Statistics assignment 4

1) A corporation takes delivery of some new machinery that must be installed and checked before
it becomes operational. The accompanying table shows a manager’s probability assessment for
the number of days required before the machinery becomes operational.
Number of days 3 4 5 6 7
Probability .08 .24 .41 .20 .07
Let, A be the event “it will be more than 4 days before the merchandise becomes operational”
and B the event “it will be less than 6 days before the merchandise becomes available”
Ans:
Let,
A denotes that, “it will be more than 4 days before the merchandise becomes operational.”
B denotes that, “it will be less than 6 days before the merchandise becomes available.”
A= [5, 6, 7] and B= [3, 4, 5]
a) P (A)= P (5)+P (6)+P (7)
=. 41 +. 20 +. 07
= .68
The probability of event A = .68
So, there will be .68 chances that it will be more than 4 days before the machinery becomes
operational.
b) P(B) = P(3) + P(4) + P(5)
= .08+. 24 +. 41
=. 73
So, there will be .73 chances that it will be less than 6 days before the machinery becomes
available.
c) The symbol of A complement is A . So, A denotes that, it will be at most 4 days before the
machinery becomes operational.
d) P ( A ) = P (S) – P (A)
= 1-.68
=. 32
e) The symbol of intersection of events A and B are AÇ B denotes is the set of all the basic
outcomes that are both in events A and B. So,
AÇ B = [5, 6, 7] Ç [3, 4, 5] = [5]
So, it will be 5 days before the machinery becomes available or operational.
f) (AÇ B) = (A) Ç (B)
=[5,6,7] Ç [3,4,5]
= [5]
P (AÇ B) = P (5)
=. 41
g) The symbol of union of events A and B are AÈ B denotes is the set of all the basic outcomes
will occur at least one of these two events. So, A È B = [5,6,7] È [3,4,5] = [3, 4, 5, 6, and 7]
So, it will be at most 7 days before the machine operational or available.
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h) P(AÈ B) = P (3)+P(4)+P(5)+P(6)+P(7)
= .08+.24+.41+.20+.07
= 1
i) No. The events A and B are not mutually exclusive. Because they have to something share
among them.
j) Yes, The events A and B are collectively exhaustive because, P (A È B)=P (S) = 1
2) A fund manager is considering investment in the stock of a health care provider. The
manager assessments of probabilities for rates of return on the stock over the next year are
summarized in the accompanying table. Let A be the event “Rate of return will be more than
10%” and B the event “Rate of return will be negative”
Rate of return Less than-
10%
-10% to 0% 0% to 10% 10% to
20%
More
than 20%
Probability .04 .14 .28 .33 .21
Ans:
Let, A = Rate of return will be more than 10%
B = Rate of return will be negative
So, A = [10% to 20%, more than 20%] and B = [Less than -10%, -10% to 0%]
a) P (A) = P (10% to 20%) + P (more than 20%)
= .33+.21 = .54
b) P (B) = P(Less than -10%) + P (-10% to 0%)
= .04+.14 = .18
c) The symbols of A complement is A . The element have of the event A that are must not have
in the A . So, A = [Less than -10%, -10% to 0%, 0% to 10% ]
d) P ( A ) = P(Less than -10%) + P (-10% to 0%) + P (0% to 10%)
= .04 + .14 + .28 = .46
e) The symbol of intersection of events A and B are AÇ B denotes is the set of all the basic
outcomes that are both in events A and B. So, AÇ B = [ ], because nothing to share among
them.
f) P(AÇ B) = P( ) = 0
g) The symbol of union of events A and B are A È B denotes is the set of all the basic outcomes
will occur at least one of these two events. So, AÈ B = [Less than -10% , -10% to 0% , 10%
to 20% , more than 20% ]
h) P (AÈ B) = P(Less than -10%) + P (-10% to 0%) + P (10% to 20%) + P (more than 20%)
= .04 + .14 + .33 + .21 = .72
i) Yes. The events A and B are mutually exclusive. Because they have nothing to share among
them, and their intersection is 0.
j) No, the events A and B are not collectively exhaustive because, their union is not equal to the
sample space. So, P (AÈ B) ¹ P (S) ¹ 1
3) A manager has available a pool of eight employees who could be assigned to a project
monitoring task. Four of the employees are women and four are men. Two of the men are
brothers. The manager is to make the assignment at random, so that each of the eight
employees is equally likely to be chosen.
Let A be the event “chosen employee is a man.”
And B the event “chosen employee is of the brothers.”
2

Ans:
Let,
A denotes that, “chosen the employee is a man.”
B denotes that, “chosen the employee is of the brother.”
a) P (A) = 4/8 = ½
b) P (B) = 2/8 = ¼
c) P (AÇ B) = 2/8 = ¼
d) P (AÈ B) = P (A) + P (B) – P (A Ç B)
= ½ + ¼ - ¼
= ½
4) In Section 3.4, we saw that if pair of events are mutually exclusive, the probability of their
union is the sum of their individual probabilities. However, this is not the case for the events
that are not mutually exclusive. Verify this assertion by considering the events A and B of
exercise 1.
Ans: The assertion by considering the events A and B of exercise 1 are not mutually exclusive
because, they have to something share. When the two events are mutually exclusive then, they have
nothing to share among them. So their intersection will be 0, AÇ B = 0. And their probability P
(AÇ B) = 0. But in exercise-1, AÇ B¹ 0, and their probability P (AÇ B) ¹ 0. In this why these two
events are not mutually exclusive.
5) A department store manager has monitored the number of complaints received per week
about poor service. The probabilities for numbers of complaints in a week, established by the
review, are shown in the table. Let A be the event “There will be at least one component in a
work.” and B the event, “There will be less than ten components in a work.”
NUMBER OF
0 1-3 4-6 7-9 10-12 More than
COMPLAINTS
12
PROBABILITY .14 .39 .23 .15 .06 .03
Ans:
Let,
A denotes that, “at least one component in a work.”
B denotes that, “less than ten components in a work.”
A= [1-3, 4-6, 7-9, 10-12, more than 12] and B= [0, 1-3, 4-6, 7-9].
a) P (A) = P (1-3)+P (4-6)+P (7-9)+P (10-12)+P (more than 12)
= .39+. 23 +. 15 +. 06 +. 03
= .86
So, there will be 53% chances to occur the events A in a week.
b) P (B) = P (0) + P (1-3) +P (4-6) +P (7-9)
= .14+. 39 +. 23 +. 15
=. 91
So, there will be 91% chances to occur the events B in a week.
c) P ( A ) = 1 – P (A)
= 1 - .86
3

= .14
d) The probability of the union A and B is;
P (AÈ B) = P (A) + P (B) – P (A Ç B)
= .86 + .91 - .77
= 1
e) The probability of the intersection A and B is:
P (AÇ B) = P (1-3) + P (4-6) +P (7-9)
= .39+. 23 +. 15
= .77
f) The events A &B are not mutually exclusive because they have something to
Share.
g) The events A & B are collectively exhaustive because P (AÈ B) equal to 1.
6) A corporation receives a particular part in shipments of 100.Research has indicated the
probabilities shown in the accompanying table for number of defective parts in a shipments.
NUMBER DEFECTIVE 0 1 2 3 More than 3
PROBABILITY .29 .36 .22 .10 .03
Ans:
Let, A denotes that “there will be less than 3 defective part in a shipment”
B denotes that “there will be more than 1 defective part in a shipment”
A =[0,1,2] & B= [2, 3, More than 3]
a) P (A)=P (0)+ P (1)+ P (2)
=. 29 +. 36 +. 22
=. 87
So, there will be 87% chances to occur the event A in a shipment.
b) P (A)=P (2)+ P (3)+ P (More than 3)
=. 22 +. 10 +. 03
= .35
So, there will be 35% chances to occur the event B in a shipment.
c) The events A & B are collectively exhaustive because P (A È B) equal to 1.
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Statistics assignment 4

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Statistics assignment 4