). Find P(X<4). M X (t )  , determine the variance of X.
2-t
PART: B
X: 0 1 2 3 4 5 6 7
P (X): 0 k 2k 2k 3k k2 2k2 7k2+k 1. (a) A random variable has the following distribution

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 35

(i) Find the value of k Find the PDF of X and evaluate𝑃(|𝑋| < 1) and 𝑃 (1/3 < 𝑋 < 4) using both the
(ii)Evaluate P(X<6) P(X≥6) CDF and PDF.
(iii)If P(X≤C)>1/2 find the minimum value of C. (b) A random variable X has the following probability distribution
( b). Find the moment generating function of an exponential random variable X -2 -1 0 1 2 3
and hence find its mean and variance.
P(X) 0.1 k 0.2 2k 0.3 3k
2. (a) If X is a Poisson variate such that P(X=2) = 9P(X=4) +90P(X=6) (i) Find k
Find (i) Mean and E(X2) (ii) Evaluate P(X<2) and P(-2<X<2)
(ii) P(X≥2) (iii) Find the PDF of X
(b) In a certain city daily consumption of electric power in millions of kilowatt (iv) Evaluate the mean of X.
hours can be treated as a random variable having Gamma distribution with 2. The probability function of an infinite discrete distribution is given
parameters  =1/2 and V=3. If the power plant of this city has a daily capacity by 𝑃(𝑋 = 𝑗) =
1
, 𝑗 = 1,2,3, … , ∞. Verify that the total probability is 1 and find
2𝑗
of 12 million kilowatt-hours. What is the probability that this power supply will
the mean and variance of the distribution. Find also P(X is even) P(X≥5) and P
be inadequate on any given day?
(X is divisible by 3).

DEC 2011
DEC 2010
PART-A
PART-A
1. A continuous random variable X that can assume any value between x=2 and
1 − 𝑒 𝑎𝑥 , 𝑥 > 0
1. If a random variable X has the distribution function 𝐹(𝑥) = {
x=5 has a density function f(x) = k(1+x). Find P(X <4). 0, 𝑥<0
2. State the probability law of Poisson distribution and also find its mean and where 𝑎 is the parameter then find𝑃(1 ≤ 𝑋 ≤ 2).
variance. 2. Every week the average number of wrong number phone calls received by a
PART: B certain mail order house is seven. What is the probability that they will receive
1. (a) The distribution function of a continuous random variable X is given by two wrong calls tomorrow?
0, 𝑥<0 PART:B
𝑥, 0 ≤ 𝑥 ≤ 1⁄2 1.(a) The distribution function of a random variable X is given by
𝐹(𝑥) = 3
1 − (3 − 𝑥), 1⁄2 ≤ 𝑥 ≤ 3 F(X) = 1- (1+X) e-x ,x≥0. Find the density function mean and variance of X.
25
{ 1, 𝑥≥3
(b) A coin is tossed until the first head occurs. Assuming that the tosses are
independent and the probability of a head occurring is ‘P’. So that the probability

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 36

that an odd number of tosses required is equal to 0.6. Can you find a value of P.  n  1 r n-r
probability P X  n     p q
 r  1
So that the probability is 0.5 that an odd number of tosses are required.
2. (a) If X is a random variable with a continuous distribution function F(X).
Prove that Y= f(x) has a Uniform distribution in (0,1). Further if find the
range of Y corresponding to the range 1≤ x ≤2.9.
b) The time (in hours) required to repair a machine is exponentially distributed
1
with parameter𝜆 = . What is the probability that the repair time exceeds 2h?
2

What is the conditional probability that a repair takes at least 10h given that its
duration exceeds 9h?

MAY 2010
PART-A
1. Obtain the mean for a Geometric random variable.
2. What is meant by memory less property? Which continuous distribution
follows this property?
PART: B
1. a). By calculating the moment generating function of a Poisson distribution
with parameter  Prove that the mean and variance of the Poisson distribution
are equal.
𝐶𝑒 −2𝑥 , 0 < 𝑥 < ∞
b) If the density function of X equals𝑓(𝑥) = { Find ‘C and
0, 𝑥 < 0
P(X>2)
2) (a) Describe the situations in which geometric distributions is used Obtain its
moment generating function.
(b) A coin having probability P of coming up heads is successively flipped until
the rth read appears. Argue that X, the number of flips required will be n≥ r with

Department of Mathematics AnjalaiAmmalMahalingam Engineering College, Kovilvenni – 614 403. Page 37