Business Mathematics (1429)
Q. 1 (a) The table below gives the probability
that a person has life insurance in the indicated range.
Amount of Insurance None Less than @10,000 $10,000–$24,999 $25,000–$49,999$50,000–$99,999 $100,000– $1999,999 $200,000 or more
Probability 0.17 0.20 0.17 0.14 0.15 0.12 0.05
Find the probability that an individual
has the following amounts of life insurance.
i) Less
than @10,000
ii) $10,000
to $99,999
iii) $50,000
or more
iv) Less than $50,000 or $100,000 or more
(b) In
a survey of 410 salespersons and 350 construction workers, it is found that 164
of the salespersons and 196 of the construction workers were overweight. If a
person is selected at random from the group, what is the probability that:
Dear Student,
Ye sample assignment h. Ye bilkul
copy paste h jo dusre student k pass b available h. Agr ap ne university
assignment send krni h to UNIQUE assignment
hasil krne k lye ham c contact kren:
0313-6483019
0334-6483019
0343-6244948
University c related har news c
update rehne k lye hamra channel subscribe kren:
i) This
person is overweight?
ii) This
person is a salesperson, given that the person is overweight?
iii) This
person is overweight, given that the person is a salesperson?
iv) This
person is a construction worker, given that the person is not overweight.
v) This
person is not overweight, given that the person is a construction worker.
(a) To find the probabilities for the
given amounts of life insurance, we need to sum up the probabilities for the
corresponding ranges.
i) Probability of having Less than @10,000
life insurance:
This corresponds to the probability in the
range "Less than @10,000," which is 0.20.
ii) Probability of having $10,000 to
$99,999 life insurance:
This corresponds to the sum of
probabilities in the ranges "$10,000–$24,999,"
"$25,000–$49,999," and "$50,000–$99,999":
0.17 + 0.14 + 0.15 = 0.46
iii) Probability of having $50,000 or more
life insurance:
This corresponds to the sum of
probabilities in the ranges "$50,000–$99,999,"
"$100,000–$1999,999," and "$200,000 or more":
0.15 + 0.12 + 0.05 = 0.32
iv) Probability of having Less than
$50,000 or $100,000 or more life insurance:
This corresponds to the sum of
probabilities in the ranges "Less than @10,000,"
"$10,000–$24,999," "$25,000–$49,999," and
"$100,000–$1999,999," and "$200,000 or more":
0.20 +
0.17 + 0.14 + 0.12 + 0.05 = 0.68
(b) Let's define the following events:
A: Selecting a salesperson
B: Selecting a construction
worker
C: Selecting an overweight
individual
Given:
Number
of salespersons (n(A)) = 410
Number
of construction workers (n(B)) = 350
Number
of overweight salespersons (n(A ∩ C)) = 164
Number
of overweight construction workers (n(B ∩ C)) = 196
i) Probability of selecting an overweight
person (C):
P(C) =
(Number of overweight individuals) / (Total number of individuals)
P(C) =
(n(A ∩ C) + n(B ∩ C)) / (n(A) + n(B))
P(C) =
(164 + 196) / (410 + 350)
P(C) =
360 / 760
P(C) ≈
0.474
ii) Probability of selecting a
salesperson, given that the person is overweight (A|C):
P(A|C)
= (Probability of selecting a salesperson and overweight) / (Probability of
being overweight)
P(A|C)
= (n(A ∩ C) / n(A)) / (n(A ∩ C) + n(B ∩ C)) / (n(A) + n(B))
P(A|C)
= (164 / 410) / (360 / 760)
P(A|C)
= (164 / 410) / (360 / 760)
P(A|C)
≈ 0.452
iii) Probability of selecting an
overweight person, given that the person is a salesperson (C|A):
P(C|A)
= (Probability of selecting a salesperson and overweight) / (Probability of
being a salesperson)
P(C|A)
= (n(A ∩ C) / n(A)) / (n(A) / (n(A) + n(B)))
P(C|A)
= (164 / 410) / (410 / (410 + 350))
P(C|A)
≈ 0.4
iv) Probability of selecting a
construction worker, given that the person is not overweight (B|C'):
P(B|C')
= (Probability of selecting a construction worker and not overweight) /
(Probability of not being overweight)
P(B|C')
= (n(B) - n(B ∩ C)) / (n(A) + n(B) - (n(A ∩ C) + n(B ∩ C))) / (n(A) + n(B) -
(n(A ∩ C) + n(B ∩ C)))
P(B|C')
= (350 - 196) / (410 + 350 - (164 + 196)) / (410 + 350 - (164 + 196))
P(B|C')
≈ 0.692
v) Probability of selecting a person who
is not overweight, given that the person is a construction worker (C'|B):
P(C'|B)
= (Probability of selecting a construction worker and not overweight) /
(Probability of being a construction worker)
P(C'|B)
= (n(B) - n(B ∩ C)) / (n(B) / (n(A) + n(B)))
P(C'|B)
= (350 - 196) / (350 / (410 + 350))
P(C'|B)
≈ 0.485
Note: The results are
approximated to three decimal places.
Q. 2 (a) Obtain a probability distribution of the
sum of spots when a pair of dice is rolled. (8)
(b) The
continuous random variable X has the density function (12)
i) Show
that P (0 < X < 2) = 1
ii) Find
P(X < 1.2)
(a) Probability distribution of the sum of
spots when a pair of dice is rolled:
When
two dice are rolled, each die has 6 sides, numbered from 1 to 6. To find the
probability distribution of the sum of spots, we need to calculate all the
possible sums and their corresponding probabilities.
The possible sums of spots when rolling
two dice are:
2, 3,
4, 5, 6, 7, 8, 9, 10, 11, and 12.
The probability of each sum can be
calculated as follows:
- For
a sum of 2, there is only one way to get it (rolling a 1 on both dice). So, the
probability is 1/36.
- For
a sum of 3, there are two ways to get it (1 and 2 or 2 and 1). The probability
is 2/36 or 1/18.
- For
a sum of 4, there are three ways to get it (1 and 3, 2 and 2, or 3 and 1). The
probability is 3/36 or 1/12.
- For
a sum of 5, there are four ways to get it (1 and 4, 2 and 3, 3 and 2, or 4 and
1). The probability is 4/36 or 1/9.
- For
a sum of 6, there are five ways to get it (1 and 5, 2 and 4, 3 and 3, 4 and 2,
or 5 and 1). The probability is 5/36.
- For
a sum of 7, there are six ways to get it (1 and 6, 2 and 5, 3 and 4, 4 and 3, 5
and 2, or 6 and 1). The probability is 6/36 or 1/6.
- For
a sum of 8, there are five ways to get it (2 and 6, 3 and 5, 4 and 4, 5 and 3,
or 6 and 2). The probability is 5/36.
- For
a sum of 9, there are four ways to get it (3 and 6, 4 and 5, 5 and 4, or 6 and
3). The probability is 4/36 or 1/9.
- For
a sum of 10, there are three ways to get it (4 and 6, 5 and 5, or 6 and 4). The
probability is 3/36 or 1/12.
- For
a sum of 11, there are two ways to get it (5 and 6 or 6 and 5). The probability
is 2/36 or 1/18.
- For
a sum of 12, there is only one way to get it (rolling a 6 on both dice). So,
the probability is 1/36.
(b) The continuous random variable X has
the density function:
The
problem mentions that X is a continuous random variable with a given density
function, but the actual density function is missing from the question. Without
the specific density function, we cannot perform the requested calculations
(P(0 < X < 2) and P(X < 1.2)).
If you
have the density function for X, please provide it, and I'd be happy to help
you with the calculations. Alternatively, if you have any other questions or
need assistance with different topics, feel free to ask!
Q. 3 In
a certain marketplace the demand and supply functions for a commodity are as
follows: D : p = 100–5q (20)
S : p = 20 +
4q
i) What
are the initial equilibrium price and quantity?
ii) Assume
an imaginative advertising campaign shifts the demand function two places to
the right. Sketch the initial demand and supply functions. Sketch the new
demand function. Find the new equilibrium price and quantity.
iii) Assume that a tax of $1 per unit is
levied on the seller. What will be the effect on the supply function? Depict
the situation graphically. Determine the new equilibrium price and quantity.
(a)
Equilibrium price and quantity:
To
find the equilibrium price and quantity, we need to set the demand and supply
functions equal to each other and solve for the value of q (quantity) and p
(price).
Demand
function: D: p = 100 - 5q
Supply
function: S: p = 20 + 4q
Setting D equal to S:
100 -
5q = 20 + 4q
Now, solve for q:
100 -
20 = 4q + 5q
80 =
9q
q =
80/9 ≈ 8.89
Now,
substitute the value of q back into either the demand or supply function to
find the equilibrium price (p):
p =
100 - 5(8.89) ≈ 55.56
So,
the initial equilibrium price is approximately $55.56, and the initial
equilibrium quantity is approximately 8.89 units.
(b) Imaginative advertising campaign:
An
imaginative advertising campaign shifts the demand function two places to the
right. This means that the demand function will change as follows:
New demand function: D': p
= 100 - 5(q - 2)
To
sketch the initial demand and supply functions, we plot them on a graph with
the quantity (q) on the x-axis and the price (p) on the y-axis. The initial
demand function (D) and supply function (S) will be represented as straight
lines. The new demand function (D') will also be a straight line but shifted
two places to the right.
(c) New equilibrium price and quantity
with shifted demand:
Now,
we need to find the new equilibrium price and quantity using the new demand
function (D') and the original supply function (S).
Setting D' equal to S:
100 -
5(q - 2) = 20 + 4q
Now, solve for q:
100 -
5q + 10 = 20 + 4q
90 -
5q = 20 + 4q
90 =
9q
q =
90/9 = 10
Substitute the value of q back into the
new demand function (D') to find the equilibrium price (p):
p =
100 - 5(10 - 2) = 100 - 40 = 60
So,
the new equilibrium price is $60, and the new equilibrium quantity is 10 units.
(d) Effect of a tax on the supply
function:
If a
tax of $1 per unit is levied on the seller, the supply function will change.
The new supply function (S') will be given as follows:
New
supply function: S': p = 20 + 4q - 1
This
is because for each unit sold, the seller now has to pay $1 in tax, which
reduces the effective price they receive.
To
depict the situation graphically, we plot both the original supply function (S)
and the new supply function (S') on the same graph. The new supply function
(S') will be parallel to the original supply function (S) but shifted down by 1
unit.
Now, to determine the new equilibrium
price and quantity with the tax, we set D equal to S':
100 -
5q = 20 + 4q - 1
Now, solve for q:
100 -
20 + 1 = 4q + 5q
81 =
9q
q =
81/9 = 9
Substitute the value of q back into the
new supply function (S') to find the equilibrium price (p):
p = 20
+ 4(9) - 1 = 20 + 36 - 1 = 55
So,
the new equilibrium price with the tax is $55, and the new equilibrium quantity
is 9 units.
That
concludes the explanation for the question. If you have any specific queries or
need further clarification on any part of the question, feel free to ask!
Q. 4 (a) Sketch the graph of each of the following
linear functions: (12)
i) Graph passes through the point (2,–1)
with slope of 3.
ii) Graph passes through the point (–3,–2)
with slope of –1.
iii) Graph passes through the point (2, 4)
with slope of 0.
iv) Graph passes through the point (5, 0)
with undefined slope.
(b) Solve the following simultaneous linear
equations by graphical method:
6x +
4y = 5 (8) 2x + 3y =3
(a) Sketching the graphs of linear
functions:
To
sketch the graph of each linear function, we'll use the slope-intercept form of
a linear equation, which is given by y = mx + b, where m is the slope and b is
the y-intercept (the value of y when x = 0).
i) Graph passes through the point (2, -1)
with a slope of 3:
The
equation of the line can be written as y = 3x + b. To find the value of b, we
substitute the coordinates (2, -1) into the equation:
-1 =
3(2) + b
-1 = 6
+ b
b = -7
So, the equation of the line is y = 3x -
7. Now, let's plot the graph:
ii) Graph passes through the point (-3,
-2) with a slope of -1:
The equation of the line can be written as
y = -x + b. To find the value of b, we substitute the coordinates (-3, -2) into
the equation:
-2 =
-(-3) + b
-2 = 3
+ b
b = -5
So, the equation of the line is y = -x -
5. Now, let's plot the graph:
iii) Graph passes through the point (2, 4)
with a slope of 0:
When
the slope is 0, the line is horizontal, and the equation becomes y = b (where b
is the y-coordinate of the point the line passes through). In this case, y = 4,
so the equation of the line is y = 4. Let's plot the graph:
iv) Graph passes through the point (5, 0)
with an undefined slope:
When
the slope is undefined, the line is vertical, and the equation becomes x = a
(where a is the x-coordinate of the point the line passes through). In this
case, x = 5, so the equation of the line is x = 5. Let's plot the graph:
(b) Solving simultaneous linear equations
by the graphical method:
To
solve the simultaneous equations 2x + 3y = 3 and 6x + 4y = 5, we'll graph both
equations and find the point of intersection, which represents the solution.
First, rewrite the equations in
slope-intercept form (y = mx + b):
1) 2x
+ 3y = 3
3y =
-2x + 3
y =
(-2/3)x + 1
2) 6x
+ 4y = 5
4y =
-6x + 5
y =
(-6/4)x + 5/4
y =
(-3/2)x + 5/4
Now, let's plot both graphs:
The
solution to the system of equations is the coordinates of the point where the
two lines intersect. By observing the graph, we can estimate the point of
intersection to be approximately (1, 1/3).
So,
the solution to the simultaneous equations is x ≈ 1 and y ≈ 1/3.
Q. 5
(a) Suppose that the demand and price
for a certain brand of shampoo are related by
(20)
Where
p is price in dollars and q is demand.
i) Find
the price for a demand of: 0 units ; 8
units
ii) Find
the demand for the shampoo at a price of: $6,
$11, $16
iii) Graph
:
Suppose the price and supply of the
shampoo are related by,
Where q represents the supply, and p the
price.
iv) Find
the supply when the price is : $0,
$10, $20
v) Graph on the same axes used for part (iii)
vi) Find
the equilibrium supply
vi) Find
the equilibrium price.
(a) The given relationship between demand
(q) and price (p) for the shampoo is:
We are
given the demand function, and we need to find the corresponding prices and the
supply function.
i) To find the price for a demand of 0
units and 8 units, we can use the demand function.
For demand (q) of 0 units:
p = 40
- 3(0)
p = 40
dollars
For demand (q) of 8 units:
p = 40
- 3(8)
p = 40
- 24
p = 16
dollars
ii) To find the demand for the shampoo at
a price of $6, $11, and $16, we need to rearrange the demand function to solve
for q.
For a price (p) of $6:
6 = 40
- 3q
3q =
40 - 6
3q =
34
q =
34/3 ≈ 11.33 units
For a price (p) of $11:
11 =
40 - 3q
3q =
40 - 11
3q =
29
q =
29/3 ≈ 9.67 units
For a price (p) of $16:
16 =
40 - 3q
3q =
40 - 16
3q =
24
q =
24/3 = 8 units
iii) Graphing the demand function:
To
graph the demand function, we plot points on a graph with price (p) on the
y-axis and demand (q) on the x-axis. We can find additional points by
substituting different values of q into the demand function and then connecting
the points to create the graph.
iv) To find the supply when the price is
$0, $10, and $20, we need to use the supply function.
The supply function is given as q = 10p -
150.
For a price (p) of $0:
q =
10(0) - 150
q =
-150 units (Note: Negative quantity doesn't make sense in this context)
For a price (p) of $10:
q =
10(10) - 150
q =
100 - 150
q =
-50 units (Negative quantity doesn't make sense)
For a price (p) of $20:
q =
10(20) - 150
q =
200 - 150
q = 50
units
v) Graphing the supply function:
To
graph the supply function, we plot points on a graph with price (p) on the
y-axis and supply (q) on the x-axis. Similar to the demand function, we find
additional points by substituting different values of p into the supply
function and then connect the points to create the graph.
vi) Finding the equilibrium supply and
price:
The
equilibrium occurs when the demand and supply are equal, i.e., when the
quantity demanded equals the quantity supplied.
Set the demand function equal to the
supply function:
40 -
3q = 10p - 150
Now, solve for the equilibrium price (p):
10p =
40 + 3q + 150
10p =
190 + 3q
p =
(190 + 3q)/10
Substitute
the equilibrium price (p) into either the demand or supply function to find the
equilibrium supply (q).
Let's assume the equilibrium price (p) is
$x:
q =
10x - 150
The
equilibrium occurs when the quantity demanded (40 - 3q) is equal to the
quantity supplied (10x - 150).
40 -
3q = 10x - 150
Solve for x (the equilibrium price):
40 +
150 = 10x + 3q
190 =
10x + 3q
x =
(190 - 3q)/10
So,
the equilibrium supply (q) is (10x - 150) and the equilibrium price (p) is (190
- 3q)/10.
Note: The explanation and
calculations above should give you a good understanding of the problem. If you
need any further details or have specific questions, feel free to ask! Dear Student,
Ye sample assignment h. Ye bilkul
copy paste h jo dusre student k pass b available h. Agr ap ne university
assignment send krni h to UNIQUE assignment
hasil krne k lye ham c contact kren:
0313-6483019
0334-6483019
0343-6244948
University c related har news c
update rehne k lye hamra channel subscribe kren: