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APPROVED BY AICTE, ACCREDITED BY NBA & NAAC – 'A' GRADE ISO 9001:2015 CERTIFIED
MALLA REDDY COLLEGE OF
ENGINEERING & TECHNOLOGY
Recognized under 2(f) and 12(B) of UGC ACT 1956
(Affiliated to JNTUH, Hyderabad, Approved by AICTE –Accredited by NBA & NAAC-“A” Grade-ISO 9001:2015 Certified)
PROBABILITY AND STATISTICS
B.Tech – II Year – I Semester
The student acquires solid skills of implementing and applying the computational methods
learnt.
The idea is that making the student comfortable with both advanced mathematical concepts and
modern computational techniques, will open a wealth of possibilities of applying mathematics to
problems of real interest.
Probability and Statistics: Course Description
This course is about the mathematics that is most widely used in the engineering core subjects.
Probability and Statistics provide an introduction to discrete and continuous probability
distributions, correlation and regression analysis, sampling distributions and sampling
inferences.Topics include the properties of both single and multiple random variables for the
discrete and continuous probability distributions, correlation and regression analysis for bivariate
as well as multivariate distributions, Estimation, sampling, testing of hypothesis for both large and
small samples.
INDEX
UNIT-I : Random Variables
UNIT-II : Probability Distributions
UNIT-III : Correlation and Regresion
UNIT-IV : Sampling and Testing of Hypothesis for Large Samples
UNIT-V : Testing of Hypothesis for Large Samples
Sampling: Definitions - Types of sampling - Expected values of sample mean and variance,
Standard error - Sampling distribution of means and variance. Estimation - Point estimation
and Interval estimation.
Testing of hypothesis: Null and Alternative hypothesis - Type I and Type II errors, Critical
region - confidence interval - Level of significance, One tailed and Two tailed test.
Large sample Tests: Test of significance - Large sample test for single mean, difference of
means, single proportion, and difference of proportions.
Small samples: Test for single mean, difference of means, paired t-test, test for ratio of variances
(F-test), Chi- square test for goodness of fit and independence of attributes.
Suggested Text Books:
i) Fundamental of Statistics by S.C. Gupta, 7
th
Edition,2016.
ii) Fundamentals of Mathematical Statistics by SC Gupta and V.K.Kapoor
iii) Higher Engineering Mathematics by B.S.Grewal, Khanna Publishers,
th
Edition,2000.
References:
i) Introduction to Probability and Statistics for Engineers and Scientists by
Sheldon M.Ross.
ii) Probability and Statistics for Engineers by Dr. J. Ravichandran.
Course Outcomes: After learning the concepts of this paper the student will be able to
independently
continuous type and compute statistical constants of these random variables.
probability distributions.
regression.
a population.
UNIT 1
RANDOM VARIABLES
Random Variable
A Random Variable X is a real valued function from sample space S to a real number R.
(or)
A Random Variable X is a real number which is determined by the outcomes of the random
experiment.
Eg:1.Tosssing 2 coins simultaneously
Sample space ={HH,HT,TH,TT}
Let the random variable be getting number of heads then
2.Sum of the two numbers on throwing 2 dice
Types of Random Variables:
1.Discrete Random Variables : A Random Variable X is said to be discrete if it takes only the
values of the set {0,1,2…..n}.
Eg:1.Tosssing a coin, throwing a dice, number of defective items in a bag.
given interval of domain.
Eg: Heights, weights of students in a class.
Discrete Probability Distribution:
Let x is a Discrete Random Variable with possible outcomes 𝑥
ଵ,
ଶ
ଷ
having probabilities
)𝑓𝑜𝑟 𝑖 = 1,2 … 𝑛 .If 𝑝(𝑥
ୀଵ
then the function 𝑝(𝑥
) is called
Probability mass function of a random variable X and {𝑥
} 𝑓𝑜𝑟 𝑖 = 1,2 … 𝑛 is called
Discrete Probability Distribution.
Eg: Tossing 2 coins simultaneously
Sample space ={HH,HT,TH,TT}
Let the random variable be getting number of heads then
Probability of getting no heads =
ଵ
ସ
, Probability of getting 1 head =
ଵ
ଶ
Probability of getting 2 heads =
ଵ
ସ
∴Discrete Probability Distribution is
Cumulative Distribution function is given by 𝐹
௫
ୀ
Properties of Cumulative Distribution function:
Mean: The meanof the discrete Probability Distribution is defined as
∑ ௫
(௫
)
సభ
∑
( ௫
)
సభ
ୀଵ
since
ୀଵ
Expectation:The Expectationof the discrete Probability Distribution is defined as
ୀଵ
In general,𝐸(𝑔(𝑥)) = ∑ 𝑔(𝑥
ୀଵ
Properties:
Variance: The variance of the discrete Probability Distribution is defined as
ଶ
ଶ
ଶ
ଶ
ଶ
Properties:
1)V(c) = 0 where c is a constant
2)V(kX) = k
ଶ
3)V(X + k) = V(X)
4)V(aX ± b) = a
ଶ
Problems
a)Find the Probability distribution of defective cars.
b)Find the Expected number of defective cars.
Sol: Number of ways to select 3 cars from 6 cars = 6
య
Let random variable X(S) = Number of defective cars = {0,1,2}
Probability ofnon defective cars =
ସ ౙ
య
ଶ ౙ
బ
ౙ
య
ଵ
ହ
Probability of one defective cars =
ସ
మ
ଶ
భ
య
ଷ
ହ
Probability of two defective cars =
ସ
ౙ భ
ଶ
ౙ మ
ౙ య
ଵ
ହ
Clearly ,p
x
୧
p
x
୧
୬
୧ୀଵ
Probability distribution of defective cars is
Expected number of defective cars =
x
୧
p
x
୧
୬
୧ୀଵ
ଵ
ହ
ଷ
ହ
ଵ
ହ
2.Let X be a random variable of sum of two numbers in throwing two fair dice. Find the
probability distribution of X, mean ,variance.
Sol: Sample space of throwing two dices is
Let X = Sum of two numbers in throwing two dice = {2,3,4,5,6,7,8,9,10,11,12}
Sol: Sample space of throwing two dices = S ={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
Let X = Maximum of two numbers in throwing two dice = {1,2,3,4,5,6,}
X Favorable cases No of
Favorable
cases
Clearly ,p
x
୧
> 0 and
p
x
୧
୬
୧ୀଵ
Probability distribution is given by
Mean = μ = x
୧
p(x
୧
୬
୧ୀଵ
Variance = V
x
୧
ଶ
p
୧
− μ
ଶ
∴ Variance = 1.99.
4.A random variable X has the following probability function
𝒊
𝒊
k 0.1 k 0.2 2k 0.4 2k
Find k ,mean, variance.
Sol: We know that∑ p(x
୧
୬
୧ୀଵ
i.e k+0.1+k+0.2+2k+0.4+2k = 1
i.e 6k+0.7 = 1 ∴ 𝑘 = 0.
Mean = μ = x
୧
p(x
୧
୬
୧ୀଵ
= k(−3) + 0.1(−2) + k(−1) + 2k( 1 ) + 2(0.4) + 3(2k) = 0.
Variance = V
x
୧
ଶ
p
୧
− μ
ଶ
= k
ଶ
ଶ
ଶ
2k
∴ Variance = 2.86.
5.A random variable X has the following probability distribution
Determine i) k ii) Mean iii) Variance.
Sol: Given probability distribution of a random variable X is
i) Since total probability of the distribution is unity i.e, ∑ 𝑃
ୀଵ
We have,
ଶ
ଶ
ଶ
ଶ
ii) Mean = 𝜇 =
ୀଵ
ଶ
ଶ
ଶ
ଶ
iii) Variance =𝜎
ଶ
ଶ
ୀଵ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
ଶ
Determine i) k ii) P (1≤ 𝒙 ≤ 𝟓) iii) P(x>3)
Sol: Given probability distribution of a random variable X is
(i)Since total probability of the distribution is unity i.e, ∑ 𝑃
ୀଵ
We have, 𝑘 + 3𝑘 + 5𝑘 + 7𝑘 + 9𝑘 + 11𝑘 = 1 ⟹ 𝑘 =
ଵ
ଷ
ii)P (1≤ 𝑥 ≤ 5) = 𝑃( 1 ) + 𝑃( 2 ) + 𝑃( 3 ) + 𝑃( 4 ) + 𝑃(5)
iii)𝑃(𝑥 > 3)=𝑃( 4 ) + 𝑃( 5 ) + 𝑃(6) = 7𝑘 + 9𝑘 + 11𝑘 = 27𝑘 = 0.
x 0 1 2 3 4 5 6 7
P(x) 0 k 2k 2k 3k 𝒌
𝟐
𝟐
𝟐
x 0 1 2 3 4 5 6 7
P(x) 0 k 2k 2k 3k 𝑘
ଶ
ଶ
ଶ
x 1 2 3 4 5 6
P(x) k 3k 5k 7k 9k 11k
x 1 2 3 4 5 6
P(x) k 3k 5k 7k 9k 11k
Continuous Probability distribution:
Let X be a continuous random variable taking values on the interval (a,b). A function f
x
is
said to be the Probability density function of x if
i) f
x
> 0 ∀ x ∈ (a, b)
ii) Total area under the probability curve is 1i. e, ∫
f(x)dx = 1.
ୠ
ୟ
iii) For two distinct numbers ‘c’ and ‘d’ in
is given by P
c < 𝑥 < 𝑑
Area under the probability curve between ordinates x = c and x = d i. e
∫ f
x
dx.
ୢ
ୡ
Note: P(c < 𝑥 < 𝑑) = P(c ≤ x ≤ d) = P(c ≤ x < 𝑑) = P(c < 𝑥 ≤ 𝑑)
Cumulative distribution function of 𝑓(𝑥) is given by
∫ f
x
dx
୶
ିஶ
i.e, f
x
ୢ
ୢ ୶
F(x)
Mean: The meanof the continuous Probability Distribution is defined as
μ = න x f
x
dx.
ஶ
ିஶ
Expectation:The Expectationof the continuous Probability Distribution is defined as
x f(x)dx.
ஶ
ିஶ
In general,E(g(x)) = ∫
g(x) f(x)dx.
ஶ
ିஶ
Properties:
1)E(X) = μ
2)E(X) = k E(X)
3 ) E(X + k) = E(X) + k
Variance: The variance of the Continuous Probability Distribution is defined as
Var(X) = V(X) = න x
ଶ
f
x
dx − μ
ଶ
ஶ
ିஶ
Properties:
1)V(c) = 0 where c is a constant
ଶ
V(X + k) = V(X)
V(aX ± b) = a
ଶ
Mean Deviation: Mean deviation of continuous probability distribution function is defined
as M.D = ∫
|x − μ| f(x)dx.
ஶ
ିஶ
Median: Median is the point which divides the entire distribution in to two equal parts. In case
of continuous distribution, median is the point which divides the total area in to two
equal parts i.e.,∫ f
x
dx = ∫ f
x
dx =
ଵ
ଶ
∀ x ∈ (a, b)
ୠ
ୟ
Mode: Mode is the value of x for which f
x
is maximum.
i.ef
ᇱ
x
= 0 and f
"
x
< 0 for x ∈ (a, b)
Problems
1.If the probability density function𝒇
𝒌
𝟏ା𝒙
𝟐
− ∞ < 𝑥 < ∞. Find the value of ‘k’ and
probability distribution function of𝐟(𝐱).
Sol: Since total area under the probability curve is 1i. e, ∫ f
x
dx = 1.
ୠ
ୟ
k
1 + x
ଶ
dx = 1.
ஶ
ିஶ
2k(tan
ି ଵ
x)
2k(tan
ି ଵ
∞ − tan
ି ଵ
∴ k =
π
Cumulative distribution function of f(x) is given by
න f(x)dx = න
𝟐
dx =
π
(tan
ି ଵ
x)
x
π
π
ି ଵ
x)].
୶
ିஶ
୶
ିஶ
ି
| 𝐱
|
Find the value of ‘c’, mean and variance.
Sol: Since total area under the probability curve is 1 i. e, ∫ f
x
dx = 1.
ୠ
ୟ
ି
| 𝐱
|
dx = 1
ஶ
ିஶ
ି 𝐱
dx = 1
ஶ
2c (
𝐞
ష𝐱
ିଵ
ஶ
∴ c =
Mean,μ = ∫
x f
x
dx =
ଵ
ଶ
x𝐞
ି
| 𝐱
|
dx = 0 since x𝐞
ି
| 𝐱
|
is an odd function.
ஶ
ିஶ
ஶ
ିஶ
Variance = V(X)
= න x
ଶ
f
x
dx − μ
ଶ
ஶ
ିஶ
න x
ଶ
ି |𝐱|
dx
ஶ
ିஶ
න 2x
ଶ
ି 𝐱
dx = [x
ଶ
ି 𝐱
) − 2x(𝐞
ି 𝐱
ି 𝐱
ஶ
𝐬𝐢𝐧𝐱
𝟐
Find mean, median, mode and𝐏(𝟎 < 𝐱 <
𝛑
𝟐
Sol:Mean =μ = ∫
x f
x
dx =
ଵ
ଶ
x
𝐬𝐢𝐧𝐱
𝟐
dx =
ଵ
ଶ
−xcosx + sinx
ଶ
ஶ
ିஶ
Let M be the Median then
න f
x
dx = න f
x
dx =
∀ x ∈
sinx
dx = න
sinx
dx =
∀ x ∈ (−∞, ∞)
Consider ∫
𝐬𝐢𝐧𝐱
𝟐
dx =
ଵ
ଶ
then (– cosx)
ଶ
Since f
x
ୱ୧୬୶
ଶ
0 , otherwise
if 0 ≤ x ≤ π
6.The diameter of ban electric cable assumed to be a continuous r.v with p.d.f
Find i)k ii)b such that P(xb).
Sol: Given probability density function of a random variable X is
(i) Since total probability of the distribution is unity i.e, ∫
𝑓(𝑥)𝑑𝑥 = 1
ஶ
ିஶ
We have ∫
ଵ
ଶ
ଷ
ଵ
(ii) Given that P(xb)
ଵ
ଵ
ଶ
ଷ
ଶ
ଷ
ଵ
ଶ
ଷ
ଶ
ଷ
మ
ଶ
య
ଷ
ଵ
ଶ
ଷ
Solving above equation,we get
b = 0.5 ( by neglecting other roots which do not belong to
Multiple Random Variables
Discrete two-dimensional random variable:
Joint probability mass function is defined as f(x, y) = P(X = x
୧
, Y = y
୨
Joint probability distribution function is defined as
ଡ଼ଢ଼
(x, y) = P(X < x
୧
, Y < y
୨
p(x
୧
, y
୨
ழ௫ ழ௬
Marginal probability mass functions of X and Y are defined as
P(X = x
୧
) = p(x
୧
) = p(x
୧
, y
୨
୨
P൫Y = y
୨
൯ = p൫y
୨
൯ = p(x
୧
, y
୨
୧
Continuous two-dimensional random variable:
Joint probability density function is defined as
f
ଡ଼ଢ଼
x, y
= P(x ≤ X ≤ x + dx, y ≤ Y ≤ y + dy)
and ∫ ∫
f
ଡ଼ଢ଼
(x, y) dxdy = 1
ஶ
ିஶ
ஶ
ିஶ
Joint probability distribution function is defined as
ଡ଼ଢ଼
(x, y) = P(X < x
୧
, Y < y
୧
f
ଡ଼ଢ଼
(x, y) dxdy
୷
ିஶ
୶
ିஶ
andf
ଡ଼ଢ଼
(x, y) =
ப
మ
ப୶ ப୷
ଡ଼ଢ଼
(x, y) ]
Marginal probability density functions of Xis defined as
f
ଡ଼
(x) = න f
ଡ଼ଢ଼
(x, y) dy
ஶ
ିஶ
Marginal probability density functions of Yis defined as
f
ଢ଼
y
=∫ f
ଡ଼ଢ଼
x, y
dx
ஶ
ିஶ
Conditional probability density function :
Conditional probability density function of X on Y is
f
ଡ଼ଢ଼
f
ଡ଼ଢ଼
x, y
f
ଢ଼
(y)
Conditional probability density function of Y on X is
f
ଡ଼ଢ଼
f
ଡ଼ଢ଼
(x, y)
f
ଡ଼
Problems
1.For the following 2-d probability distribution of X and Y
Find i) P(X≤ 𝟐, 𝐘 = 𝟐) ii)𝐅 𝐗
iii)P(Y=3) iv) P(X< 𝟑, 𝐘 ≤ 𝟒) v)𝐅
𝐲
Sol: Given
i)P(X≤ 2, Y = 2) =P(X= 1, Y = 2)+ P(X= 2, Y = 2)
ii) F
ଡ଼
p(x
୧
, y
୨
୨
p൫x
୧
, y
୨
୨
iii) P(Y=3) =
p(x
୧
, y
୨
୧
iv) P(X< 3, Y ≤ 4) =P(X< 3, Y = 1)+ P(X< 3, Y = 2)+ P(X< 3, Y = 3)
v)F
୷
2.Suppose the random variables X and Y have the joint density function defined by
Find i)c ii)P(X>3,Y>2) iii) P(X>3)
