probability and statistics

Probability and Statistics
UNIT-I
1. Define the terms:
(i) Sample space and sample point (ii) Mutually exclusive events (iii) Classical, Mathematical or Apriori definition of
probability
Soln.: (i) Set S of all equally likely outcomes of a random experiment is called sample space, e.g. in rolling of a die,
S={1,2,3,4,5,6}. Each element of S is sample point.
(ii) A number of events A1, A2,…, An are said to be mutually exclusive when occurrence of any one of the events
excludes the occurrence of all other events. Alternatively, the events are called disjoint events. Thus, for
mutually exclusive events A1, A2,…, An, we have A1∩A2∩…∩An = ∅.
(iii) Let S be the set of all equally likely outcomes of a random experiment, and A be any event (a subset of S)
associated with the experiment. Then the probability of occurrence of A, denoted by P(A) defined as
P(A)=
𝑛(𝐴)
𝑛(𝑆)
=
number of outcomes favourable to 𝐴
Exhaustive number of cases in 𝑆
.
For example, in a die rolling experiment S={1, 2, 3, 4, 5, 6}, event A of showing up odd numbers is A={1, 2, 3}.
Therefore P(A)=n(A)/n(S)=3/6=0.5
2. Write Axiomatic definition of probability. Show that (i) P(Φ) = 0 (ii) P(A’)=1-P(A)
Definition: Probability P of event A is a real-valued function P: A→*0, 1+ satisfying the following axioms:
(i) 0<P(A)<1; (ii) P(S) = 1, S is sample space.
(iii) If A1 and A2 are mutually exclusive events then P(A1UA2)=P(A1) + P(A2)
Proofs:
(a) P(SUΦ)=P(S) or P(S)+P(Φ)=P(S) or P(Φ)=0.
(b) P(AUA’)=P(S) or P(A)+P(A’)=1 So P(A’)=1- P(A)
3. Show that (i) P(A∩B’)= P(A)-P(A∩B) (ii) P(A’∩B)= P(B)-P(A∩B)
Proofs: (i) From the Venn diagram, we see that
A = (A∩B’)U (A∩B).
Since here (A∩B’) ∩ (A∩B) =Φ,
we use axiom A3 of axiomatic definition of probability,
to get P(A)=P(A∩B’)+P(A∩B)
Or P(A∩B’)=P(A)-P(A∩B) QED
(ii) Similarly (A∩B)∩(A’∩B)=Φ, and B=(A∩B)U(A’∩B).
So, P(B)=P(A∩B) +P(A’∩B) by A3. Thus, P(A’∩B)= P(B)-
P(A∩B) QED
4. If B A
 then show that (i) P(A∩B’)= P(A)-P(B) (ii)
P(B)< P(A).
Sol.: (i) We have P(A∩B’) = P(A)-P(A∩B)
From Venn diagram, B = A∩B.
Therefore P(A∩B’) = P(A) - P(B).
Alternatively A=(A∩B’)UB where (A∩B’)∩B= Φ.
By A3, P(A)=P(A∩B’)+P(B ). So P(A∩B’)=P(A)-P(B)
(ii) By axiom A1 of probability definition, we have
P(A∩B’) > 0. Using result (i) we get P(A)-P(B) > 0.
Therefore, P(B)< P(A).
5. State and prove addition theorem for two events.
Addition Theorem: For any two non-disjoint events A and B,
P(AUB)=P(A)+P(B)-P(A∩B)Proof: From the diagram:
A=(A∩B’)U (A∩B) (1)
B=(A∩B)U(A’∩B) (2)
Here (A∩B’)∩(A∩B)∩(A’∩B)=Φ.

But AUB=(A∩B’)U(A∩B)U(A’∩B). So, P(AUB)= P(A)+P(A’∩B) (3)
[By axiom A3 and Eq.(1)]
From Eq.(2): P(B)=P(A∩B) +P(A’∩B) by A3.
Or P(A’∩B)= P(B)-P(A∩B) (4)
Using Eq.(4) in (3): P(AUB)=P(A)+P(B)-P(A∩B)
6. Assume that in a nuclear accident 30% of the workers are exposed to LD50 and die; 40% of the workers die; and
68% are exposed to LD50 or die. What is the probability that a randomly selected worker is exposed to the LD50?
Use a Venn diagram to find the probability that a randomly selected worker is exposed to the LD50 but does not
die. Find the probability that a randomly selected worker is not exposed to the LD50 but dies.
Soln. 6: Let E: Worker is exposed to LD50,
D: Worker dies.
Given: P(E∩D)=.30, P(D)=.40, P(EUD)=.68
P(EUD)= P(E) + P(D) - P(E∩D)
So, P(E)= P(EUD) - P(D) + P(E∩D)=.68-.40+.30=.58,
P(E∩D’)= P(E)-P(E∩D)=.58- .30 = .28,
P(E’∩D)= P(D)-P(E∩D)=.40- .30 = .10
7. Assume that the engine component of a spacecraft
consists of two engines in parallel. If the main engine is 95% reliable, the backup is 80% reliable and the engine
component as a whole is 99% reliable, what is the probability that both engines will be operable? Use a Venn
diagram to find the probability that the main engine will be fail but backup will be operable. Find the probability
that the backup engine will fail but the main engine will be operable. What is the probability that the engine
component will fail?
Sol: Let M: Main engine functioning; B: Backup engine
functioning. Given P(M)=0.95, P(B)=0.80, P(MUB)=0.99.
To find P(M∩ B), P(𝑀 ∩ 𝐵), 𝑃(𝑀 ∩ 𝐵) and P(𝑀 ∪ 𝐵).
So, (i) P(M∩ 𝐵) = 𝑃 𝑀 + 𝑃 𝐵 − 𝑃(𝑀 ∪ 𝐵)
=0.95+0.80-0.99=0.76.
(ii) P(𝑀 ∩ 𝐵)=P(B) - P(𝑀 ∩ 𝐵) = 0.80-0.76=0.04,
(iii) 𝑃(𝑀 ∩ 𝐵) )=P(M) - P(𝑀 ∩ 𝐵)=0.95-0.76=0.19
(iv) P(𝑀 ∪ 𝐵)= 1-P(𝑀 ∪ 𝐵)=1-0.99=0.01.
8. Assume that in a particular military exercise involving two units, Red and Blue, there is a 60% chance that the
Red unit will successfully meet its objectives and a 70% chance that the blue unit will do so. There is an 18%
chance that only the Red unit will be successful. What is the probability that both units will meet their objectives?
What is the probability that one or the other but not both of the units will be successful?
Sol. 8: Let R: Red unit will successfully meet its objectives;
and B: Blue unit will successfully meet its objectives
Given: P[R]= 0.6, P[B]=0.7, P[R∩ 𝐵]=P[R]-P[R∩B]=0.18,
(i)P[R∩B] =P[R]-P[R∩ 𝐵] =0.6-0.18=0.42
(ii) P(RUB)- P[R∩B]= P(R)+P(B)- 2P[R∩B] =0.6+0.7-2(0.42)=0.42.
9. When a computer goes down, there is a 75% chance that it is due to an overload and a 15% chance that is due to
software problem. There is an 85% chance that it is due to an overload or a software problem. What is the
probability that both of these problems are at fault? What is the probability that there is a software problem but
no overload?
Sol. 9: Let A1: Computer goes down due to overload;
A2: Computer goes down due to software problem
Given P(A1)=0.75; P(A2)=0.15; P(A1U A2)=0.85;
(i) P[A1∩ A2]= P[A1]+P[ A2]-P[A1 UA2]=.75+.15-.85=.05
(ii) P(𝐴1 ∩ 𝐴2)=P(𝐴2) - P(𝐴1 ∩ 𝐴2) = 0.15-0.05=0.10.
10. Define Conditional probability. State and prove multiplication rule.
Sol.10: Probability of occurrence of an event A (say) given that another event B (say) has already occurred is
called Conditional probability, denoted by P(A/B) and is read as probability of A given B. It is defined as

P(A|B)=n(A∩B)/n(B). Similarly, P(B|A)=n(A∩B)/n(A)
Obviously P(A|B)=[n(A∩B)/n(S)]/[n(B)/n(S)]=P(A∩B)/P(B). So, P(A∩B)= P(A|B) P(A).
This is called multiplication rule of probability.
Proceeding in same way as above with P(B|A)=n(A∩B)/n(A), we have P(A∩B)= P(B|A) P(B).
11. In studying the causes of power failures, these data have been gathered: 5% are due to transformer damage,
80% are due to line damage, 1% involves both problems. Based on these percentages, approximate the
probability that a given power failure involves
(i) Line damage given that there is a transformer damage
(ii) Transformer damage given that there is line damage
(iii) Transformer damage but not line damage
(iv) Transformer damage given that there is no line damage
(v) Transformer damage or line damage
Sol.11: Let T: Transformer damage causes power failure
L: Line damage causes power failure
Given: P(T)=0.05, P(L)=0.80, P(T∩L)=0.01,
Find P(L|T), P(T|L), P(T∩L’), P(T|L’) and P(TUL).
(i) P(L|T)=
𝑃(𝐿∩𝑇)
𝑃(𝑇)
=.01/.05=0.20,
(ii) P(T|L)=
𝑃(𝑇∩𝐿)
𝑃(𝐿)
=.01/.80=.0125
(iii) P(T∩L’) = P(T)-P(T∩L)=0.05-0.01=0.04
(iv) P(T|L’) )=
𝑃(𝑇∩𝐿′ )
𝑃(𝐿′ )
=.04/(1-0.80)=.04/.20=0.20
(v) P(TUL)= P(T)+ P(L)-P(T∩L)= 0.05+0.80-0.01=0.84.
12. In a study of waters near power plants and other industrial plants that release waste water into the water
system it was found that 5% showed signs of chemical and thermal pollution, 40% showed signs of chemical
pollution, and 35% showed evidence of thermal pollution. Assume that the results of the study accurately reflect
the general situation. What is the probability that a stream that shows some thermal pollution will also show
signs of chemical pollution? What is the probability that a stream showing chemical pollution will not show signs
of thermal pollution?
Sol.12: Let C: water system showed signs of chemical pollution
T: water system showed signs of thermal pollution
Given: P(C∩T)=0.05, P(C)=0.40, P(T)=0.35
(i) P(C|T)=
𝑃(𝐶∩𝑇)
𝑃(𝑇)
=.05/.35=.1429
(ii) P(T’∩ 𝐶) = P(C)-P(C∩T)=0.40-0.05=0.35
(iii) P(T’|C) )=
𝑃(𝑇′ ∩𝐶)
𝑃(𝐶)
=0.35/0.40=0.35/0.40=0.875
13. Assume that in a nuclear accident 30% of the workers are exposed to LD50 and die; 40% of the workers die; and
68% are exposed to LD50 or die. What is the probability that a randomly selected worker is exposed to the LD50?
Use a Venn diagram to find the probability that a randomly selected worker is exposed to the LD50 but does not
die. Find the probability that a randomly selected worker is not exposed to the LD50 but dies. (i) What is the
probability that a randomly selected worker will die given that he is exposed to LD50 (ii) what is the probability
that a randomly selected worker will not die given that he is exposed to LD50? (iii) What is the probability that a
randomly selected worker will die given that he is not exposed to the LD50?
Sol.13: Let E: Worker is exposed to LD50, D: Worker dies.
Given : P(E∩D)=.30, P(D)=.40, P(EUD)=.68
P(EUD)= P(E) + P(D) - P(E∩D)
So, P(E)= P(EUD) - P(D) + P(E∩D)=.68-.40+.30=.58,
P(E∩D’)= P(E)-P(E∩D)=.58- .30 = .28,
P(E’∩D)= P(D)-P(E∩D)=.40- .30 = .10
(i) P(D|E)=
𝑃(𝐷∩𝐸)
𝑃(𝐸)
=.30/.58 =15/29 =.51
(ii) P(D’|E)=
𝑃(𝐷′ ∩𝐸)
𝑃(𝐸)
=.28/.58=14/29=.48

(iii) P(D’|E)=
𝑃(D∩𝐸′ )
𝑃(𝐸′ )
=.10/(1.00-0.58)=.10/.42=.238
14. Define the Independence of two events. Consider the experiment of drawing a card from a well-shuffled deck of
52 cards. The event A1 denote drawing a spade card and the event A2 denote an honor (10, J, Q, K, A). Are the
events A1 and A2 are independent?
Sol.: P(A1)=13/52 and P(A2)=20/52=5/13; P(A1| A2)=(13/52)(20/52)=5/52. Also, P(A1∩ A2)=5/52.
15. Let A and B be events such that P(A) =0.5, P(B)=0.7, what must be P(A∩B) equal for A and B to be independent?
Sol.15: For A and B to be independent, we must have
P(A∩B) =P(A) P(B) =.5(.7) =.35
16. Let A and B be events such that P(A) =0.6,, P(B)=0.4 and P(AUB)= 0.8. Are A and B independent?
Sol. 16: P(A∩B)=P(A)+P(B)-P(AUB) =.6+.4-.8=.2
P(A) P(B)=.6(.4)=.24, Thus, P(A∩B)≠ P(A) P(B)
So A and B are not independent
17. The probability that a unit of blood was donated by a paid donor is 0.67. If the donor was paid, the probability of
contracting serum hepatitis from the unit is 0.0144. If the donor was not paid, the probability of contracting
serum hepatitis from the unit is 0.0012. A patient receives a unit of blood. What is the probability of the patient’s
contracting serum hepatitis from this source?
Sol.17: Let T: a unit blood was donated by a paid donor,
H: a person contracts hepatitis from a unit of blood.
Given P(D)=.67, P(H|D)=.0144, P(H|D’)=.0012, P(H)=P(H∩D)+P(H∩D’)=P(H|D) P(D)+P(H|D’)P(D’)
=(.0144)(.67)+(.0012)(.33)= .010044
18. *Assume that there is a 50% chance of hard drive damage if a power line to which a computer is connected is hit
during an electrical storm. There is a 5% chance that an electrical storm will occur on any given summer day in a
given area. If there is a 0.1% chance that the line will be hit during a storm, what is the probability that the line
will be hit and there will be hard drive damage during the next electrical storm in this area?
Sol.: Given : P[D|(H∩E)]=.50, P(E)=.05, P(H∩E)=.001
To find P*(H∩D)|E+ we proceed as given below.
P*(H∩D)|E+ =P*H∩D∩E)/P(E)
= P*D|(H∩E)+.P(H∩E)/P(E)
=.5×.001/.05 = 0.10
19. A foundry is producing cast iron parts to be used in the automatic transmissions of trucks. There are two crucial
dimensions to the part, A and B. Assume that if the part meets specifications on dimension A then there is a 98%
chance that it will also meet specifications on dimension B. There is a 95% chance that it will also meet
specifications on dimension A and a 97% chance that it will meet specifications on dimension B. A part is
randomly selected and inspected. What is the probability that it will meet specifications on both dimensions?
Sol.: Let A: part meets specifications on dimension A
and B : part meets specifications on dimension AB.
Given : P(B|A)=.98, P(A)=.95, P(B)=.97
P[meeting specifications on both dimensions ]
=P(A∩B)=P(B|A) P(A) =.98(.95)
20. The use of plant appearance in prospecting for ore deposits is called geobotanical prospecting. One indicator of
copper is a small mint with a mauve-colored flower. Suppose that, for a given region, there is a 30% chance that
the soil has a high copper content and 23% chance that the mint will be present there. If the copper content is
high, there is a 70% chance that the mint will be present.
(i) Find the probability that the copper content will be high and the mint will be present
(ii) Find the probability that the copper content will be high given that the mint is present.
Sol. 20 : Let C : Soil has high copper content
and M : Soil has some mint content.
Given : P(C)=.30, P(M)=.23, P(M|C)=.70,
(i) P(M∩C)=P(M|C) P(C) =.70(.30)=.21
(ii) P(C|M)=P(C∩M)/ P(M) =.21/.70=.30
21. The blood type distributions in the United States are 41% of type A, 9% of type B, 4% of type AB and 46% of type
O. It is estimated that during World War-II, 4% of inductees with type O blood were typed as A; 88% of those with

type A were correctly typed; 4% with type B blood were typed as A; 10% with type AB were typed as A. A soldier
was wounded and brought to surgery. He was typed as having type A blood. What is the probability that I his
true blood type?
Sol. 21: Let B1, B2, B3, B4 be the person to have type of blood as A, B, AB, O respectively; and A be typed as to
have blood of type A.
Here, B1, B2, B3, B4 are mutually exclusive and their union is S. (Conditions for Bayes theorem)
Given: P(B1)=.41, P(B2)=.09, P(B3)=.04, P(B4)=.46
P(A|B4)=.04, P(A|B1)=.88, P(A|B2)=.04, P(A|B3)=.10
Here total probability P(A)= P(A|B1)P(B1)
+P(A|B2) P(B2)+P(A|B3) P(B3) +P(A|B4) P(B4)
=.88(.41)+.04(.09)+.10(.04)+.04(.46)= .3868
By Bayes’ formula, P(B1|A) = P(A|B1) P(B1) /P(A)
=.88(.41)/.3868=.3608/.3868=.933
22. A computer center has three printers A, B, and C, which print at different speeds. Programs are routed to the first
available printer. The probability that a program is routed to printers A, B and C are 0.6, 0.3 and 0.1 respectively.
Occasionally a printer will jam and destroy the printout. The probability that printers A, B and C will jam are 0.01,
0.05 and 0.04 respectively. Your program is destroyed when a printer jams. What is the probability that printer A
is involved? Printer B involved? Printer C involved?
23. In a bolt factory machines A, B and C manufacture 25%, 35% and 40% of the total. Of their output 5%, 4% and
2% are defective bolts. A bolt is drawn at random from the product and is found to be defective. What are the
probabilities that it was manufactured by machines A, B and C.
Sol. 23: Total probability
P(D)= P(A) P(D|A)+ P(B) P(D|B)+ P(C) P(D|C)
= .0125+.0140+.0080= .0345
So, P(A|D)= P(A) P(D|A)/P(D)
=.0125/.0345= 0.3623
P(B|D)= P(B) P(D|B)/P(D)
=.0140/.0345= 0.4058
P(C|D)=P(C) P(D|C)/P(D)=.0080/.0345 = 0.2318
24. Define a discrete random variable and discrete density function? State necessary and sufficient conditions for a
function to be a discrete density.

Sol.: Discrete random variable and discrete density function: A random variable (RV or rv) say X is discrete if it
can take a finite or countably infinite number of possible values at X= x or xi for which there exists a function
f(x) = P(X=x) or f(xi ) = P(X= xi ) and
called discrete density function if and only if (necessary and sufficient condition)
1. f(x) > 0 or f(xi ) > 0; 2.
25. Define distribution function or cumulative distribution in discrete case and its properties?
26. Let density for X , the number of grafts that fail in a series of five trials, is given the following table:
x 0 1 2 3 4 5
( )
f x 0.7 0.2 0.05 0.03 0.01 (5)
f
(a) Find (5)
f
(b) Find the table for F
(c) Use F to find the probability that at most three grafts fail; that at least two grafts fail.
(d) Use F to verify the probability of exactly three failures is 0.03.
27. Let X denote, the number of holes that for can be drilled per bit. The density for X is given the following table:
x 1 2 3 4 5 6 7 8
( )
f x 0.02 0.03 0.05 0.2 0.4 0.2 0.07 (8)
f
(a) Find f(8).
(b) Find the table for F.
(c) Use F to find the probability that a randomly selected bit can be used to drill between three and five holes
inclusive.
(d) Find ( 4)
p X  and ( 4)
p X  . Are these probabilities the same?
1
)
(
or
1
)
( 
 
 
n
i
x
x
x
i
x
al l
x
f
,
x
f

Sol.27: Let X be number of holes drilled per bit.
(a) Here f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)=1
or F(7)+ f(8)=1 or .92+ f(8)=1 or f(8)=1.-0.92=0.08
(b) Distribution function F(x) is given by
X 1 2 3 4 5 6 7 8
f(x) 0.02 0.03 0.05 0.20 0.40 0.20 0.02 0.08
F(x)0.020.050.100.300.700.900.921.00
(c) P(random selected bit can be used to drill between three and five holes inclusive)
=P(3<X<5) =F(5)- F(3)=.7-.1=.6
(d) P(X<4) =F(4)=.3 and P(X<4)=F(3)= .1. No, P(X<4) =.3 and P(X<4)= .1 are not the same?
28. Let X denote the number of attempts that must be made to gain access to the computer. The distribution
function of X is given the following table:
x 0 1 2 3 4 5 6
( )
F x 0.05 0.15 0.35 0.65 0.85 0.95 1.00
(a) Find the probability density function
(b) Find probabilities for at most three attempts and at least three attempts to gain access to the computer
Sol. 28: (a) By using f (0)=F(0); f (1)=F(1)-F(0),…. , f (n)=F(n)-F(n-1), we obtain the Probability density f(x) as
X 0 1 2 3 4 5 6
F(x)0.050.150.350.650.850.951.00
f(x) 0.050.100.200.300.200.100.05
(b) P(at most three attempts to access to computer)
= P(X<3) = F(3) = .65 as given in the table.
P(at least three attempts to access to computer =P(X>3) = 1-F(2)=1-.35=.65
29. Define expectation and variance of a discrete random variable.
30. A drug is used to maintain a steady heart rate in patients who have suffered a mild heart attack. Let X denotes
the number of heart beats per minute obtained per patient
x 40 60 68 70 72 80 100
( )
f x 0.01 0.04 0.05 0.80 0.05 0.04 0.01
Find the average heart beat of the patients and also find the variance of heart beats
Sol. 30: As shown in the following table, we get the average heart beat as E(x)=Σ x f(x) = 70.
And variance as σ2
= Var(X)=E(X2
)-µ2
=Σx2
f –(Σxf)2
=4926.4-4900= 26.4
Therefore, Standard Deviation σ= 5.138
X 40 60 68 70 72 80 100
f(x) 0.01 0.04 0.05 0.80 0.05 0.04 0.01 Total
xf 0.4 2.4 3.4 56.0 3.6 3.2 1.0 70.0

x
2
f
16.0144.0231.23920.0259.2256.0 100.0 4926.4
31. Let X and Y be the independent random variables with E(X) =3, E(Y) =10, E(X2
) =25, E(Y2
)=164,
then find E(3X + Y- 8), E(2X -3Y+7), V(X) and V(Y), V(3X + Y- 8), V(2X -3Y+7).
Hint: Since X and Y be the independent random variables, E(XY)=E(X) E(Y).
Sol.31: a) E(3X+Y-8)=3E(X)+E(Y)-8 =3×3 +10-8 =11
b) E(2X+3Y+7)=2E(X)+3E(Y)+7 =2×3+3×10+7 =43
c) V(X) =E(X2
)-[E(X)]2
=25-32
=16
d) V(Y) =E(Y2
)-[E(Y)]2
=164 - 102
=64
e) V(3X+Y-8)=E[(3X+Y-8)2
]-[E(3X+Y-8)]2
=E[9X2
+Y2
+6XY-48X-16Y+64]-112
[From sol.(a) ]
=9E[X2
]+E[Y2
] +6E[X] E[Y]-48E(X)-16E(Y)+64] -121
=9(25)+164+6× 3(10)-48(3)-16(10)+64-121= 208
f) V(2X+3Y+7)=E[(2X+3Y+7)2
]-[E(2X+3Y+7)]2
=E[4X2
+9Y2
+12xy+28X+42Y+49]-432
[From sol.(b) ]
=4E[X2
]+9E[Y2
] +12E[X] E[Y]+28E[X]+42E[Y]+49-432
=4×25+9×164+12× 3×10+28×3+42×10+49 -1849= 2489 -1849=640

Unit-2
Practise questions
1. Define Geometric Distribution and its mean, variance and Moment generating function.
Sol. : Geometric G(p): f(x)=pqx-1
; x=1,2,3,...,∞;0<p<1.
mX(t)=pet
/(1-qet
),t<-lnq; E(X)=1/p, V(X)=q/p2
.
2. Define Binomial Distribution and its mean, variance and Moment generating function.
Sol. : Binomial B(n, p): f(x)=n
Cx px
qn-x
; x=0,1,2,3,...,n;
mX(t)=(pet
+q)n
; E(X) = n p, V(X)= n p q; q=1-p.
3. Define Hyper-geometric Distribution and its mean, variance and Moment generating function.
Sol. : Hyper-geometric H(N, n, r): f(x)= r
Cx
N-r
Cn-x/N
cn, where max[0,n-(N-r)]<x<min(n, r);
E(X) = nr/N, V(X)=(nr/N)(N-r)(N-n)/[N2
(N-1)].
4. Define Poisson distribution and its mean, variance and Moment generating function.
Sol. : Poisson P(k): f(x)=e-k
kx
/x!, x=0,1,2,3,...,∞;
mX(t)=exp[k(et
-1)]; E(X)=k, V(X)=k.
5. Define continuous density function.
Continuous random variable and continuous density function: A random variable (RV or
rv) say X is continuous if it can take a infinite many possible values at X= x for which
there exists a function f(x) = P(x<X<x=dx) dx for x called continuous density function if
and only if (nacs)
1. f(x) > 0 2.
6. Define Gamma and Chi-Square distributions.
Sol. : Gamma γ(α, β): f(x)=xα-1
e-x/β
/[Γ(α)βα
]; Continuous distribution
mX(t) = (1-βt)-α
, t<1/β; E(X)=αβ, V(X)= αβ 2
.
Chi square X2
γ is a special form of the Gama distribution with α= γ/2 and β=2, for γ>0 as degree of
freedom (d.o.f.).
pdf : f(x)=x(γ/2)-1
e-x/2
/[Γ(γ/2)2γ/2
];
mX(t) = (1-t)-γ/2
, t<1/2; E(X)=γ, V(X)=2γ.
Note: X2
r is used to denote that point linked to a chi-squared random variable such that
P[X2
γ > χ2
r ]= r. i.e. χ2
r is the point such that area to its right is r.
Technically speaking we should write χ2
r,γ , as the value of the point does depend on both the
probability desired and degrees of freedom linked the random variable.
7. Consider the random variable X, the number of trials needed to obtain the first zero from a set
of 0,1,2,3,4,5,6,7,8,9 numbers. Find its moment generating function, mean and variance of the
random variable X.
Sol.7: Here RV X follow Geometric distribution; X~G(p);
probability of one success p=1/10=.1,
q=1-p =9/10 = .9.
pdf : f(x)=pqx-1
=.1(.9)x-1
; x=1,2,3,...,∞.
mgf: mX(t)=pet
/(1-qet
)=.1et
/(1-.9et
),t<-ln(.9);
E(X)=1/p=1/.1=10, V(X)=q/p2
=.9/(.1)2
.=90.
8. The random variable X, the number of radar signals properly identified in a 30 minute period,
is a binomial random variable with parameters n=10 and p=1/2. Find its mean, variance and
Moment generating function.
Sol.8: Given X~B(n, p), n=10 and p=1/2=.5, q=.5
Mean E(X)= n p =10(.5)=5;
Variance V(X)= n p q =10(.5)(.5) = 2.5;
Mgf : mX(t)=(pet
+q)n
=(.5et
+.5)10
.




1
)
( dx
x
f

9. A foundry ships engine blocks in lots of size 20. Before accepting a lot 3 items are selected and
tested. Suppose that a given a given lot contains 5 defective items. Find its distribution
function, moment generating function, mean and variance of the random variable X.
Sol.9: Let X: number of defectives. Here X ~H(N,n,r); Given: Total number of items N=20, Size
of (items with a trait) defectives r=5, sample size drawn n=3.
f(x)= r
Cx
N-r
Cn-x/N
cn= 5
Cx
15
C3-x/N20
c3
E(X)=nr/N= 3(5)/20=.75,
V(X)=(nr/N)(N-r)(N-n)/[N2
(N-1)]
=(3×5/2)(20-5)(20-3)/[202
(20-1)]=153/304.
10. Let X be a Poisson random variable with parameter k. Prove that its mean, variance, Moment
generating function are 𝐸 𝑋 = 𝑘, 𝑉𝑎𝑟𝑋 = 𝑘, 𝑚𝑥 𝑡 = 𝑒𝑘 𝑒𝑡−1
respectively.
Sol.10: mgf: mX(t) =E(etX
) =ΣxetX
f(x) =ΣxetX
e-k
kx
/x!
=Σxe-k
etX
kx
/x! = Σxe-k
(ket
)x
/x! = e-k
Σx (k et
)x
/x!
= e-k
[1+ket
+(k et
)2
/2! + (k et
)3
/3! + …….]
= e-k
[1+u +u2
/2! +u3
/3! + …….]; [Take u(t)= ket
]
=e-k
eµ(t)
= eµ(t)-k
.
So µ=E(X)=mx’(t)|t=0=e-k
eµ(t)
u’(t)|t=0 [ Here u’(t)= k et
]
=e-k
eµ(t)
(ket
)|t=0= e-k
ek
k = k [Since µ(0)= k and e0
=1]
E(X2
)=mX”(t)|t=0 =e-k
[eu(t)
(ket
)+eu(t)
ket
(ket
)|t=0
=e-k
[ek
(k)+ek
k (k)]=k+k2
. So,V(X)=E(X2
)-µ2
=k+k2
-k2
=k.
11. A discrete random variable has moment generating function 𝑚𝑥 𝑡 = 𝑒2 𝑒𝑡−1
a) Find E[X] and E[X2
].
b) Find σ2
and σ.
Sol.11: Comparing the given m.g.f. with that of a Poisson RV X, that is 𝒎𝒙 𝒕 = 𝒆𝟐 𝒆𝒕−𝟏
we get the Poisson parameter k=2.
µ=E(X)=mX’(t)|t=0=e-2
eµ(t)
u’(t)|t=0; where u(t)=2et
=e-2
eu(t)
(2et
)|t=0=2; as u(0)=2.
E(X2
)=mX”(t)|t=0 =e-2
[eµ(t)
(2et
)+eµ(t)
2et
(2et
)|t=0
=e-2
[e2
(2)+e2
2 (2)]=2+22
=6.
So, σ2
=V(X)=E(X2
)-µ2
=6-22
=2; σ=21/2
=1.41421
12. A binomial random variable has mean 5 and variance 4. Find the values of n and p that
characterize the distribution of this random variable.
Sol. 12: Given X~ B(n, p); E(X)=np=5 (1)
Also, V(X)=npq=4. So, 5q=4 [Using Eq. (1)]
or 1-p=4/5 or p=1/5=.2; So q=1-.2=.8.
Using p=.2 in Eq.(1), we get
n(.2)=5 or n=5/.2=25.
Probability density function of RV X is given by

f(x)=n
Cx px
qn-x
= 25
Cx (.2)x
(.8)25-x
x=0,1,2,3,...,25.
13. Let X be a Poisson random variable with parameter k=10.
a) Find the expression for the density for X.
b) Find E[X], Var X, σx.
c) Find P[X<4].
d) Find P[X>4].
e) Find P[X>4].
f) Find P[4<X<9].
Sol.: a)Poisson pdf: f(x)=e-k
kx
/x!=e-10
10x
/x!,x=0,1,…
b) E[X]=k=10; Var X =k=10. So, σx=(10)1/2
=3.1623;
c) P[X<4]=P(0)+ P(1)+ P(2)+ P(3)+ P(4)
=e-10
+e-10
10+e-10
102
/2!+e-10
103
/3!+e-10
104
/4!
=e-10
[1933/3]=0.029 c) P[X>4]=1-P[X<4]=0.971
(e) P[X>4]=P[X=4]+P[X>4]=e-10
103
/3!+.971 =0.0076+.971=0.9786.
(f) P[4<X<9]=P(4)+P(5)+P(6)+P(7)+P(8)+P(9)
=e-10
104
[1/4!+10/5!+100/6!+103
/7! +104
/8! +105
/9!] =0.3225
14. Let X be a Gamma random variable with α =3, β = 4.
a) What is the expression for the density for X?
b) What is the Moment generating function for X?
c) Find µ, σ2
, σ
Sol.14: (a) pdf: f(x)=xα-1
e-x/β
/[Γ(α)βα
]
=x2
e-x/4
/[Γ(3)43
]=[x2
e-x/4
]/128. (Since Γ(3)=2)
(b) mgf: mX(t) =(1-βt)-α
=(1-4t)-3
, t<1/4;
(c) µ=E(X)=αβ=3×4=12;
(d) σ2
=V(X)= αβ2
=3×42
=48 Therefore, σ =(48)1/2
= 6.928
15. Let X be a gamma random variable with parameters α and β. Use the moment generating
function to find E[X], E[X2
]. Use these expectations to show that Var[X] = α β 2
Sol. 15: mgf for Gamma RV X is given by
mX(t) = (1-βt)-α
, t<1/β; E(X)=αβ, V(X)= αβ2
.
E[X] = mX’(t)|t=0 =-α(1-βt)-α-1
(-β)|t=0 = αβ(1-βt)-α-1
|t=0 = αβ
E[X2
]= mX”(t)|t=0 = αβ(-α-1)(1-βt)-α-2
(-β)|t=0 = αβ2
(α+1)(1-βt)-α-2
|t=0 = αβ2
(α+1)
Var[X] = E[X2
]-µ2
= αβ2
(α+1) – (α β)2
= αβ2
.
16. Consider a chi-squared random variable with 10 degree of freedom. Find the value of χ2
0.05,
pdf, mgf, mean and variance.
Sol.16: By definition the area to the right of the point is 0.05 and the area to its left is 0.95. The
Column probabilities in chi-square table give the area to the left of the point listed. So to find χ2
0.05,
we look in row 10 and column 0.95 and see that χ2
0.05=18.3.
pdf: f(x)=x0.5γ-1
e-x/2
/[Γ(0.5γ)2 0.5γ
];
=x4
e-x/2
/[Γ(5)2 5
] =x4
e-x/2
/768;
mX(t)=(1-2t)-0.5γ
=(1-2t)-5
,t<0.5; E(X)= γ=10, V(X)= 2γ=20.
17. Let X denotes the number of runs conducted to produce an unacceptable law. Assume that the
runs are independent in the sense that the outcome of one run has no effect on that of any other.
Assume that the probability that a given lot is unacceptable is 0.05.
a) Verify that X is Geometric. What is success in this experiment? What is the numerical
value of p.
b) What is the exact expression for the density for X.?
c) What is the exact expression for the m.g.f. for X.?
d) What are the numerical values of E[X], E[X2
], σ.
Sol. 17: Given: X=number of runs to produce an unacceptable law; Success is selecting an
unacceptable lot. Numerical value of p=.05, So, q=.95

a) For X to be geometric, we need its pdf as
f(x)=pqx-1
=.05(.95)x-1
<1; x=1,2,3,...,∞ (1)
Σx f(x)=Σx .05(.05)x-1
= .05Σx .95x-1
=.05/(1-.95)=1
b) Exact expression for pdf of X is given by (1).
c) Expression for mgf of X is mX(t) = pet
/(1-qet
) = .05et
/(1-.95et
).
d) E(X)=1/p=1/.05=20,
E(X2
)=(1+q)/p2
=1.95/(.05)2
=780
σ2
= Var X= q/p2
=.95/(.05)2
=380
So, σ =(380) ½
= 19.4936
18. Geophysicists determine the age of a zircon by counting the number of uranium fission tracks
on a polished surface. A particular zircon is of such an age that the average number of tracks
per square centimetre is five. What is the probability that a 2-cm square sample of this zircon
will relieve at most three tracks, thus leading to an underestimation of the age of the material?
Sol. 18: Given the average number of tracks per square centimeter is λ=5; measurement
unit in square centimeter, the sample of 2 square centimeter is taken so s=2.
Therefore,k=λs=10; P(X=x)=f(x)=e-10
(10)x
/x!.
P(X<3)=P(0)+ P(1)+ P(2)+ P(3)
=e-10
+e-10
10+e-10
102
/2!+e-10
103
/3!
=e-10
[1+10+50+500/3]
=e-10
[3+30+150+500]/3 =e-10
[683]/3
= 4.54×10-5
×227.667= 0.01034
19. The white blood cell count of a healthy individual can average as low as 6000 per cubic
millimeter of blood. To detect a white cell deficiency, a 0.001 cubic millimeter drop of blood is
taken and the number of blood cells X is found. How many white cells are expected in a healthy
individual ? If at most two are found, is it an evidence of a white cell deficiency ?
Sol.19 : Given λ=6000 average number of occurrences of event, here number of white cells
per cubic mm; the continuous “interval” involved is a drop of blood of size s=0.001 cubic
mm. Thus X is a Poisson RV with k=λs=6000× 0.001=6. Therefore, E(X) = k =6, that is,
average number of white cells expected in a healthy individual is 6.
P(X=x)= f(x)=e-6
6x
/x!.
P(at most two white cells are found) = P(X<2)
=P(X=0)+P(X=1)+P(X=2)
=e-6
+e-6
6+e-6
(62
)/2!
= e-6
(1+6+18) =25 e-6
=0.062.
20. A machine fills 1000 bottles of coke in an hour. Each hour a sample of 20 bottles is randomly
selected and amount of content is checked. Let X be the number of bottles selected that are
underfilled. Suppose that during a particular hour 100 underfilled bottles are sampled. Find the
probability that at least 3 underfilled bottles will be among those sampled.
Sol.: Given: N=1000, Sample size n=20,
trait unfilled, r=100. X~H(N,n,r)=H(1000,20,100)
So P(X=x)= f(x)= r
Cx
N-r
Cn-x/N
cn.
Parametrs: N=1000, n=20, r=100.
f(x)= r
Cx
N-r
Cn-x/N
cn. , N-r=1000-100=900
P(at least 3 bottles are underfilled)
= 1-P(X<3)=1-P(X<2) =1-[P(X=0)+P(X=1)+P(X=2)]
= 1-[100
C0
900
Cn-x+20
C1
900
C19+100
C2
900
C18]/1000
c20.
= 0.3224
Note: Here n/N=20/1000 <.05. So, P(X>3) can be approximated by B(n, p) with n=20,
p=r/N=100/1000=0.1, Thus P(X>3)=1-P(X<3)=1-P(X<2)==1-[P(X=0)+P(X=1)+P(X=2)]=.323
21. Let X denote the length in minutes of a long-distance telephone conversation. The density for X
is given by 10
1
( ) ; 0
10
x
f x e x

  . Find (i) ( 10)
p X  (ii) (5 20)
p X
  (iii) moment
generating function (iv) mean (v) variance.
Sol.21: Here β=10. (i) P(X>10)=(1/10) ) ∫0
10
e- x/β
dx

=(1/β)/(-1/β) e- x/β
|0
10
=-[e-10
- e- 0
] =1-e-10
.
(ii) P(5<X<10) )=(1/β) ) ∫5
10
e- x/β
dx
=[(1/β)/(-1/β)] e- x/β
|0
10
=-[e-10/β
– e-5/β
] = e- 0.5
– e-1
.
(iii) MGF =mX(t)=(1- βt)-1
=(1- 10t)-1
(iv) μ=E(X)= dmX(t)/dt|t=0=-(1- βt)-2
(-β)|t=0=10.
(v) E(X2
)= d2
mX(t)/dt2
|t=0=2(1- βt)-3
(-β)2
|t=0=2(102
)=200.
Therefore, Var X = E(X2
)- [E(X)]2
=2β2
-β2
=β2
=102
=100.
22. Find moment generating function of Exponential distribution and then find mean and variance.
Sol. 22: Exponential distribution of a continuous R.V. X is given by f(x) = (1/β) e-x/β
, x>0.
MGF of X is given by mX(t)=E(etX
)=∫0
∞
etx
(1/β) e-x/β
dx = (1/β) ) ∫0
∞
e- (1/β-t)x
dx= -1/(1-βt) e- (1/β-t)x
|0
∞
= -1/(1-βt)[0-1]=1/(1- βt) =(1- βt)-1
. Thus, mX(t)=(1- βt)-1
.
μ=E(X)= dmX(t)/dt|t=0=-(1- βt)-2
(-β)|t=0=β.
E(X2
)= d2
mX(t)/dt2
|t=0=2(1- βt)-3
(-β)2
|t=0=2β2
.
)- [E(X)]2
=2β2
-β2
=β2
23. The spontaneous flipping of a bit stored in a computer memory is called a “Soft fail”. Let X
denote the time in millions of hours before the first soft fail is observed. Suppose the density
for X is given by ( ) ; 0
x
f x e x

  . Find the moment generating function, mean and
variance.
Sol. 23: A general exponential distribution of a continuous R.V. X is given by f(x) = (1/β) e-x/β
.
Here β =1. MGF of X is given by mX(t)=E(etX
)=∫0
∞
e tx
e-x
dx
= ∫0
∞
e- (1-t)x
dx= -1/(1-t) e- (1-t)x
|0
∞
= -1/(1-t)[0-1]=1/(1-t) =(1-t)-1
. Thus, mX(t)=(1-t)-1
.
μ=E(X)= dmX(t)/dt|t=0=-(1-t)-2
(-1)|t=0=1.
E(X2
)= d2
mX(t)/dt2
|t=0=2(1-t)-3
(-1)2
|t=0=2.
)- [E(X)]2
=2-1 =1.
24: If the probability density-function of a random variable X is given by
f(x)=k(1-x2
), for 0<x<1; f(x)=0, otherwise. Find the value of k and the probabilities that X will take a
value (i) between 0.1 and 0.2; (ii) greater than 0.5.
Sol.: ∫0
1
k(1-x2
)dx=1 or k(x-x3
/3)|0
1
=1 or k=3/2
P(0.1<X<0.2)= 1.5∫.1
.2
(1-x2
)dx=1.5(x-x3
/3)|0.1
0.2
=1.5[(.2-(.2)3
/3)-.1+ (.1)3
/3]= 0.1963.
P(X>0.5) = ∫.5
1
k(1-x2
)dx= k(x-x3
/3)|0.5
1
=1.5(1-1/3-.5+(.5)3
)= 0.3125.
Exercise 21: A computer terminal can pick up an erroneous signal from the keyboard that does not
show up on the screen. This creates a silent error that is difficult to detect. Assume that for a particular
keyboard the probability that this will occur per entry is 1/1000. In 5000 entries find the probability
that (i) No silent error (ii) At least 2 silent errors (iii) At most 3 silent errors occur.
25: Let X be the number of silent errors out of 5000 entries. Given n=5000, p=.001. Since n→∞ and
p→0, but np is finite, one can use the Poisson distribution with parameter k given by
k = n p = 5000(.001)=5.
(i) P(no silent error)=P(X=0)≈e-5
50
/0!= e-5
= 0.0067
(ii) P(at least two silent errors) = P(X>2) =1-P(X<1)≈1-[P(X=0)+P(X=1)]=1-[.0067+.0067(5)]
=1-.0067(6)= 0.9598
(iii) P(At most 3 silent errors)=P(X<3)=P(X=0)+ P(X=1)+ P(X=2)+ P(X=3)
= e-5
50
/0! + e-5
51
/1!+ e-5
52
/2!+ e-5
53
/3! = e-5
[1 + 5+ 25/2+ 125/6]= e-5
(39.33333) = 0.265
Note: Here X~B(n, p). So B(n, p) may also be used but calculation could be little difficult on calculator.

UNIT-3
PRACTICE QUESTIONS
1. Find the mean and variance of Normal distribution.
2. Let X denote the number of grams of Hydrocarbons emitted by an automobile per mile.
Assuming that X is normal with mean and S. D. are 1 and 0.25 grams. Find the
probability that randomly selected automobile will emit between 0.9 and 1.54?
Sol.: Given µ=1, σ =0.25.
Therefore Z=(X-µ)/σ=(X-1)/0.25
And P(0.9<X<1.54) = P(0.9-1<X-1<1.54-1)
= P(-0.1/0.25<(X-1)/0.25<0.54/0.25)
= P(-0.4<Z<2.16) = P(Z<2.16)-P(Z<-0.4)
= FZ(2.16)- FZ(-0.4) = 0.9846 – 0.3446 =0.64
3. Let X denote the number of BTU’s of petroleum products used with mean 153 million
BTU’s and standard deviation 25 million BTU’s. Use chebyshev’s inequality then
approximate what percentage of the population used between 128 and 178 million
BTU’s. Approximate wthat percentage of the population used in excess of 228 millon
BTU’s?
Sol.: Given µ=153, σ =25. So, Z=(X-µ)/σ=(X-153)/25
(i) P(128<X<178)=P(128-153<X-153<178-153)
=P(-25/25<(X-153)/25<25/25) =P(|Z|<1)>1-1/12
, So k=1
by Chebyshev’s inequality P[|Z|<k] > 1- 1/k2
& P(-1<Z<1)>0
So, P(-1<Z<1)= FZ(1 )- FZ(-1)=0.8413-0.1587= 0.6826
P(X>228) = P(X-153>228-153)=P(X-153>75)
= P(Z>3)= P(3<Z<∞)= P(-∞< Z< ∞)-P(-∞<Z<3)
= 1- FZ(3) = 1– 0.9987 =0.0013.
4. Among diabetic, the fasting blood glucose level X may be assumed to be approximately

normally distributed with mean 106 milligrams and S. D. 8 milligrams.
a) Find the probability that randomly selected diabetic will have blood glucose level
between 90 and 122 mg.
b) Find  
120mg

X
P
c) Find the point that has the property that 25% of all diabetic have a fasting glucose
level of this value or lower.
Sol. : Given µ=106, σ =8.
So, Z=(X-µ)/σ =(X-106)/8
a) P(90<X<122)=P[(90-106)/8<Z<(122-106)/8]
=P(-2<Z<2)=F(2)-F(-2)=0.9772-0.0228=0.9544
b) P[X < 120]=P[(X-µ)/σ <(120-106)/8]
=P[Z<1.75]= FZ(1.75) = 0.9599.
c) Let a be the value of fasting glucose level below which 25% of all diabetic fall.
Then P(X<a) =0.25 or P[Z<(a-106)/8]=0.25
From Gauss table, we get P[Z<-0.67]=0.25,
So, (a-106)/8=-0.67 i.e. a= 106-0.67(8) =100.64
5. If X denotes the amount of radiation that can be absorbed by an individual before death
ensues. Assume that that X is normal with a mean of 500 roentgens and S.D. of 150
roentgens. Above what dosage level will only 5% of dosage exposed survive?
Sol.: Given µ=500, σ =150. Given P(X>a)=5%
where ‘a’ is the dosage level of death to occur.
So P[(X-500)/150>(a-500)/150]= .05
Or P[Z>z0]=0.05 where z0=(a-500)/150]
Or P[Z<z0]=1-0.05=0.95 [since P[Z<z0]+P[Z>z0]=1]
So, (a-500)/150=1.645 (From table: F(0.95)=1.645
Solving for a we get a=500+150(1.645)= 746.75
6. State and Prove Chebyshev’s inequality

7. The safety record of an industrial plant is measured in terms of M, the total staffing -
hours worked without a serious accident. Past experience indicates that M has a mean of
2 million with a standard deviation of 0.1 million. Use Chebyshev’s inequality to find
the next serious accident to occur within the next 1.6 million staffing-hours.
Sol.: Given µ=2, σ =0.1. So, Z=(X-µ)/σ=(X-2)/0.1
P(1.6<X<2.4)=P[(1.6-2)/.1<Z<(2.4-2.0)/.1|] =P(|Z|<4)
So, P(|Z|<4)>1-1/42
= 0.9375 & k=4
or P(1.6<X<2.4)>.9375 (1). By inequality: P[|Z|<k] >1- 1/k2
P(1.6<X<2.4)=1-{P[X<1.6]+P[X>2.4]}.
Putting it in (1), we get 1-{P[X<1.6]+P[X>2.4]}> .9375
P[X<1.6]+P[X>2.4]<1-0.9375<0.0625 So, P[X<1.6]< 0.0625

8. A study is performed to investigate the connection between maternal smoking during
pregnancy and birth defects in children. Of the mothers studied, 40% smoke and 60%
do not. When the babies were born, 20 were found to have some sort of birth defect. Let
X denotes the number of children whose mother smoked while pregnant. What is the
probability that 12 or more of the affected children had mothers who smoked?
Sol.: Given n=20, p=0.4. So np=20(.4)=8
For Gaussian approximation, we should have n and p such that either p<0.5 and np> 5
or p>0.5 and npq>5.
Here the same are valid because p=0.4<0.5 and µ= np=8>5.
So, σ=[npq]1/2
=[8(.6)]1/2
=4.81/2
=2.19
P[X>12]=P[Y>11.5]= P[Z>(11.5-8)/2.19] (by using half unit correction)
=P[Z>1.59]= 1-P[Z<1.59]=1-0.9441 =0.0559
9. Let X denote the time in hours needed to locate and correct the problem in the software
that governs the timing of traffic lights in the down town area of a large city. Assume
that X is normally distributed with mean 10 hours and variance 9.
a) Find the probability that the next problem will require at most 15 hours to find and
correct.
b) The fastest 5% of repairs take at most how many hours to complete.
Sol.: Given µ=10, σ =3. X~N(10, 6).
So, Z=(X-µ)/σ=(X-10)/6 and Z~N(0, 1).
(i) P(at most 15 hours to find and correct) =P(X<15)=P[Z<(15-10)/3]=P(Z<1.67)=0.9525
Let ‘n’ be the number of hours to complete fastest 5% repairs. Then P(X<n)=5%=0.05.
Or, P[Z<(n-10)/3]=0.05 where z0=(n-10)/3.
So P[Z<z0]=0.05
From normal table, z0=1.65.
So (n-10)/3=-1.65 or n=10-3(1.65)=5.05 hours
10. Let X be a random variable with density   x
=
x
fx 2 , 1
0 <
x
< and g(x)=Y=3x+6 then find
 
y
fy .
Sol.: Here fY(y) is given by fY(y)=fX[g-1
(y)]|(d/dy) g-1
(y)|
where y=g (x) is strictly monotonic and differentiable and so solving it, we get x=g-1
(y) .
We solve y=3x+6 for x to get x=g-1
(y) and find that x=(y-6)/3=g-1
(y).
So, (d/dy)g-1
(y)=(d/dy)[(y-6)/3]=1/3; fY(y)=2x|x=(y-6)/3 (1/3)=2[(y-6)/3](1/3),
Thus fY(y)=2(y-6)/9, 6<y<9.
11. Find each of the following
a)  
1.57

Z
P b) P[Z<1.57] c) P[Z= 1.57] d) P[Z>1.57] e)  
1.75
1.25 

 Z
P
Sol.11: a) P[Z<1.57]=FZ(1.57)=0.9418
b) P[Z<1.57]=FZ (1.57)=0.9418 ;
c) P[Z=1.57]=0;

d) P[Z>1.57]=1-P[Z<1.57] =1-FZ (1.57) =1-0.9418=0.0582
e) P[-1.25<Z<1.75]= P[Z<1.75]-P[Z<-1.25]=FZ(1.75)-FZ (-1.25)=0.9599-.1056 = 0.8543
12. Let X be a random variable with density   2
4
1
x
x xe
=
x
f

, 0

x and let 2
2
1


 x
y .
Find the density for y.
Sol.12: Solving y=-0.5x+2 for x we get x=4-2y=g-1
(y). So, (d/dy) g-1
(y)=-2.
Putting these in fY(y)=fX[g-1
(y)] |(d/dy) g-1
(y)|, we get
fY(y)=0.25(4-2y) e-(4-2y)/2
|-2| =0.25(2)(2)(2-y)e-(2-y)
=(2-y)e-(2-y)
where y<2. (Since x>0 gives 4-2y>0 i.e. y<2.)
Thus fY(y) =(2-y)e-(2-y)
, y<2.
13. Find each of the following
a) The point such that   95
.
0



 z
Z
z
P
b) The point such that   90
.
0



 z
Z
z
P
c) Z 0.90
Sol.: In Normal table, we find F(z)=P(Z<z).
So, P[Z<-z]+P[-z<Z<z]+P[Z>z]=1
or P[-z<Z<z]=1-{P[Z<-z]+P[Z>z]}
a) 0.95 =1-2P[Z<-z] [Since P[Z<-z]=P[Z>z]]
or P[Z<-z]=.025. From Normal table, z=1.96
b) Similarly 0.90 =1-2P[Z<-z] Or P[Z<-z]=.05
So, from Normal table, z=1.645.
c) zr is the z value for which probability to its right is r
i.e. P[Z>zr]= r. To find P[Z>z0.90]=0.90. Here r=0.90),
So, 1- P[Z<z0.90]=0.90 Or, P[Z<z0.90]=1-0.90 =0.10
So from Normal table is z0.90=-1.28
14. Let X be a random variable with density   x
x e
=
x
f 
, x>0 and let Y= x
e . Find the
density for Y.
Sol.14 : Here fY(y) is given by fY(y)=fX[g-1
(y)]|(d/dy) g-1
(y)|
where Y=g(x) is strictly monotonic and differentiable.
To get g-1
(y) we solve y= ex
for x and find that x=log(y)=g-1
(y). So, (d/dy) g-1
(y)=1/y;
So the density for y is given by fY(y)= e-log(y)
(1/y) =(1/y)(1/y)=1/y2
, y>1.

15. Verify the normal probability rules.
Sol.: Let X be normally distributed with parameters µ and σ. Then as per Normal
probability rule, we have
P[-σ<X-µ<σ]= 0.68; P[-2σ<X-µ<2σ]= 0.95
and P[-3σ<X-µ<3σ]= 0.997.
Taking Z= (X-µ)/σ, from normal table values, it follows (as also given in the figure) that
P[-1<Z<1]=F(1)-F(-1)=.8413-.1587= 0.6826
P[-2<Z<2]=F(2)-F(-2)=.9772-.0228= 0.9544
P[-3<Z<3]=F(3)-F(-3)=.9987-.0013= 0.997
15. Let C denote the temperature in degrees Celsius to which a computer will be subjected in
the field. Assume that C is normally distributed over the interval (15, 21). Let F denote
the field temperature in degrees Fahrenheit so that F = (9/5)C+32. Find the density of F.
Sol.: Here X=C and Y=F , X is uniformly distributed over the interval. So
fX(x)=fC(x)=1/6 for 15<x<21
Solving y=(9/5)x+32 for x, we get x=(5/9)(y-32)=g-1
(y)
So, (d/dy) g-1
(y)=5/9.
Putting it in fY(y)= fX[g-1
(y)]|(d/dy) g-1
(y)|, we get fY(y)=(1/6)|5/9|=5/54 where 59<y<69.8.
Thus pdf of Y=F is fF(y) =5/54 , 59<y<69.8.
16. Assume that the probability of a serious measurement error is 0.05. a serious of 150
independent measurements are made. Let X denote the number of serious errors
made.
a) In finding the probability of making at least one serious error.
b) Approximate the probability that at most three serious errors will be made.
Sol.: Given p=0.05; n=150; q=1-p=1-0.05=0.95.
µ=np=150× 0.05=7.5, σ2
=7.5×0.95=7.125, So σ =2.6693
(a) P(at least one serious error)=P(X>1)
=P[Z>(1-7.5)/2.6693] =P[Z> -2.435]
=1- P[Z< -2.44]=1-.0073=0.9927
(b) P(at most three serious errors)=P(X<3)
=P[Z<(3-7.5)/2.6693] =P[Z< - 1.686 ]=0.0455
17. Let X be binomial with n=20 and p=0.3. use the normal approximation to approximate
each of the following.
a)  
3

X
P ; b)  
4
=
X
P ; c)  
4

X
P ; d)  
6
3 
 X
P .
Sol.: Normal approximation of binomial gives
µ= np=20(0.3) =6 and σ=[20(.3).7]1/2
=2.04939
a) P[X<3] =P [(X-µ)/σ<(3-6)/2.04939]=P[Z<-1.46]=F(-1.46)=0.0721
b) P[X=4] =0 [Since ∫4
4
.f(x) dx=0]
c) P[X>4] =P[(X-µ)/σ>(4-6)/2.049]=P[Z>-0.976]
=1- P[Z<-0.98]=1-F(-0.98)=1-0.1635=0.837
d) P[3<X<6]=P[(3-6)/2.049<Z<(6-6)/2.324]=P[-1.46<Z<0]=.5-P[Z<-1.46]=0.5-0.072=0.4279.

UNIT -IV
1. For the following bivariate probability distribution find (i) marginal distributions of X and Y (ii)
( 1, 2)
p X Y
  (iii) ( 2, 2)
p X Y
 
XY 0 1 2 3
0 0.840 0.030 0.020 0.010
1 0.060 0.010 0.008 0.002
2 0.010 0.005 0.004 0.001
Sol.: (i) Marginal distribution of X is
fX(x) =Σall y fXY(x, y) for given x; fX(0) =.84+.030+.020+.010=.90,
fX(1)=.060+.010+.008+.002=.080; fX(2)=.010+.005+.004+.001=.020
The same are shown as last column of following table.
Similarly, marginal distribution of Y is
fY(y) =Σall x fXY(x, y) for given y (as shown by last row in table.
fY(0) =.84+.06+.01=.91, fY(1)=.030+.010+.005=.045
fY(2)=.020+.008+.004=.032, fY(3)=.010+.002+.001=.013,
(ii) P[X<1, Y=2]=P[0,2]+P[1,2]=.020+.008=.028;
(iii) P[X<2,Y<2]= P[0,0]+P[0,1]+P[1,0]+P[1,1]+P[2,0]+P[2,1]
=.840+.030+.060+.010+.010+.005=0.955 (as shown by differently highlighted entries in table).
XY 0 1 2 3 fX(x)
0 0.840 0.030 0.020 0.010 .900
1 0.060 0.010 0.008 0.002 .080
2 0.010 0.005 0.004 0.001 .020
fY(y) .910 .045 .032 .013 1.000
2. For the following bivariate probability distribution. (i) Find E(X) (ii) Find E(Y) (iii) Verify
E(X+Y)=E(X)+E(Y) (v) Find ( )
E XY (vi) Find Cov (X, Y).
XY 0 1 2 3
0 0.840 0.030 0.020 0.010
1 0.060 0.010 0.008 0.002
2 0.010 0.045 0.032 0.013
Sol. Q.2: (i) E(X)= Σall x Σall y xfXY (x, y)
=0(.840)+0(.030)+0(.020)+0(.010)+1(.060)+1(.010)
+1(.008)+1(.002)+2(.010)+2(.005)+2(.004)+2(.001)=0.12,
or E(X)=Σall x xfX (x)=0(.9)+1(.08)+2(.02)=0.12
(ii) E(Y)=Σall y yfY (y)=0(.91)+1(.085)+2(.06)+3(.025)=0.28;
(iii) E(X+Y) = Σall x Σall y (x+y) fXY (x, y)
=(0+0)(0.84)+(0+1)(0.03)+(0+2)(0.02)+(0+3)(0.01)
+(1+0)(0.06)+(1+1)(0.01)+(1+2)(0.008)+(1+3)(0.002)
+(2+0)(0.01)+(2+1)(0.045)+(2+2)(0.032)+(2+3)(0.013)=0.40.
Also E(X)+E(Y)=0.12+0.28=0.40. So E(X+Y)=0.40=E(X)+E(Y)
XY 0 1 2 3 fX(x)
0 0.840 0.030 0.020 0.010 .900
1 0.060 0.010 0.008 0.002 .080
2 0.010 0.045 0.032 0.013 .100
fY(y) 0.910 0.085 0.600 0.025 1.00
0

(iv) E(XY) = Σall x Σall y xy fXY (x, y) =0(0)(.840)+0(1)(.030)+0(2)(.020)+0(3)(.010)
+1(0)(.060)+1(1)(.010)+1(2)(.008)+1(3)(.002) +2(0)(.010)+2(1)(.005)+2(2)(.004)+2(3)(.001)=0.064.
(v) Cov (X, Y) = E(XY)-E(X) E(Y) = 0.064-0.12(0.148) = 0.04624
Since Cov(X,Y)≠0, R.V.’s X and Y are not Independent.
3. Let X denote the number of “do loops” in fortran program and Y the number of runs needed for a
novice to debug the program. Assume that the joint density for (X, Y) is given table.
x/y 1 2 3 4
0 0.059 0.100 0.050 0.001
1 0.093 0.120 0.082 0.003
2 0.065 0.102 0.100 0.010
3 0.050 0.075 0.070 0.020
(a) Find the probability that a randomly selected program contains at most one do loop and
requires at least two runs to debug the program.
(b) Find E[XY].
(c) Find the marginal densities for X and Y. Use these to find the mean and variance for both X
and Y.
(d) Find the probability that a randomly selected program requires at least two runs to debug
given that it contains exactly one do loop.
(e) Find ( , )
Cov X Y . Find the correlation between X and Y. Based on the observed value of  , can
you claim that X and Y are not independent? Explain.
Sol.3: a) P[at most one do loop and at least two runs]
= P[X<1,Y>2]=P[0,2]+P[0,3]+P[0,4]+P[1,2]+P[1,3]+P[1,4]
=.100+.050+.001+.120+.082+.003=0.356
b) E(XY)=E(XY) = Σall x Σall y xy fXY (x, y)
= 0(1)(.059)+0(2)(.100)+0(3)(.050)+0(4)(.001)+1(1)(.093)+1(2)(.120)+1(3)(.082)+1(4)(.003)
+2(1)(.065)+2(2)(.102)+2(3)(.100)+2(4)(.010)+3(1)(.050)+3(2)(.075)+3(3)(.070)+3(4)(.020)=3.279
c) Marginal density of X is fX(x) =Σall y fXY(x, y)
for given x
fX(0) =.059+.100+.050+.001=0.210
fX(1)=.093+.120+.082+.003=0.298
fX(2)=.010+.005+.004+.001=0.215
Similarly, fY(y) =Σall x fXY(x, y) for given y.
fY(1)=0.267;fY(2)=0.397; fY(3)=0.302;
fY(4)=0.034
d) P[Y>2|X=1]= Σ y>2 fY=y|x=1(y)= Σ y>2 fXY(1, y)/fX(1)
= [fXY(1, 2)+fXY(1, 3)+fXY(1, 4)]/ fX(1) = (.120+.082+.003)/.298=0.6879.
e) Correlation ρXY=Cov(X, Y)/[Var X Var Y]1/2
E(X)=ΣxxfX(x)=0(.210)+1(.298)+2(.277)+3(.225)=1.527
E(Y)=ΣyyfY(y)=1(.267)+2(.397)+3(.302)+4(.034)=2.103
E(X2
)=Σxx2
fX(x)=02
(.210)+12
(.298)+22
(.277)+32
(.225)=3.431
E(Y2
)=Σxy2
fY(y)=12
(.267)+22
(.397)+32
(.302)+42
(.034)=5.117
So, Var X= E(X2
)-[E(X)]2
=3.431-(1.527)2
=1.0993
and Var Y= E(Y2
)-[E(Y)]2
=5.117-(2.103)2
=0.6944.
x/y 1 2 3 4 f
X
(x)
0 0.059 0.100 0.050 0.001 0.210
1 0.093 0.120 0.082 0.003 0.298
2 0.065 0.102 0.100 0.010 0.277
3 0.050 0.075 0.070 0.020 0.215
f
Y
(y) 0.267 0.397 0.302 0.034 1.000

Cov (X, Y)= E(XY)-E(X) E(Y) =3.279-1.527(2.103)=.0677
ρXY=Cov(X, Y)/[Var X Var Y]1/2
=.0677/[(1.0993)(0.6944)]1/2
=0.0775
4. Let X denote the temperature ( )
C
 and let Y denote the time in minutes that it takes for the
diesel engine on an automobile to get ready to start. Assume that the joint density for ( , )
X Y is
given by ( , ) (4 2 1); 0 40, 0 2
f x y c x y x y
      
(i) Find the value of c that makes this a density.
(ii) Find the probability that on a randomly selected day temperature will exceed 20 C
 and it
will take at least one minute for the car to be ready to start.
(iii) Find the marginal densities for X and Y .
(iv) Find the probability that on a randomly selected day it will take at least one minute for
the car to be ready to start.
(v) Find the probability that on a randomly selected day the air temperature will exceed 20 C

(vi) Are X andY independent?
Sol. 4: (i) ∫0
2
∫0
40
c(4x+2y+1)dx dy =1 or c ∫0
2
[4x2
/2+2yx+x]x=0
40
dy =1
or c ∫0
2
[4(1600)/2+2y(40)+40-0]dy =1 or c ∫0
2
[3200+80y+40]dy =1
or c [3240y+80y2
/2] 0
2
=1 or c [3240(2)+80(22
)/2-0] =1
or c [6480+160] =1 or c=1/6640.
(ii) P[X>20 and Y>1] = ∫0
2
∫20
40
c(4x+2y+1 )dx dy = ∫1
2
[4x2
/2+2yx+x]x=0
40
dy
= c ∫1
2
[4(1600)/2+2y(40)+40-{2(400)+2y(20)+20}] dy
= c ∫1
2
[3200+80y+40-{800+40y+20}] dy = c ∫1
2
[2420-40] dy = c [2420y-40y2
/2]1
2
= c [2420(2-1)-20(4-1)] = c [2420-60] =2360/6640=59/166 =0.355
(iii) fX(x)=∫0
2
c(4x+2y+1) dy =c(4xy+2y2
/2+y)|y=0
2
=c[4x(2)+4+2-0]=(8x+6)/6640
fY(y)=∫0
40
c(4x+2y+1) dx =c[4x2
/2+2yx+x)]|x=0
40
=c[2(40)2
+2y(40)+40-0]=(3240+80y)/6640
(iv) P(Y>1) = ∫1
2
c(3240+80y)dy=c(3240y+80y2
/2)|1
2
=c[3240(2)+160-{3240+40}]=3360/6640 =0.506
(v) P[X>20] = ∫20
40
c(8x+6 )dx =[8x2
/2+6x]|20
40
= c[8(40)2
/2+6(40)-{4(20)2
/2+6(20)}]
= c[4(1600)+240-{1600-120}]= c [4800+120] = c [4920]=4920/6640=123/166 =0.741
(vii) fX(x) fY(y)=c(8x+6)c(3240+88y) ≠ c(4x+2y+1) = fXY (x, y). So X and Y are not independent.
5. The joint density for ( , )
X Y is given by
3 3
( , ) 0 2, 0 2
16
x y
f x y x y
    
(i) Find the marginal densities for X andY .
(ii) Are X andY independent?
(iii) Find ( 1)
p X 
(iv) Find ( 1, 1)
p X Y
 
Sol.5: (i) fX(x)=∫-∞
∞
fXY(x,y)dy = ∫0
2
(x3
y3
/16) dy=(x3
y4
/(16×4)|y=0
2
= x3
24
/(16×4)=x3
/4, 0<x<2

Similarly fY(y)=∫0
2
(x3
y3
/16) dx=(x4
y3
/(16×4)|x=0
2
= 24
y3
/(16×4)=y3
/4, 0<y<2
(ii) fX(x) fY(y)= x3
/4 y3
/4= x3
y3
/16=fXY (x, y). So X and Y are independent.
(iii) P(X<1)=∫0
1
(x3
/4)dx=[x4
/(4×4)|]0
1
=1/16..
(iv) P(X<1,Y<1)=∫0
1
∫0
1
(x3
y3
/16) dx dy)= ∫0
1
(x3
/4)dx ∫0
1
(y3
/4)dy=[x4
/(4×4)|]0
1
[y4
/(4×4)|]0
1
=1/(16)2
=1/256.
6. The joint density for ( , )
X Y is given by ( , ) 0, 0
x y
f x y xye e x y
 
  
(i) Find the marginal densities for X andY .
(ii) Cov(X,Y)
(iii) Are X andY independent?
(iv) Find ( 1)
p X 
Sol.6: (i) fX(x)= ∫-∞
∞
fXY(x,y)dy= xe-x
∫0
∞
ye-y
dy =xe-x
[ye-y
/(-1)-(1)e-y
/(-1)2
] y=0
∞
[using ∫uvdy=uv1-u’v2]
= xe-x
[-ye-y
-e-y
] y=0
∞
=xe-x
[-0-0-{-0-1}]= xe-x
.
fY(y)= ∫-∞
∞
fXY(x,y)dx= ye-y
∫0
∞
xe-x
dx =ye-y
[xe-x
/(-1)-(1)e-x
/(-1)2
] x=0
∞
[using ∫uv dx=uv1-u’v2]
= ye-y
[-xe-x
-e-x
] x=0
∞
=ye-y
[-0-0-{-0-1}]= ye-y
.
Thus fX(x)= xe-x
, x>0, fY(y)= ye-y
, y>0.
(ii) Cov(X,Y)=E(XY)-E(X)E(Y)
E(X)= ∫-∞
∞
x fX(x)dx =∫0
∞
x xe-x
dx
=[x2
e-x
/(-1)-(2x)e-x
/(-1)2
+(2)e-x
/(-1)3
] x=0
∞
[ using ∫uv dx=uv1-u’v2+u”v3, u= x2
, v=e-x
]
=[-0-0-0-{0-0-2}] =2.
Similarly, E(Y)= ∫-∞
∞
y fY(y)dy =∫0
∞
y ye-y
dy
=[y2
e-y
/(-1)-(2y)e-y
/(-1)2
+(2)e-y
/(-1)3
] y=0
∞
[ using ∫uv dy=uv1-u’v2+u”v3, u= y2
, v=e-y
=[-0-0-0-{-0-0-2}]=2.
E(XY)= ∫-∞
∞
xy fXY(x,y)dxdy =∫0
∞
∫0
∞
xye-x
e-y
dxdy=[∫0
∞
xe-x
dx][∫0
∞
ye-y
dy]=2×2=4
Therefore Cov(X,Y)=E(XY)-E(X)E(Y)=4-2×2=0
(iii) fX(x) fY(y)= xe-x
ye-y
= xye-x
e-y
= fXY (x, y). So X and Y are independent.
(iv) P(X<1)= ∫0
1
fX(x)dx∫0
1
xe-x
dx =[ xe-x
/(-1)-e-x
/(-1)2
]|0
1
=[-e-1
-{e-1
-1}]= -2e-1
+1=0.264241.
[ using ∫uv dx=uv1-u’v2+u”v3, u= x2
, v=e-x
, u’=2x, u”=2, v1= e-x
/(-1), v2=e-x
/(-1)2
,v3=e-x
/(-1)3
]
7. Show that Var(X+Y) =V(X)+V(Y)+2 Cov(X,Y). What is about when X and Y independent?
Sol.7 : Since V(Z)=E(Z2
)-[E(Z)]2
by putting Z=(X+Y) in it, we get
Var(X+Y)=E[(X+Y)2
]-[E(X+Y)]2
(1)
In the above E[(X+Y)2
]=E[X2
+Y2
+2XY] Or E[(X+Y)2
]=E[X2
]+E[Y2
]+2E[XY] (2)
Also, [E(X+Y)]2
=[E(X)+E(Y)]2
Or [E(X+Y)]2
=[E(X)]2
+[E(Y)]2
+2E(X)E(Y) (3)
Using eq. (2) and (3) in eq.(1), we get
Var(X+Y) =E[X2
]+E[Y2
]+2E[XY]-[E(X)]2
-[E(Y)]2
-2E(X)E(Y)
=E[X2
]-[E(X)]2
+E[Y2
]-[E(Y)]2
+2{E[XY]-E(X)E(Y)}=V(X)+V(Y)+2Cov(X, Y).
So Var(X+Y)= Var(X)+Var(Y)+2Cov(X,Y).
when X and Y are independent, then Cov(X,Y)=0, and therefore Var(X+Y)= Var(X)+Var(Y).

8. Let 1 2 3 4 5
, , , ,
X X X X X be a random sample from a binomial distribution with n=10 and p unknown.
(i) Show that 𝑋/10 is an unbiased estimator for p (ii). Estimate p based on these data: 3, 4, 4, 5, 6.
Sol. (i) Let X be random variable which follows binomial distribution with parameters n and p.
Then E[X] =p. Replacing E[X] by its first moment estimator M1=Σi=1
n
Xi/n=𝑋, we get the equation
𝑋=n𝑝, where 𝑝 is the estimator of p. On solving the above equation, we get this gives 𝑝= 𝑋/n.
(ii) 𝑋=(3+4+4+5+6)/5 =22/5 =4.4. So 𝑝 = 𝑋/10= 0.44. E(𝑝)=E(𝑋/10)=
1
10
𝐸(𝑋𝑖)/5
5
𝑖=1 =
1
10
5𝜇
5
= 𝑝
9. Show that 2 2
1
1
( )
1
n
i
n
S X X
n 
 

 is an unbiased estimator of population variance.
Proof.: 𝑬 𝒔𝟐
= 𝑬
(𝑿𝒊−𝑿)𝟐
𝒏−𝟏
𝒏
𝒊=𝟏 =
𝟏
𝒏−𝟏
𝑬 {𝑿𝒊 − 𝝁 + 𝝁 − 𝑿}
𝟐
𝒏
𝒊=𝟏 =
𝟏
𝒏−𝟏
𝑬 (𝑿𝒊
− 𝝁 − (𝑿 − 𝝁)}
𝟐
𝒏
𝒊=𝟏
=
𝟏
𝒏−𝟏
𝑬[ {(𝑿𝒊 − 𝝁)
𝟐
− 𝟐(𝑿𝒊 − 𝝁)(𝑿 − 𝝁) + (𝑿 − 𝝁)
𝟐
}
𝒏
𝒊=𝟏 ]
[Using (U-V)2
=U2
-2UV+V2
, U=(𝑿𝒊 − 𝝁),V=(𝑿 − 𝝁)]
=
𝟏
𝒏−𝟏
𝑬[ (𝑿𝒊 − 𝝁)
𝟐
𝒏
𝒊=𝟏 − 𝟐(𝑿 − 𝝁) (𝑿𝒊 − 𝝁)
𝒏
𝒊=𝟏 + (𝑿 − 𝝁)
𝒏
𝒊=𝟏
𝟐
]
=
𝟏
𝒏−𝟏
𝑬[ (𝑿𝒊 − 𝝁)
𝟐
𝒏
𝒊=𝟏 − 𝟐 𝑿 − 𝝁 𝒏( 𝑿𝒊 − 𝒏𝝁)
𝒏
𝒊=𝟏 /𝒏 + (𝑿 − 𝝁)
𝟐
𝒏
𝒊=𝟏 ]
=
𝟏
𝒏−𝟏
𝑬[ (𝑿𝒊 − 𝝁)
𝟐
𝒏
𝒊=𝟏 − 𝟐𝒏 𝑿 − 𝝁
𝟐
+ 𝒏(𝑿 − 𝝁)
𝟐
]=
𝟏
𝒏−𝟏
𝑬[ (𝑿𝒊 − 𝝁)
𝟐
𝒏
𝒊=𝟏 − 𝒏 𝑿 − 𝝁
𝟐
]
=
𝟏
𝒏−𝟏
[ 𝑬(𝑿𝒊 − 𝝁)
𝟐
𝒏
𝒊=𝟏 − 𝒏𝑬 𝑿 − 𝝁
𝟐
]= =
𝟏
𝒏−𝟏
[ 𝝈𝟐
𝒏
𝒊=𝟏 − 𝒏𝑬 𝑿 − 𝝁
𝟐
]
=
𝟏
𝒏−𝟏
[ 𝝈𝟐
𝒏
𝒊=𝟏 − 𝒏
𝝈𝟐
𝒏
=
𝟏
𝒏−𝟏
[𝒏𝝈𝟐 − 𝝈𝟐] = 𝝈𝟐
. [Since X1, X2, .., Xn is a random sample with S.D. =σ,
𝐸(𝑋𝑖 − 𝜇)2
=σ2
for each i = 1, 2, 3,…,n and 𝐸 𝑋 − 𝜇 2
= 𝑉𝑎𝑟 𝑋 =
𝜎2
𝑛
].
This proves that E [s2
] = σ2
.
10. Explain point estimation. In normal distribution sample mean X is an unbiased estimator of
population  .
Sol. 10: Point estimation is a method of determining a point estimator , a statistic, to give a representative
value of corresponding population parameter.
Proof : Sample mean X is an unbiased estimator of population mean μ.
Let X1, X2,…, Xn be a sample of size n taken from a population of mean μ. Let the sample mean be X .
Then E( X ) =E[(1/n)(X1+ X2+… +Xn)] =(1/n) E(X1+ X2+… +Xn)
= (1/n)( μ+ μ+…+ μ n times) = (1/n) nμ = μ.
Maximum Likelihood Estimators:
Let the population have a probability density f (x, θ) with a parameter θ to be estimated. If there are n
independent observations X1, X2,…, Xn, then the joint density (also called as likelihood) is given by
L= f (x1, θ) f (x2, θ)… f (xn, θ) or ln L= ln[ f (x1, θ)]+ ln[ f (x2, θ)]+…+ ln[ f (xn, θ)]
The maximum likelihood is given by
𝜕𝐿
𝜕𝜃
= 0
or
1
𝑓(𝑥1,𝜃)
𝜕𝑓(𝑥1,𝜃)
𝜕𝜃
+
1
𝑓(𝑥2,𝜃)
𝜕𝑓(𝑥2,𝜃)
𝜕𝜃
+ ⋯ +
1
𝑓(𝑥𝑛 ,𝜃)
𝜕𝑓(𝑥𝑛 ,𝜃)
𝜕𝜃
= 0.
The solution for this equation for θ in terms of xk is known as maximum likelihood estimator of θ.
11. Obtain the maximum likelihood estimators of  and 2
 from a normal population.
Sol.11: For a normal distribution: 𝑓 𝑥𝑘, 𝜇 =
1
2𝜋𝜎2
𝑒− 𝑥𝑘 −𝜇 2/(2𝜎2)
,
L= f (x1, θ) f (x2, θ)… f (xn, θ) =(2𝜋𝜎2
)−𝑛/2
𝑒
−
(𝑥𝑘−𝜇 )2
2𝜎2
𝑛
𝑘=1
Or 𝑙𝑛𝐿 = −
𝑛
2
ln 2𝜋𝜎2
−
1
2𝜎2 (𝑥𝑘 − 𝜇)2
𝑛
𝑘=1 .

Setting
𝜕𝐿
𝜕𝜇
= 0 gives 0-
1
2𝜎2
2 𝑥𝑘 − 𝜇 −1 = 0
𝑛
𝑘=1 .
Or 𝑥𝑘 − 𝜇 = 0
𝑛
𝑘=1 , or 𝑥𝑘 = 𝑛𝜇
𝑛
𝑘=1 ,
or 𝜇 = 𝑥𝑘
𝑛
𝑘=1 .
Therefore, maximum likelihood estimator of μ for a normal distribution is the sample mean.
Setting
𝜕𝐿
𝜕𝜎2
= 0 gives −
𝑛
2
1
2𝜋𝜎2
2𝜋-
−1
2(𝜎2)2
𝑥𝑘 − 𝜇 2
= 0
𝑛
𝑘=1 .
or – 𝑛+
1
𝜎2
𝑥𝑘 − 𝜇 2
= 0
𝑛
𝑘=1 , or 𝜎2
=
1
𝑛
𝑥𝑘 − 𝜇 2
𝑛
𝑘=1 .
Therefore, maximum likelihood estimator of 𝜎2
for a normal distribution is the sample variance.
12. Assume that X follows the exponential distribution with parameter  . Find the maximum
likelihood estimate for 
Sol.12: Since X follows exponential distribution with parameter β, we have f (x)=(1/β)e-x/β
.
Let X1, X2,…, Xn be a random sample of n independent observations. Then, the joint density is
L= f (x1, θ) f (x2, θ)… f (xn, θ) =
1
𝛽𝑛 𝑒−(𝑥1+𝑥2+⋯+𝑥𝑛 )/𝛽
Or ln L= n ln β-(x1+ x1+…+ xn)/β.
𝑑 𝑙𝑛 𝐿
𝑑𝛽
= 0 or
1
𝐿
𝑑 𝐿
𝑑𝛽
= 0 gives 𝑛
1
𝛽
− 𝑥1 + 𝑥2 + ⋯ +𝑥𝑛 (−
1
𝛽2
)=0.Thus β = (x1+ x1+…+ xn)/n

1
UNIT V (questions for practice)
1. *A low-noise transistor for use in computing products is being developed. It is claimed that the mean
noise level will be below the 2.5 dB level of products currently in use.
a. Set up the appropriate null and alternative hypothesis for verifying the claim.
b. A sample of 16 transistors yields 𝑥 = 1.8 with 𝑆 = 0.8. Do you think that 𝐻0 should be
rejected?
c. Explain, in the context of this problem, what conclusion can be drawn concerning the noise
level of these transistors. If you make a Type I error, what will have occurred? What is the
probability that you are making such an error?
Sol. 1: H0: μ=2.5, H1: μ<2.5 (Left tailed test)
Given 𝑋=1.8, s=0.8, n=16. Test statistic is:
T=(𝑋-μ)/(s/√n)=(1.8-2.5)/(.8/√30)=-3.5.
From Ttable 3.5 lies between 2.947 and 3.733 for γ =16-1=15; P[T15<2.94)=0.995, P[T15<3.73)=0.999
So P[T15 < -2.94)=0.005, P[T15 < -3.73)=0.001.
Thus 0.001<P<0.005. P seems to be very small. Yes, we reject H0.
Also |Tcomp|=3.5>TTable=2.947 at α=0.005. So we reject H0 at LOS α=0.005.
c) Conclusion: Mean noise level is below 2.5 db. If a type I error is made, we shall assume that new
product reduces noise when, in fact, it does not.
d) At LOS α=5%=.05, P[t15>.05]=.05. So P[t15<0.05]=.95. This gives t15=1.753 from T table.
From T table: 3.5 lies between 2.947 and 3.733 for dof γ =16-1=15; |Tcomp|=3.5>ttable =1.753 at α=0.05.
Therefore, we reject H0 at LOS α=0.05.
2. In order to be effective, reflective highway signs must be picked up by the automobile’s headlights. To
do so at long distances requires that the beams be on “high.” A study conducted by highway engineers
reveals that 45 of 50 randomly selected cars in a high-traffic-volume area have the headlights on low
beam.
a. Find a point estimate for 𝑝, the proportion of automobiles in this type area that use low beams.
b. Find a 90% confidence interval on 𝑝.
c. How large a sample is required to estimate 𝑝 to within 0.02 with 90% confidence?
Sol.2: a) Given n=50, pe=45/50=0.9 is the point estimate of population proportion p. we have two-tailed
test.
b) 90% Confidence interval of p is : pe+zα/2√[peqe/n].
or pe+z0.05√[peqe/n] [Since 100(1-α)%=90% gives α=0.1]
Here z0.05 is the value for which P[Z>z0.05]=0.05,
i.e. P[z< z0.05]=0.95. So z0.05=1.645 [See normal table]
Thus, pe+z0.05√[peqe/n] gives 0.9+1.645√[0.9(0.1)/50]
Or 0.9+0.0698 or [0.83021, 0.96979] is 90% confidence interval of population proportion p.
c) Given d=0.02. Sample size for estimating p for given prior estimate pe is n=z2
α/2 pq/d2
=(1.645)2
0.9(0.1)/(0.02) 2
=609
3. *A new computer network is being designed. The makers claim that it is compatible with more than
99% of the equipment already in use.
a. Set up the null and alternative hypothesis needed to get evidence to support this claim.
b. A sample of 300 programs is run, and 298 of these run with no changes necessary. That is, they
are compatible with the new network. Can 𝐻0 be rejected?
c. What practical conclusion can be drawn on the basis of your test?
Sol.3: a) H0: p=0.99, H1: p>.99 (Right tailed test)
b) Given pe=298/300= 0.9934, p0=0.99, n=300.
Test statistic is: Z=(pe-p0)/√[p0q0/n] =(0.993 -0.99)/ √[0.99(0.01)/300]=0.57
From normal table: P[Z<0.57]=0.7157. So P[Z>0.57]=1-0.7157=0.2843= P in right tail.
Since this P value is quite significant, H0 can not be rejected.
Note: If the above P value had been less than 0.05, we could reject H0 at test level α=0.05.

2
c) Here we are not able to show that the new network is compatible with more than 99% of the equipment
already in use.
4. *It is thought that over 60% of the business offices in the United States have a mainframe computer as
part of their equipment.
a. Set up the appropriate null and alternative hypothesis for supporting this claim.
b. Find the critical point for an 𝛼 = 0.05 level test
c. When data are gathered, it is found that 233 of the 375 offices studied have mainframe
computers. Can 𝐻0 be rejected at the 𝛼 = 0.05 level?
d. Explain in the context of this problem, the practical consequences of making the type of error to
which you are subject.
Sol.4: a) H0: p=0.6, H1: p>.6 (Right tailed test)
b) For an α=0.05 level test, critical point zα=z0.05 is the value for which P[Z>z0.05]=0.05. i.e. P[z<
z0.05]=0.95. So z0.05=1.645 [See normal table]
c) Given pe=233/375=0.6213, p0=0.6, n=375.
Test statistic is: Z=(pe-p0)/√[p0q0/n] =(0.6213-0.6)/√[0.6(0.4)/375]=0.843
Since Z=0.843<z0.05=1.645, H0 can be rejected at test level α=0.05. The only option left is to accept H0. In
doing so, we may commit Type II error.
d) We cannot show that more than 60% of the business offices in the US have mainframe if, in reality, it is
true.
5. *The relationship between consumption and household income was studied, yielding the following
data on household income X (in units of 1000/year) and energy consumption Y (in units of 108
Btu/year)
Energy
consumption(Y)
Household income (X)
1.8 20.0
3.0 30.5
4.8 40.0
5.0 55.1
6.5 60.3
7.0 74.9
9.0 88.4
9.1 95.2
a. Estimate the linear regression equation 𝜇𝛾 𝑥 = 𝛽0 + 𝛽1𝑥.
b. If 𝑥 = 50 (house hold income of 50,000), estimate the average energy consumed for households
of this income. What would your estimate be for a single household?
c. How much would you expect the change in consumption to be if any household income
increases 2000/year (2 units of 1000)?
d. How much would you expect consumption to change if any household income decreases
2000/year?
Sol. 5: (a) Given sample size n=8, xi and yi are
given as in the table. From the table we find
Σxi=464.4, Σyi, =46.2, Σyi
2
=315.34, Σxi
2
= 32089.96
and Σxiyi =3173.17.
Using these values in expressions for line intercept b0
with y axis and line slope b1, we get
b1=[nΣi xiyi –(Σi xi)(Σi yi )]/[nΣi xi
2
– (Σi xi)2
]
= 0.0957; b0= Σi yi /n - b1 Σi xi /n= 0.2177.
Therefore using β0= 0.2177 and β1=0.0957 in μY|x = β0+ β1 x we get
linear regression equation μY|x = 0.2177+0.0957x (1)
(b) If x=50, eq. (1) gives μY|x =0.2177+0.0957(50) =5.0043.
(c) Change in consumption for increase of 2 units is given by
ΔμY|x =0.0957Δx=0.0957(2)= 0.1915.
Y X Y2
X2
XY
1.80 20.00 3.24 400.00 36.00
3.00 30.50 9.00 930.25 91.50
4.80 40.00 23.04 1600.00 192.00
5.00 55.10 25.00 3036.01 275.50
6.50 60.30 42.25 3636.09 391.95
7.00 74.90 49.00 5610.01 524.30
9.00 88.40 81.00 7814.56 795.60
9.10 95.20 82.81 9063.04 866.32
46.20464.40 315.34 32089.96 3173.17
=Σyi =Σxi =Σyi
2
=Σxi
2
=Σxiyi

3
(d) Similarly decrease in consumption for decrease of 2 units is given by ΔμY|x =0.0957Δx=0.0957(2)=
0.1915.
6. A machinist is making engine parts with axle diameters of 0.700 inch. A random sample of 10 parts shows
a mean diameter of 0.742 inch with a standard deviation of 0.040 inch. Construct 95% confidence limits for
true mean axle diameter.
Sol. 6: Sample size n=10, Xbar=0.742, s=0.040.
For small sample n<30, 100(1-α)% confidence interval is given by 𝑋+tα/2 s/√n. The 95% confidence limits for
two tailed test α=0.05, we need tα/2= t0.05 value from T table for dof =n-1=10-1=9. t0.05 is such that
P[T>t0.05]=0.025 or P[T<t0.05] =1-0.025 =0.975, we get t0.05]=2.262. Confidence limits are
0.742+2.262(0.04)/√10 or 0.742+ 0.0286 or (0.7134, 0.7706)
Q.7: *Metal conduits or hollow pipes are used in electrical wiring. In testing 1-inch pipes, these data are
obtained on the outside diameter (in inches) of the pipe:
Assume that sampling is from a normal distribution and find (i)
95% confidence interval (ii) 90% confidence interval on the mean
outside diameter of pipes of this type.
Sol. 7: Given sample size n=20 is a small sample as n<30 . The
100(1-α)% confidence interval is given by +tα/2 s/√n. The 90%
confidence limits for two tailed test α=0.10, we need tα/2= t0.05
value from T table for dof =n-1
=20-1=19. t0.05 is such that P[T>t0.05]=0.05
or P[T<t0.05] =1-0.05 =0.95, we get t0.05=1.729.
Sample mean and sample S.D. Calculation for outside diameter (in inches) of the pipe:
No. 1 2 3 4 5 6 7 8 9 10
X 1.281 1.288 1.292 1.289 1.291 1.293 1.293 1.291 1.289 1.288
(x-E(x))
2
0.000074 0.000003 0.000006 0.000000 0.000002 0.000012 0.000012 0.000002 0.000000 0.000003
No. 11 12 13 14 15 16 17 18 19 20 Σxi/n=
X 1.287 1.291 1.290 1.286 1.289 1.286 1.295 1.296 1.291 1.286 =1.289
(x-E(X))2
0.000007 0.000002 0.000000 0.000013 0.00000 0.00001 0.000029 0.00004 0.000002 0.00001 0.0000123
= s2
For given sample size n=20, mean and variance are calculated as in above given table are
𝑋 =Σxi/n=1.2896, s2
=Σ(xi- 𝑋 )2
/(n-1)
=0.0000123 gives s=0.0035, so confidence limits are given by 1.2896+1.729(.0035)/√19 or 1.2896+0.00139 or (1.2882,
1.29099).
8. Find the correlation between X and Y for the following data.
Enzyme
Level (X)
95 110 118 124 145 140 185 190 205 222
Detoxification
Level (Y)
108 126 102 121 118 155 158 178 159 184
Sol.8: n=10, Σy=1409, Σx=1534, Σy2
=206319, Σx2
=252684, Σxy=226463.
To find: ρXY=cXY /sXsY, cXY = Σ xy/n–(Σx)(Σy)/n2
=226463/10-1534(1409)/100 =1032.24.
sX=[Σx2
/n– (Σx)2
/n2
]1/2
=[252684/10-(1534)2
/100]1/2
=41.6754
sY=[Σy2
/n– (Σy)2
/n2
]1/2
=[206319/10-(1409)2
/100]1/2
= 27.9122;
So ρXY=cXY /sXsY = 1032.24/[41.6754(27.912)]ρXY=0.887375
9. *Find the correlation between X and Y for the following data.
Percentage of
copper (X)
.01 .03 .01 .02 .10 .08 .12 .15 .10 .11
Rockwell
hardness rate (Y)
58 66 55 63 58 57 69 70 65 62
1.281 1.288 1.292 1.289 1.291
1.293 1.293 1.291 1.289 1.288
1.287 1.291 1.290 1.286 1.289
1.286 1.295 1.296 1.291 1.286
y x y
2
x
2
xy
108 95 11664 9025 10260
126 110 15876 12100 13860
102 118 10404 13924 12036
121 124 14641 15376 15004
118 145 13924 21025 17110
155 140 24025 19600 21700
158 185 24964 34225 29230
178 190 31684 36100 33820
159 205 25281 42025 32595
184 222 33856 49284 40848
14091534206319252684226463
Σy
i
Σx
i
Σy
i
2
Σx
i
2
Σx
i
y
i

4
Sol.9: n=10, Σy=623, Σx=0.73, Σy2
=39057, Σx2
=0.0769, Σxy=46.83.
To find: ρXY=cXY /(sXsY)
cXY = Σ xy/n–(Σx)(Σy)/n2
=46.83/10-0.73(623)/100 =0.1351.
sX=[Σx2
/n– (Σx)2
/n2
]1/2
=
[0.0769/10-(0.73)2
/100]1/2
=0.0486
sY=[Σy2
/n– (Σy)2
/n2
]1/2
=[39057/10-(623)2
/100]1/2
= 4.9406 ;
So ρXY=cXY /(sXsY)
= 0.1351/[0.0486(4.9406)]=0.56276
10. *The following data represent carbon dioxide (CO2) emissions from
coal-fired boilers (in units of 1000 tons) over a period of years between 2010 and 2016. The independent
variable (year) has been standardized to yield the following table:
Year (x) 1 2 3 4 5 6 7
CO2 emission 910 680 520 450 370 380 340
b. Estimate the average CO2 emission from coal-fired boilers for the year 2018?
Sol. 10: (a) Given sample size n=7, xi and yi are given
as in the table. From the table we find Σxi=28,
Σyi=3650, Σyi
2
=2160300, Σxi
2
=140 and Σxiyi =12140.
Using the above values, we get b1=[Σxy/n-(Σx/n)(Σy/n)]/
[Σx2
/n-(Σx/n)2
]=cxy/sX
2
=-87.857;
b0 = Σy/n - b1 Σx/n= 872.86 .
Using β0= 872.86 and
β1=-87.857 in μY|x = β0+ β1 x,
we get linear regression equation
μY|x = 872.86 -87.857x (1)
(b) For year 2018, we have x=9, for which
eq. (1) gives μY|x= 872.86 -87.857×9 = 82.143.
11. *Compute the least square regression equation of Y on X and estimate the blood pressure when the age is
45 years.
Age (x) 36 38 42 47 53 60 65
Blood pressure
(Y)
118 115 125 128 147 140 150
Sol.11: For given sample size n=7, xi and yi are as in the table,
we find Σxi=341, Σyi=923, Σyi
2
=122867, Σxi
2
=17347 and Σxiyi =45825.
Using the above values: b1=cxy/sX
2
=[Σxy/n-(Σx/n)(Σy/n)]/[Σx2
/n-(Σx/n)2
] =1.1717;
b0 = Σy/n - b1 Σx/n= 74.78 .
Using β0= 74.78 & β1=1.1717 in μY|x = β0+ β1 x,
we get linear regression equation
μY|x =74.78+1.1717x (1)
(b) For age 45 years, we have x=45.
So eq. (1) gives μY|x= 74.78+1.1717× 45
= 127.5 as blood pressure at 45 years .
y x y2
x2
xy
58 0.01 3364 0.0001 0.58
66 0.03 4356 0.0009 1.98
55 0.01 3025 0.0001 0.55
63 0.02 3969 0.0004 1.26
58 0.10 3364 0.0100 5.80
57 0.08 3249 0.0064 4.56
69 0.12 4761 0.0144 8.28
70 0.15 4900 0.0225 10.50
65 0.10 4225 0.0100 6.50
62 0.11 3844 0.0121 6.82
623 0.73 39057 0.0769 46.83
Σyi Σxi Σyi
2
Σxi
2
Σxiyi
Y X X2 XY
910 1 1 910
680 2 4 1360
520 3 9 1560
450 4 16 1800
370 5 25 1850
380 6 36 2280
340 7 49 2380
Σy
i
Σx
i
Σx
i
2 Σx
i
y
i
3650 28 140 12140
Y X X2
XY
118 36 1296 4248
115 38 1444 4370
125 42 1764 5250
128 47 2209 6016
147 53 2809 7791
140 60 3600 8400
150 65 4225 9750
Σyi Σxi Σxi
2
Σxiyi
923 341 17347 45825

5
12. In 1980 the Bureau of Labour Statistics conducted a study of 1000 minor eye injuries received by
workers in the workplace. The study revealed that 600 of the workers involved were not wearing eye
protection at the time of the injury. It also revealed that 900 of the injuries received could have been
prevented through the proper use of protective eyewear. Assume that current conditions in the
workplace have not changed substantially from those encountered in 1980 relative to the use of the
eye protection.
a) Find a 90% confidence interval on the proportion of workers who receive minor eye injuries this
year that will not be wearing eye protection at the time of the injury.
b) Find a 95% confidence interval on the proportion of minor eye injuries occurring this year that
could be prevented through the proper use of protective eyewear.
Sol. 12: a) Let the proportion of workers receiving minor eye injuries this year by not wearing eye
protection: p1. Then p1=600/1000=0.6.
90% confidence interval is: pe + zα/2√[peqe/n].
or p1 +z0.05√[p1q1/n] [Since 100(1-α)%=90% gives α=0.1]
Here z0.05 is the value for which P[Z>z0.05]=0.05,
i.e. P[z< z0.05]=0.95. So z0.05=1.645 [See normal table]
Thus, p1+z0.05√[p1q1/n] gives 0.6+1.645√[0.6(0.4)/1000]
Or 0.6+0.0254 or [0.5746, 0.6254] is 90% confidence interval of population proportion p.
b) Let the proportion of workers preventable from eye injuries this year: p2. Then p2=900/1000=0.9.
95% confidence interval is: pe + zα/2√[peqe/n],α=0.05
i.e. p2 +z0.025√[p2q2/n]=0.9+ 1.96√[0.9(0.1)/1000]=0.9+ 0.0186=(0.8814, 0.9186)
[Since P(Z< z0.025)=1-.025 gives z0.025=1.96 from normal table for P=0.975]
13. Ozone is a component of smog that can injure sensitive plants even at low levels. In 1979 federal
Ozone standard of 0.12 ppm was set. It is thought that the Ozone level in air currents over New
England exceeds this level. To verify this contention, air samples are obtained from 30 monitoring
stations setup across the region
a. Set up the appropriate null and alternative hypothesis for verifying the contention
b. What is the critical point for a 𝛼 = 0.01 level test based on a sample of size 30?
c. When the data are analysed, a sample mean of 0.135 and sample standard deviation of 0.03 are
obtained. Use this data to test 𝐻0. Can 𝐻0 be rejected at the 𝛼 = 0.01 level? What does this
mean in a practical sense?
Sol.13: a) H0: μ=0.12, H1: μ>0.12(Right tailed test)
b) For a α=0.01 (significance) level test and sample size n= 30, the critical point Zα=Z.01 is point on which
right side has an area P[Z>Z.01]=.01 and thus P[Z<Z.01]=.99=F(.99) in T table for d.o.f. ϒ=n-1=29. Thus
Z.01=2.462.
c) Given Xbar=0.135 and s=0.03, n=30. Test statistic is:
T=(Xbar-μ)/(s/√n)=(.135-.120)/(.03/√30)=2.7386.
Since T=2.738>Ttable=2.462 for ϒ=29, we reject H0 at level of significance.
d) Practically it means that RV X at least follows approximately normal distribution (Since T distribution
is approximate normal distribution with steep rise and fall of F value).
14. A battery-operated digital pressure monitor is being developed for use in calibrating pneumatic
pressure gauges in the field. It is thought that 95% of the readings it gives lie within 0.01 lb/in2
of the
true reading. In a series of 100 tests, the gauge is subjected to a pressure of 10,000 la/in2
. A test is
considered to be a success if the reading lies within 10,000 ± 0.01 lb/in2.
We want to test
𝐻0: 𝑝 = 0.95
𝐻1: 𝑝 ≠ 0.95
At the 𝛼 = 0.05 level
a. What are the critical points for the test?
b. When the data are gathered, it is found that 98 of the 100 readings sere successful. Can 𝐻0 be
rejected at the 𝛼 = 0.05 level? To what type error are you now subject?

6
Sol.14: a) Given H0: p=0.95; H1: p≠0.95.(Two-tailed test)
At the level α=0.01, a critical point zα/2 =z0.005 [Since P(Z>z0.005)=0.005, or P(Z<z0.005)=1-.005 =.995
which gives z0.005=2.57 [from normal table] So the critical points are +z0.005=+2.57
c) Given pe=98/100=0.98, p0=0.95, n=100.
Test statistic: Z=(pe-p0)/√[p0q0/n]=(0.98-0.95)/√[0.95(0.05)/100]=1.38
Since Zcalc=1.38<zα/2=z0.005=2.57, H0 cannot be rejected at test level α=0.01 of two tailed test. The only
option left is to accept H0. The only option remains is to reject H1. In doing so, we would commit
Type II error.
15. Recent research indicates that heating and cooling commercial buildings with ground water-source
heat pumps is economically sound. The crucial random variable being studied is the water
temperature. A sample of 15 wells in the state of California yields a sample standard deviation of
0
7.5 F . Find 90% confidence interval on S.D in temperature of wells in California.
Sol. 15: 100(1-α)% confidence limits on σ2
are: 𝐿 = 𝑛 − 1 𝑠2
/𝜒𝛼/2
2
and 𝑈 = 𝑛 − 1 𝑠2
/𝜒1−𝛼/2
2
.
For 90% confidence limits, α =0.10 Sample variance is given by s2
=
1
𝑛−1
(𝑋𝑖 − 𝑋)2
𝑛
𝑖=1 =0.0011.e 1.
𝑃 𝑋19
2
> 𝜒0.025
2
= 0.025, or 𝑃 𝑋19
2
≤ 𝜒0.025
2
= 1 − .025=.975.
So 𝜒0.025
2
=32.5 from Chi-square table.
To find: 𝜒1−𝛼/2
2
= 𝜒1−0.025
2
= 𝜒0.975
2
, 𝑃 𝑋19
2
> 𝜒0.975
2
= 0.975, or 𝑃 𝑋19
2
≤ 𝜒0.975
2
=1-0.975= 0.025.
Therefore 𝜒0.975
2
L= 𝑛 − 1 𝑠2
/𝜒𝛼/2
2
=(20-1)0.0011/32.9=0.000635;
U= 𝑛 − 1 𝑠2
/𝜒1−𝛼/2
2
=(20-1)0.0011/8.91= 0.00235.
So 95% confidence limits for σ2
are 0.000635 and 0.00235.
Therefore, 95% confidence limits for σ are √0.000635= 0.025199 and √0.00235= 0.048477.
16. When programming from a terminal, one random variable of concern is the response time in seconds.
The following data are obtained for one particular installation. Construct 95% confidence limits for
2

1.48 1.26 1.56 1.48 1.57
1.30 1.43 1.43 1.55 1.61
1.51 1.64 1.37 1.47 1.65
1.49 1.51 1.51 1.60 1.46
Sol.16: 100(1-α)% confidence limits on σ2
are: 𝐿 = 𝑛 − 1 𝑠2
/𝜒𝛼/2
2
and 𝑈 = 𝑛 − 1 𝑠2
/𝜒1−𝛼/2
2
. For 95%
confidence limits, α =.05. Sample variance is given by s2
=
1
𝑛−1
(𝑋𝑖 − 𝑋)2
𝑛
𝑖=1 =0.0011 as in Table 1.
𝑃 𝑋19
2
> 𝜒0.025
2
= 0.025, or 𝑃 𝑋19
2
≤ 𝜒0.025
2
=1-0.025=0.975. So 𝜒0.025
2
To find: 𝜒1−𝛼/2
2
= 𝜒1−0.025
2
= 𝜒0.975
2
, 𝑃 𝑋19
2
> 𝜒0.975
2
= 0.975, or 𝑃 𝑋19
2
≤ 𝜒0.975
2
=1-0.975=0.025.
Therefore 𝜒0.975
2
L = 𝑛 − 1 𝑠2
/𝜒𝛼/2
2
=(20-1)0.0011/32.9=0.000635; U= 𝑛 − 1 𝑠2
/𝜒1−𝛼/2
2
=(20-1)0.0011/8.91= 0.00235.
So 95% confidence limits for σ2
are 0.000635 and 0.00235. Therefore, 95% confidence limits for σ are
√(0.000635)= 0.025199 and √0.00235= 0.048477.
Table for calculating sample variance s2
S. No. 1 2 3 4 5 6 7 8 9 10
X
1.48 1.26 1.56 1.48 1.30 1.43 1.43 1.55 1.51 1.64
(X-𝑿)2
0.0001 0.043472 0.0084 0.0001 0.028 0.001 0.00148 0.0066 0.002 0.0294
X 11 12 13 14 15 16 17 18 19 20 𝑿
(X-𝑿)2
1.37 1.47 1.49 1.51 1.51 1.60 1.48 1.26 1.56 1.48 1.469
0.0097 0.000002 0.0005 0.0017 0.002 0.017 0.00013 0.0435 0.008 0.0001 0.0216
s2
= 0.0011

7
17. The relationship between energy consumption and household income was studied, yielding the
following data on household income X (in units of 1000/year) and energy consumption Y (in units of
108
Btu/year)
Energy consumption(Y) Household income (X)
1.8 20.0
3.0 30.5
4.8 40.0
5.0 55.1
6.5 60.3
7.0 74.9
9.0 88.4
9.1 95.2
Test 𝐻0: 𝛽0 = 2 at the 0.01 level of significance.
18. The following data represent carbon dioxide (CO2) emissions from coal-fired boilers (in units of 1000
tons) over a period of years between 1965 and 1977. The independent variable (year) has been
standardized to yield the following table:
Year (x) 0 5 8 9 10 11 12
CO2 emission 910 680 520 450 370 380 340
b. Is there a significant linear trend in CO2 emission over this time span? That is, test 𝐻0: 𝛽1 = 0
at the 0.01 level of significance
c. Would it be wise to use the estimated regression line to estimate the average CO2 emission
from coal-fired boilers for the year 2000?
19. Given X and Y data as below.
Enzyme level x 95 110 118 124 145 140 185 190 205 222
Detoxification
level y
108 126 102 121 118 155 158 178 159 184
Test 𝐻0: 𝜌 = 0.8 at the 𝛼 = 0.05 level of significance.
20. These data are obtained in the random variables x, the percentage copper of a sample and its Rockwell
hardness rating y:
X Y
0.01 58.0
0.03 66.0
0.01 55.0
0.02 63.2
0.10 58.3
0.08 57.9
0.12 69.3
0.15 70.1
0.10 65.2
0.11 62.3
a. Plot a scatter gram for the above data.
b. Find a point estimate for 𝜌.
c. Find a 95% confidence interval for 𝜌 and discuss your conclusion.

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