Course: 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.?
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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: (10)
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.
Let's
tackle each part of the question step by step.
### Part (a):
i)
Probability that an individual has less than $10,000 of life insurance:
\[
P(\text{Less than } \$10,000) = 0.17 \]
ii)
Probability that an individual has between $10,000 and $99,999 of life
insurance:
\[
P(\$10,000 \text{ to } \$99,999) = 0.20 + 0.17 + 0.14 = 0.51 \]
iii) Probability that an individual has
$50,000 or more of life insurance:
\[
P(\$50,000 \text{ or more}) = 0.15 + 0.12 + 0.05 = 0.32 \]
iv) Probability that an individual has
less than $50,000 or $100,000 or more of life insurance:
\[
P(\text{Less than } \$50,000 \text{ or } \$100,000 \text{ or more}) =
P(\text{Less than } \$50,000) + P(\$100,000 \text{ or more}) = 0.17 + 0.32 =
0.49 \]
### Part (b):
Let's define some probabilities first:
- \(
P(\text{Overweight}) \) is the probability that a randomly selected person is
overweight.
- \(
P(\text{Salesperson}) \) is the probability that a randomly selected person is
a salesperson.
- \(
P(\text{Construction worker}) \) is the probability that a randomly selected
person is a construction worker.
Given:
-
Number of salespersons (S) = 410
-
Number of construction workers (C) = 350
-
Number of overweight salespersons (OS) = 164
-
Number of overweight construction workers (OC) = 196
i) Probability that a randomly selected
person is overweight:
\[
P(\text{Overweight}) = \frac{164 + 196}{410 + 350} = \frac{360}{760} =
\frac{9}{19} \]
ii) Probability that a randomly selected
person is a salesperson, given that the person is overweight:
\[
P(\text{Salesperson | Overweight}) = \frac{164}{164 + 196} = \frac{164}{360} =
\frac{41}{90} \]
iii) Probability that a randomly selected
person is overweight, given that the person is a salesperson:
\[
P(\text{Overweight | Salesperson}) = \frac{164}{410} = \frac{82}{205} \]
iv) Probability that a randomly selected
person is a construction worker, given that the person is not overweight:
\[
P(\text{Construction worker | Not Overweight}) = \frac{350 - 196}{760 - 360} =
\frac{154}{400} = \frac{77}{200} \]
v) Probability that a randomly selected
person is not overweight, given that the person is a construction worker:
\[
P(\text{Not Overweight | Construction worker}) = \frac{350 - 196}{350} =
\frac{154}{350} = \frac{22}{50} = \frac{11}{25} \]
That
covers all the probabilities for each part of the question. Let me know if you
need further clarification on any part!
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)
elsewhere
for x
for x
x
x
f x 1 2
0 1
0
( ) 2
i) Show that P (0 < X < 2) = 1
ii) Find P(X < 1.2)
Let's handle each part of the question:
### Part (a):
To
obtain a probability distribution of the sum of spots when a pair of dice is
rolled, we'll consider all possible outcomes when two dice are rolled and
calculate the probability for each sum.
The
sum of spots on two dice can range from 2 to 12.
Here's the probability distribution:
| Sum
of Spots | Probability |
|--------------|-------------|
|
2 | 1/36 |
|
3 | 2/36 |
|
4 | 3/36 |
|
5 | 4/36 |
|
6 | 5/36 |
|
7 | 6/36 |
|
8 | 5/36 |
|
9 | 4/36 |
|
10 | 3/36 |
|
11 | 2/36 |
|
12 | 1/36 |
### Part (b):
The
density function for the continuous random variable X is given as:
\[
f(x) = \begin{cases}
x & \text{for } 1 \leq x < 2 \\
2 - x & \text{for } 0 \leq x < 1
\\
0 & \text{elsewhere}
\end{cases}
\]
i) To
show that \( P(0 < X < 2) = 1 \), we need to find the integral of the
density function over the interval (0, 2).
\[ P(0
< X < 2) = \int_{0}^{1} (2 - x) dx + \int_{1}^{2} x dx \]
\[ =
\left[2x - \frac{x^2}{2}\right]_{0}^{1} + \left[\frac{x^2}{2}\right]_{1}^{2} \]
\[ =
\left[2 - \frac{1}{2}\right] + \left[2 - \frac{1}{2}\right] = 1 \]
ii) To
find \( P(X < 1.2) \), we integrate the density function over the interval
(0, 1.2).
\[ P(X
< 1.2) = \int_{0}^{1.2} (2 - x) dx \]
\[ =
\left[2x - \frac{x^2}{2}\right]_{0}^{1.2} \]
\[ =
\left[2(1.2) - \frac{(1.2)^2}{2}\right] - \left[2(0) - \frac{(0)^2}{2}\right]
\]
\[ =
2.4 - \frac{1.44}{2} = 2.4 - 0.72 = 1.68 \]
This
covers both parts of the question. Let me know if you need further
clarification!
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.
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: (8)
2x + 3y =3
6x + 4y = 5
Let's handle each part of the question:
### Part (a):
#### i) Graph passes through the point
(2,–1) with a slope of 3:
We
know that the slope-intercept form of a linear equation is \( y = mx + b \),
where \( m \) is the slope and \( b \) is the y-intercept.
Given the point (2, -1) and the slope of
3, we can find the equation:
\[ y -
(-1) = 3(x - 2) \]
\[ y +
1 = 3x - 6 \]
\[ y =
3x - 7 \]
#### ii) Graph passes through the point
(–3,–2) with a slope of –1:
Following the same procedure:
\[ y -
(-2) = -1(x - (-3)) \]
\[ y +
2 = -1(x + 3) \]
\[ y +
2 = -x - 3 \]
\[ y =
-x - 5 \]
####
iii) Graph passes through the point (2, 4) with a slope of 0:
When
the slope is 0, the line is horizontal. So, the equation is just \( y = 4 \).
#### iv) Graph passes through the point
(5, 0) with undefined slope:
If the
slope is undefined, the line is vertical. For a line passing through (5, 0),
the equation is just \( x = 5 \).
Now,
let's sketch these graphs.
### Part (b):
To
solve the simultaneous linear equations by the graphical method, we'll graph
both lines and find the point where they intersect.
The given equations are:
\[ 2x
+ 3y = 3 \]
\[ 6x
+ 4y = 5 \]
We can rewrite these equations in
slope-intercept form to graph them:
\[ y =
\frac{-2}{3}x + 1 \]
\[ y =
\frac{-3}{2}x + \frac{5}{4} \]
Then,
we'll plot these lines on the graph and find the point of intersection.
Let me
know if you need further clarification or assistance with the graphing!
Q. 5 (a) Suppose that the demand and price
for a certain brand of shampoo are related by
p q
4
5
16 (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 :
p )q
4
5
16
(
Suppose
the price and supply of the shampoo are related by,
p q
4
3
Where
q represents the supply, and p the price.
iv)
Find the supply when the price is : $0, $10, $20
v)
Graph
p q
4
3
on the
same axes used for part (iii)
vi)
Find the equilibrium supply
vi)
Find the equilibrium price.
Total
Marks: 100 Pass Marks (BA Old): 40
Pass Marks: (BA AD): 50
ASSIGNMENT
No. 2
Let's handle each part of the question:
### Part (a):
#### i) Find the price for a demand of: 0
units ; 8 units
To find the price for a given demand, we
substitute the demand (q) into the demand function:
- For \( q = 0 \):
\[ p =
16 - \frac{20}{5} \times 0 = 16 \]
- For
\( q = 8 \):
\[ p =
16 - \frac{20}{5} \times 8 = 16 - 16 = 0 \]
####
ii) Find the demand for the shampoo at a price of: $6, $11, $16
To
find the demand for a given price, we rearrange the demand function:
\[ q =
\frac{16 - p}{\frac{20}{5}} = 4(16 - p) \]
- For
\( p = 6 \):
\[ q =
4(16 - 6) = 4 \times 10 = 40 \]
- For
\( p = 11 \):
\[ q =
4(16 - 11) = 4 \times 5 = 20 \]
- For
\( p = 16 \):
\[ q =
4(16 - 16) = 4 \times 0 = 0 \]
####
iii) Graph: \( p = \frac{16 - q}{\frac{4}{5}} \)
We can
graph this equation by plotting points for different values of \( q \) and then
connecting them to form the graph.
####
iv) Find the supply when the price is: $0, $10, $20
To
find the supply for a given price, we substitute the price (p) into the supply
function:
- For
\( p = 0 \):
\[ q =
\frac{4}{3} \times 0 = 0 \]
- For
\( p = 10 \):
\[ q =
\frac{4}{3} \times 10 = \frac{40}{3} \]
- For
\( p = 20 \):
\[ q =
\frac{4}{3} \times 20 = \frac{80}{3} \]
####
v) Graph: \( p = \frac{q}{\frac{4}{3}} \)
We'll
graph this equation on the same axes used for part (iii).
####
vi) Find the equilibrium supply
The
equilibrium supply occurs when the quantity supplied equals the quantity
demanded. So, we set the two supply and demand functions equal to each other
and solve for \( q \):
\[
4(16 - p) = \frac{q}{\frac{4}{3}} \]
\[ 3
\times 4(16 - p) = q \]
\[ 48
- 12p = q \]
####
vii) Find the equilibrium price
To
find the equilibrium price, we substitute the equilibrium supply value into
either the supply or demand function and solve for \( p \).
This
covers all parts of the question. Let me know if you need further clarification
or assistance with any part!
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:
JUST
5 BULLET POINTS WITHOUT ANY HEADINGS AND SUB BULLET POINTS