Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Title: Probability distributions
Description: This file is a pdf for the topic "probability distributions" from the subject " probability and statistics"
Description: This file is a pdf for the topic "probability distributions" from the subject " probability and statistics"
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
Lecture Notes
H
AT
Prepared by
IT
H
A
N
S
15MA301-Probability & Statistics
Dr
...
ATHITHAN
O
F
Assistant Professor
TE
S
Department of Mathematics
Faculty of Engineering and Technology
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
LE
C
TU
R
E
N
O
Kattankulathur-603203, Kancheepuram District
...
Probability & Statistics
S
...
1 Binomial Distribution
...
1
...
1
...
2 Mean and Variance of Binomial Distribution
...
2 Poisson Distribution
...
2
...
1
...
2 Mean and Variance of Poisson Distribution
...
3 Geometric Distribution
...
3
...
1
...
2 Mean and Variance of Geometric Distribution
...
3
...
...
...
...
...
...
...
...
...
...
...
H
Standard Continuous Distributions
2
...
2
...
1 Moment Generating Function of U NIFORM D ISTRIBUTION
...
1
...
2
...
2
...
1 Moment Generating Function of E XPONENTIAL D ISTRIBUTION
2
...
2 Mean and Variance of Exponential Distribution
...
2
...
2
...
2
...
1 MGF of Normal Distribution
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
7
7
7
8
8
8
9
9
10
10
A
N
S
1
Example Worked out Problems
4
Exercise/Practice/Assignment Problems
...
...
...
...
...
...
...
...
...
Page 1 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
Unit-2
S
...
LE
C
TU
R
E
N
O
TE
S
O
F
AT
H
IT
H
A
N
S
? Functions of Random Variables
Page 2 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
1
1
...
ATHITHAN
Standard Discrete Distributions
Binomial Distribution
The Probability Mass Function (P
...
F
...
1
...
S
Note that
x = 0, 1, 2,
...
1
...
...
ATHITHAN
Second Moment = E(X 2 ) = MX00 (0)
MX00 (t) = np[et (q + pet )n−1 + et (n − 1)(q + pet )n−2 pet ]
MX00 (0) = np[1 + (n − 1)p] = np + n2 p2 − np2
S
Var(X) = E(X 2 ) − [E(X)]2
= np + n2 p2 − np2 − n2 p2 = np − np2
= np(1 − p) = npq
N
Poisson Distribution
A
1
...
AT
P (X = x) =
IT
H
The Probability Mass Function (P
...
F
...
2
...
...
2
...
ATHITHAN
t
MX (t) = eλ(e −1)
A
AT
H
Var(X) = E(X 2 ) − [E(X)]2
= λ2 + λ − λ2 = λ
IT
H
Second Moment = E(X 2 ) = MX00 (0)
t
t
MX00 (t) = (λet )2 eλ(e −1) + λet eλ(e −1)
MX00 (0) = λ2 + λ
N
S
First Moment = Mean = E(X) = MX0 (0)
t
MX0 (t) = λet eλ(e −1)
MX0 (0) = λ
O
F
Note that for Poisson Distribution, Mean=Variance= λ
...
3
N
O
A random variable X is said to follow Geometric distribution, if it assumes only non-negative
values and its probability mass function is given by
R
TU
where q = 1 − p
...
, 0 < p < 1
P (X = x) = q x−1 p, x = 1, 2, 3,
...
Page 5 of 27
https://sites
...
com/site/lecturenotesofathithans/home
1
...
1
Probability & Statistics
Moment Generating Function of Geometric Distribution
S
...
3
...
]
= p(1 − qet )−1 [since(1 − x)−1 = 1 + x + x2 +
...
3
...
Page 6 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
∞
X
P (X > r) =
S
...
]
1
r
= q p
= qr
1−q
(1)
(2)
P (X > s + k, X > s)
P (X > s)
(s+k)
q
=
= q k = P (X > k)
qs
S
Now
(3)
H
Uniform or Rectangular Distribution
F
2
...
1
...
...
1
...
ATHITHAN
ebt − eat
(b − a)t
MX (t) =
First Moment = Mean = E(X) = MX0 (0) doesn’t exist
The moment-generating function is not differentiable at zero
...
2
3
(b − a)2
Variance=E(X 2 ) − [E(X)]2 =
...
2
R
Moment Generating Function of E XPONENTIAL D ISTRIBUTION
TU
2
...
1
E
N
O
TE
S
A continuous random variable X is said to follow an exponential distribution with parameter
λ > 0 if its probability density function is given by
(
λe−λx , when , x > 0
f (x) =
0,
otherwise
LE
C
The MGF of Exponential distribution is given by
Z∞
tX
MX (t) = E(e ) =
x=0
Z∞
=
etx f (x)dx
etx λe−λx dx
0
Z∞
= λ
e−(λ−t)x dx
0
∞
e−(λ−t)x
= λ
−(λ − t) 0
λ
=
= λ(λ − t)−1
λ−t
Page 8 of 27
https://sites
...
com/site/lecturenotesofathithans/home
2
...
2
Probability & Statistics
Mean and Variance of Exponential Distribution
S
...
2
...
Z∞
P (X > r) =
λe−λx dx
x=r
e−λx
= λ
λ
∞
= e−λr
(4)
r
Now
P (X > s + k, X > s)
P (X > s)
−λ(s+k)
e
= e−λk = P (X > k)
=
e−λs
P (X > s + k/X > s) =
Page 9 of 27
(5)
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
2
...
ATHITHAN
Normal Distribution
Normal distribution is the most important continuous probability distribution in statistics both
from practical and theoretical point of views
...
(6)
e
−z 2
2
dz
(8)
O
−∞
N
MGF of Normal Distribution
MX (t) = eµt+
t2 σ 2
2
R
E
2
...
1
O
TE
S
This Φ(z) is called the standard normal curve which is bell shaped and symmetrical about the
line z = 0
...
C
Note 2
...
LE
1
...
In the binomial distribution with parameters n and p, when n is very large and p is nearly
1
the binomial approaches normal
...
Let X N (µ, σ )
...
For any x1 , x2 this integral
x1
cannot be evaluated in closed form and so it should be evaluated by numerical methods
...
So it is convenient to have a fixed table which could be used for all such
X −µ
computations
...
σ
Page 10 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
S
...
N
S
Table 1: Hints to solve the problems based on distributions
Attribute
Binomial Dist
...
Geometric Dist
...
of Sam- n < 10
n > 10
There is no information about
ples ‘n’
samples
Probability of p > 10%
p < 10%
0
F
Example Worked out Problems
O
3
AT
H
IT
H
A
Remark 2
...
However in most of the cases which are applicaple to B INOMIAL we may use
P OISSON distribution and vice versa
...
We may solve the
problems on discrete type when we consider the hints given in Table 1
...
TE
S
Example: 1
...
O
Hints/Solution: Here
TU
R
E
N
p = P (X = 3 heads or 3 tails in tossing 3 coins)
1 1
1
+ =
=
8 8
4
3
and q = 1 − p =
...
i
...
The first 4 trials resulted in 1 success and 3 failures
...
256
∴ Required Probability =
4
Example: 2
...
The probabilities
3
5
of their hitting the target at each shot are and respectively
...
Page 11 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
S
...
Then
X follows a geometric distribution given by
P (X = r) =
p1 q1r−1
r−1
2
3
...
=
5
5
Let Y denote the number of trials required by Q to get his first success
...
; r = 1, 2, 3,
...
)
=
=
AT
r=1
∞
X
P (X = r) × P (Y = r + 1, r + 2,
...
...
7
35
7
r=1
k=1
LE
C
TU
R
E
r−1 2
∞
X
4
3
7
=
...
4
35 r=1 35
35 1 − 35
31
Example: 3
...
If the produced items are sent to the market as packets of 20, find the number
of packets containing
1
...
exactly 2
3
...
Hints/Solution: We assume the Random Variable (R
...
) X follows the Poisson distribution
...
M
...
) of Poisson distribution is given by
Page 12 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
P (X = x) =
e−λ λx
,
x!
S
...
5
= 1
...
Here p = 5% = 0
...
368(2) = 1 − 0
...
264 u 26%
S
O
F
AT
H
P (X = 2) = {P (X = 2)}
e−1 12
=
2!
= 0
...
184 u 18%
E
N
O
TE
P (X < 2) = {P (X = 0) + P (X = 1)}
−1 0
e 1
e−1 11
=
+
0!
1!
−1
= e {1 + 1} = 0
...
736 u 74%
LE
C
TU
R
Note 3
...
To solve the above example we may use both the Binomial and Poisson distribution
...
The mean of a binomial distribution is 20 and standard deviation is 4
...
Hints/Solution: Let (n,p) be the parameters of the binomial distribution,
√
then mean =np and standard deviation= npq
...
...
ATHITHAN
1
∴ (1) ⇒ n
...
5
H
A
Hints/Solution: Let (n,p) be the parameters of the binomial distribution,
then mean =np and variance= npq
...
Comment on the following ”The mean of a binomial distribution is 3 and variance
is 4”
...
TE
S
O
So, there is no binomial distribution with this data, the statement is false
...
Six dice are thrown 729 times
...
Let X denote the number of success when 6
dice are thrown
...
∴ P (X = x) =n Cx px q n−x , x = 0, 1, 2, · · · , n
LE
C
Given n = 6 and p =probability of getting 5 or 6 =
1
2
=
6
3
1
2
=
3
3
x 6−x
2
1
6
∴ P (X = x) = Cx
, x = 0, 1, 2, · · · , 6
3
3
∴ q =1−p=1−
Page 14 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
S
...
729
F
AT
Hints/Solution: Let X denote the number of heads
...
∴ P (X = x) =n Cx px q n−x , x = 0, 1, 2, · · · , n
H
IT
Example: 7
...
12
2
TU
R
E
N
O
TE
∴ q =1−p=1−
LE
C
∴ the number of times to get 8 heads and 4 tails appearing in 256 sets= 256 × 0
...
72 ≈
31times
...
Ten coins are thrown simultaneously
...
Let X denotes number of heads
...
∴ P (X = x) =n Cx px q n−x , x = 0, 1, 2, · · · , n
1
1
Now p=probability of getting a head=
...
...
ATHITHAN
Required P (X ≥ 7) = P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)
10
10
10
10
1
1
1
1
10
10
10
10
+ C8
+ C9
+ C10
= C7
2
2
2
2
10
1
=
[120 + 45 + 10 + 1] = 0
...
A discrete random variable X has moment generating function MX (t) =
5
1 3 t
...
+ e
4 4
n
Hints/Solution: We know that moment generating function MX (t) = q + pet −−−−−(1)
5
1 3 t
...
F
3
1
∴ n = 5, p = , q =
...
0879
= C2
=
4
4
4
512
C
TU
R
E
∴ P (X = 2) = C2
N
5
O
3 1
15
and V ar [X] = npq = 5
...
=
4
4
LE
Example: 10
...
Find the probability of getting 6 heads 10 times
...
∴ λ = np = 6400 ×
1
= 100
...
...
ATHITHAN
Example: 11
...
5
...
some demand is refused
...
no car is used
...
no demand is refused
...
5 (1
...
5 (1
...
5 (1
...
5)
= 1 − e−1
...
5 (1
...
5
2
= 1 − [0
...
334 + 0
...
191
S
Hints/Solution: Given λ = np = 1
...
P (X > 2) = 1 − P (X ≤ 2)
(1
...
5
=e
= 0
...
P (X = 0) = e−1
...
P (X ≤ 2) = [P (X = 0) + P (X = 1) + P (X = 2)]
1
2
0
−1
...
5)
−1
...
5)
−1
...
5)
= e
+e
+e
0!
1!
2!
2
(1
...
5 + e−1
...
5) + e−1
...
233 + 0
...
251] = 0
...
Buses arrive at a specified bus stop at 15 minutes intervals starting at 7 am
...
e
...
00,7
...
30, and so on
...
00 and 7
...
for a bus
(b) at least 12 min for a bus
...
Then X is uniformly distributed over (0,30), i
...
, f (x) = , 0 < x < 30
...
if he arrives at the stop between 7
...
15 or 7
...
30
...
...
ATHITHAN
= P (10 < X < 15) + P (25 < X < 30)
Z30
Z15
1
1
1
dx +
dx =
=
30
30
3
∴ Required probability
25
10
S
(b) The passenger will have to wait at least 12 min
...
00 and
7
...
15 and 7
...
N
= P (0 < X < 3) + P (15 < X < 18)
Z18
Z3
1
1
1
dx +
dx =
=
30
30
5
H
A
∴ Required probability
AT
H
IT
15
0
TE
S
O
F
Example: 13
...
Find the probability
that the daily sales will be
(a) between 2500 and 3000 PC
(b) at least 4000 PC’s
N
O
Hints/Solution: Let X denote the no
...
1
Then X is uniformly distributed over (2000,5000), i
...
, f (x) =
, 2000 < x < 5000
...
C
TU
∴ Required probability
= P (2500 < X < 3000)
3000
Z
1
1
=
dx =
3000
6
LE
2500
(b) The probability that the company sells at least 4000 PC’s
...
...
ATHITHAN
Example: 14
...
Find the probability that these tires
will last
(a) at least 20,000 km
(b) at most 30,000 km
Hints/Solution: Let X denote the mileage obtained with the help of the tire, then f (x) =
x
1
e− 40,000 , x > 0
...
= P (X ≥ 20, 000)
Z∞
x
1
=
e− 40,000 dx = e−0
...
6065
...
75 = 0
...
40, 000
F
AT
∴ Required probability
H
(b) The probability that the tire last at most 30,000 km
...
The time required (in hours) to repair a machine is exponentially distributed
with mean 2 hours
...
R
1
1
= 2 =⇒ λ =
...
2
C
TU
Hints/Solution: Given mean of exponential distribution
LE
(a) The probability that the time exceeds 2 hours
...
0 = 0
...
2
2
(b) The probability that the time takes at least 10 hours given that its duration already exceeds
6 hours
...
0 = 1 − 0
...
8650
...
...
ATHITHAN
Example: 16
...
d 3
...
P (X ≥ 19)
2
...
5 < X < 19)
3
...
24
A
N
S
Hints/Solution: Given µ = 16 and σ = 3
...
When X = 19, Z =
3
0 0
...
5 − P (0 < z < 1)
= 0
...
3413
= 0
...
5 − 16
= 1
...
When X = 12
...
5
TE
19 − 16
=1
3
∴ P (12
...
16 < Z < 1)
= P (−1
...
16) + P (0 < z < 1)
= 0
...
3413
-1
...
3710
= 0
...
P (X > K) = 0
...
3710
0 0
...
16
z
LE
C
TU
R
E
N
O
When X = 19, Z =
∴ P (X > K) = 0
...
24
⇒ P (0 < Z < z1 ) = 0
...
24 = 0
...
26 and z1 = 0
...
7 ⇒ K = 3(0
...
1
∴
3
3
Page 20 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
S
...
In a test on 2000 electric bulbs, it was found that the life of a particular make
was normally distributed with an average life of 2040 hours and standard deviation of 60 hours
...
more than 2100 hours
2
...
more than 1920 hours but less than 2160 hours
...
X −µ
X − 2040
Put Z =
=
σ
60
2100 − 2040
=1
1
...
341352300
1
?
H
y
IT
H
A
∴ P (X ≥ 2100) = P (Z ≥ 1)
= 0
...
5 − 0
...
1587
z
0
...
5
60
TE
S
2
...
1587 ≈ 159
...
5
y
0 0
...
5
?
E
?
N
O
∴ P (X ≤ 1950) = P (Z ≤ −1
...
5 − P (0 < z < 1
...
5 − 0
...
0668
z
R
0
...
0668 ≈ 134
...
When X = 1920, Z =
1920 − 2040
= −2
60
2160 − 2040
=2
60
∴ P (1920 < X < 2160) = P (−2 < Z < 2)
= P (−2 < Z < 0) + P (0 < z < 2)
= P (0 < z < 2) + P (0 < z < 2)
0
...
4772)
-2
0
2
?
y
= 0
...
9544 ≈ 1908
...
...
ATHITHAN
Example: 18
...
Find
the mean and s
...
of the distribution
...
31 and P (X > 64) =
= 0
...
31
100
⇒ 45 − µ = −σz1
0
...
31
z
ϕ(z)dz = 0
...
5
y z=00
...
5
0
IT
H
∴ z1 = 0
...
495σ − − − − − (1)
64 − µ
= +z2 , ∵ P (X < 64) = 0
...
42
AT
0
...
341352300
z=0
Z2
O
−z1
F
0
...
31
H
When X = 64, Z =
0
LE
C
TU
R
E
N
O
TE
S
∴ z2 = 1
...
405σ − − − − − (2)
Solving (1) and (2), we get µ = 49
...
...
ATHITHAN
Exercise/Practice/Assignment Problems
1
...
The mean and variance of a binomial distribution are 8 and 6
...
IT
H
3
...
d
...
Determine the distribution
...
Out of 800 families with 4 children each, how many families would be expected to have
(1) 2 boys and 2 girls (2) atleast one boy (3) at most 2 girls (4) children of both sexes
...
Out of 2000 families with 4 children each, how many would you expect to have (1) at
least 1 boy (2) 2 boys (3) 1 or 2 girls (4) no girls?
TE
S
O
6
...
If he sells pins in
boxes of 100 and guarantees that not more than 4 pins will be defective
...
N
O
7
...
TU
R
E
8
...
01
...
The MGF of a random variable X is e3(e −1) , find P (X = 1)
...
Out of 500 companies with 4 secretaries each, how many companies would be expected
to have
(a) 2 Men and 2 Women executives
(b) More than one men executives
(c) At least 2 women executives
(d) Executives of both genders
(Assume both men and woment have equal probabilities)
11
...
Find: (i) P (5 ≤
X ≤ l0) , (ii) P (10 ≤ X ≤ 15), (iii) P (X ≥ 15), (iv) P (X ≤ 5)
...
...
ATHITHAN
12
...
The top 15% of the students receive A’s and the bottom 10% receive F’s
...
13
...
14
...
15
...
What is the probability that among 700 text books taken from the shipment contains
S
(a) At most 5 imperfect bindings
N
(b) At least 5 imperfect bindings
H
A
(c) All books with perfect bindings
IT
16
...
What is the probability that the person talks
AT
H
(a) more than 8 min
(b) between 4 and 8 min
S
O
F
17
...
Find the probability that a student selected at random has secured a total of
TE
(a) 180 and above
O
(b) 135 or less
TU
R
E
N
18
...
If in a particular year 5% of the students scored
distinction and 10% of the students failed, then what percentage of students who have
scored first and second class?
LE
C
19
...
f 5 18 28 12 7 6 4
20
...
f 142 156 69 27 5 1
21
...
f 2 14 20 34 22 8
22
...
If the probability
that the target is shot on any one of the shot 0
...
(a) What is the probability that the target would be hit on the sixth attempt?
Page 24 of 27
https://sites
...
com/site/lecturenotesofathithans/home
Probability & Statistics
(b) What is the probability that it takes odd number of shots?
S
...
If the probability that an applicant for a driving license will pass the road test on any given
trial is 0
...
What is the probability that he will pass the test
(a) on the fourth trial
(b) in fewer than 5 trials
(c) takes more than 4 trials
A
N
S
24
...
30 am
...
15 am the train has not
yet arrived, what is the probability that he will have to wait at least 10 more additional
minutes?
H
IT
H
25
...
What are the expected values of the roots of the equation? Assume
that b is uniformly distributed in the permissible range
...
The daily consumption of milk in excess of 20,000 gallons is approximately exponentially
distributed with mean 3000
...
What is the
probability that of two days selected at random, the stock is insufficient for both the days?
TE
S
O
27
...
Find the probability that such a watch will
(i) will have to be set in less than 24 days and (ii) not have to set in at least 180 days
...
If X is normally distributed with mean 8 and s
...
Find
N
O
(a) P (5 ≤ X ≤ 10)
R
(c) P (X ≥ 15)
E
(b) P (10 ≤ X ≤ 15)
TU
29
...
What
are the mean and standard deviation of the distribution?
LE
C
30
...
D
...
Estimate the number of students whose marks will be
(a) between 60 and 75
...
(c) less than 65 marks
...
s@ktr
...
ac
...
...
...
ATHITHAN
O
TE
S
O
F
AT
H
IT
H
A
N
S
Probability & Statistics
LE
C
TU
R
E
N
Figure 1: Values of e−λ
...
...
ATHITHAN
LE
C
TU
R
E
N
O
TE
S
O
F
AT
H
IT
H
A
N
S
Probability & Statistics
Figure 2: Area under Standard Normal Curve
...
Title: Probability distributions
Description: This file is a pdf for the topic "probability distributions" from the subject " probability and statistics"
Description: This file is a pdf for the topic "probability distributions" from the subject " probability and statistics"