Question Details
Aspect |
Details |
Programme Title |
Bachelor of Commerce B.Com |
Course Code |
BCOC – 134 |
Course Title |
BUSINESS MATHEMATICS AND STATISTICS |
Assignment Code |
BCOC-134 |
University |
Indira Gandhi National Open University (IGNOU) |
Type |
Free IGNOU Solved Assignment |
Language |
English |
Session |
JAN 2024 – DEC 2024 |
Submission Date |
31st March for July session, 30th September for January session |
BCOC-134 Solved Assignment
Section-A
Q. 1 A dataset representing the monthly sales (in thousands of dollars) for a small business over the past 12 months:
20,22,18,25,21,23,19,24,20,22,26,21
Q. 1 A dataset representing the monthly sales (in thousands of dollars) for a small business over the past 12 months:
20,22,18,25,21,23,19,24,20,22,26,21
Calculate the range, variance, and standard deviation for the given dataset of monthly sales. Interpret the results in the context of the business’s sales variability.
Q. 2 Given the following national income model
Q. 2 Given the following national income model
Find Y Y and C.
Q. 3 Discuss the various functions related to business and economics.
Q. 4 What do you mean by maxima or minima of a function? State the meaning of absolute minimum of a function. Explain the steps for finding maxima and minima of a function.
Q. 5 The manager of a departmental store compiled information on 200 accounts receivable which were delinquent. For each account he has noted the number of days passed after the due date. He then grouped the data as shown in the following frequency distribution. Determine the median using two methods.
Q. 3 Discuss the various functions related to business and economics.
Q. 4 What do you mean by maxima or minima of a function? State the meaning of absolute minimum of a function. Explain the steps for finding maxima and minima of a function.
Q. 5 The manager of a departmental store compiled information on 200 accounts receivable which were delinquent. For each account he has noted the number of days passed after the due date. He then grouped the data as shown in the following frequency distribution. Determine the median using two methods.
Section-B
Q. 6 A manufacturer earns Rs. 5500 in the first month and Rs. 7000 in the second month. On plotting these points, the manufacturer observes a linear function may fit the data.
i. Find the quadratic function that fits the data.
ii. Using the model make a prediction of the earning for the fourth week.
Q. 7 You are given the profit function of a business activity and asked to offer your suggestion on the rate of change of profit. What would you do?
Q. 8 Bank pays compound interest at the rate of5% 5 \% p.a. XYZ deposited a principal amount of RS. 5,000 in bank for 4 years. Find the interest that XYZ will receive.
Q. 9 Explain the difference between Karl Pearson’s correlation co-efficient and spearsman’s rank correlations co-efficient. Under what situations, in the latter preferred to the former?
Q. 10 Explain briefly the additive and multiplicative models of time series. Which of these models is more commonly used and why?
Q. 6 A manufacturer earns Rs. 5500 in the first month and Rs. 7000 in the second month. On plotting these points, the manufacturer observes a linear function may fit the data.
i. Find the quadratic function that fits the data.
ii. Using the model make a prediction of the earning for the fourth week.
Q. 7 You are given the profit function of a business activity and asked to offer your suggestion on the rate of change of profit. What would you do?
Q. 8 Bank pays compound interest at the rate of
Q. 9 Explain the difference between Karl Pearson’s correlation co-efficient and spearsman’s rank correlations co-efficient. Under what situations, in the latter preferred to the former?
Q. 10 Explain briefly the additive and multiplicative models of time series. Which of these models is more commonly used and why?
Section-C
Q. 11 Write short notes on the following:
(a) Factorization
(b) Harmonic Mean
Q. 12 Differentiate between the following:
(a) Descriptive and Inferential statistics
(b) Absolute measures and relative measures of dispersion
Q. 11 Write short notes on the following:
(a) Factorization
(b) Harmonic Mean
Q. 12 Differentiate between the following:
(a) Descriptive and Inferential statistics
(b) Absolute measures and relative measures of dispersion
Expert Answer
Section-A
Question:-01
A dataset representing the monthly sales (in thousands of dollars) for a small business over the past 12 months:
20,22,18,25,21,23,19,24,20,22,26,21
Calculate the range, variance, and standard deviation for the given dataset of monthly sales. Interpret the results in the context of the business’s sales variability.
Answer:
To calculate the range, variance, and standard deviation for the given dataset of monthly sales, we’ll follow these steps:
Given Dataset:
20, 22, 18, 25, 21, 23, 19, 24, 20, 22, 26, 21
Step 1: Calculate the Range
The range is the difference between the maximum and minimum values in the dataset.
- Maximum Value: 26
- Minimum Value: 18
Range = Maximum Value – Minimum Value = 26 – 18 = 8
Step 2: Calculate the Mean
The mean (average) is the sum of all the values divided by the number of values.
Mean (μ) = (20 + 22 + 18 + 25 + 21 + 23 + 19 + 24 + 20 + 22 + 26 + 21) / 12 = 261 / 12 ≈ 21.75
Step 3: Calculate the Variance
Variance measures how much the numbers in the dataset differ from the mean.
-
Subtract the mean from each number in the dataset and square the result:
- (20 – 21.75)² = 3.0625
- (22 – 21.75)² = 0.0625
- (18 – 21.75)² = 14.0625
- (25 – 21.75)² = 10.5625
- (21 – 21.75)² = 0.5625
- (23 – 21.75)² = 1.5625
- (19 – 21.75)² = 7.5625
- (24 – 21.75)² = 5.0625
- (20 – 21.75)² = 3.0625
- (22 – 21.75)² = 0.0625
- (26 – 21.75)² = 18.0625
- (21 – 21.75)² = 0.5625
-
Calculate the average of these squared differences:Variance (σ²) = (3.0625 + 0.0625 + 14.0625 + 10.5625 + 0.5625 + 1.5625 + 7.5625 + 5.0625 + 3.0625 + 0.0625 + 18.0625 + 0.5625) / 12 ≈ 5.91
Step 4: Calculate the Standard Deviation
The standard deviation is the square root of the variance.
Standard Deviation (σ) = √5.91 ≈ 2.43
Interpretation:
- Range: The range of 8 indicates that the sales values vary by up to $8,000 over the year.
- Variance: The variance of 5.91 shows that the sales figures are spread out around the mean, with most monthly sales figures within a certain range from the average.
- Standard Deviation: The standard deviation of 2.43 means that, on average, the monthly sales figures deviate from the mean by about $2,430. A smaller standard deviation would indicate more consistency in sales, while a larger one indicates more variability.
In conclusion, the business experiences moderate variability in its monthly sales, with most months’ sales figures being close to the average.
Question:-02
Given the following national income model
Find Y Y and C.
Answer:
To solve for Y Y (national income) and C C (consumption), we can substitute the given equations into each other and solve the resulting equation.
Given Equations:
Y=C+I Y = C + I C=5+(3)/(4)Y C = 5 + \frac{3}{4} Y I=10 I = 10
Step 1: Substitute C C and I I into the first equation
Substitute the values of C C and I I into the equation for Y Y :
Step 2: Simplify the equation
Combine the constant terms:
Step 3: Solve for Y Y
To isolate Y Y , subtract (3)/(4)Y \frac{3}{4} Y from both sides:
This simplifies to:
Multiply both sides by 4 to solve for Y Y :
Step 4: Find C C
Now that we have Y=60 Y = 60 , substitute this value back into the equation for C C :
Calculate the value:
Final Answer:
Y=60 Y = 60 (National Income)C=50 C = 50 (Consumption)
Question:-03
Discuss the various functions related to business and economics.
Answer:
In business and economics, various mathematical functions are used to model and analyze different aspects of economic behavior, business operations, and market trends. Here’s a discussion of some of the key functions:
1. Linear Functions
- Description: A linear function represents a relationship where the change in the dependent variable is proportional to the change in the independent variable.
- Application: In business, linear functions are often used in cost and revenue analysis, where the total cost or revenue is a linear function of the quantity produced or sold. For example,
R(x)=px R(x) = px , whereR(x) R(x) is the revenue andp p is the price per unit, is a linear revenue function.
2. Quadratic Functions
- Description: A quadratic function represents a parabolic relationship where the dependent variable changes at an increasing or decreasing rate as the independent variable changes.
- Application: Quadratic functions are used in profit maximization and cost minimization problems. For instance, the profit function
P(x)=ax^(2)+bx+c P(x) = ax^2 + bx + c can show how profits vary with changes in production levels.
3. Exponential Functions
- Description: Exponential functions model situations where the rate of change of a quantity is proportional to the current amount of the quantity.
- Application: Exponential functions are frequently used in finance to model compound interest, population growth, and the depreciation of assets. For example, the future value of an investment can be modeled by
F(t)=P(1+r)^(t) F(t) = P(1 + r)^t , whereP P is the principal amount,r r is the interest rate, andt t is time.
4. Logarithmic Functions
- Description: The logarithmic function is the inverse of the exponential function and is used to model the time required to reach a certain level of growth.
- Application: Logarithmic functions are used in economics to model learning curves, where the efficiency of production improves over time, and in demand analysis, where the elasticity of demand can be expressed as a logarithmic function.
5. Cobb-Douglas Production Function
- Description: This is a specific form of a production function that shows the relationship between inputs (like labor and capital) and the output produced.
- Application: The Cobb-Douglas function
Q=A*L^( alpha)*K^( beta) Q = A \cdot L^\alpha \cdot K^\beta is used to model the outputQ Q in terms of laborL L and capitalK K , whereA A is a constant, andalpha \alpha andbeta \beta are the output elasticities of labor and capital, respectively.
6. Cost Functions
- Description: Cost functions represent the cost incurred by a business in producing a certain level of output.
- Application: These functions help in understanding how costs change with different levels of output and are used in decision-making related to pricing, production, and budgeting. The typical cost function is
C(Q)=F+VQ C(Q) = F + VQ , whereC(Q) C(Q) is the total cost,F F is fixed costs, andV V is variable costs per unit of outputQ Q .
7. Utility Functions
- Description: Utility functions represent a consumer’s preference ranking over a set of goods and services.
- Application: In economics, utility functions are used to analyze consumer behavior, particularly in understanding how consumers allocate their income among different goods to maximize their utility or satisfaction. A common utility function is
U(x,y)=x^( alpha)*y^( beta) U(x, y) = x^\alpha \cdot y^\beta , wherex x andy y are quantities of goods, andalpha \alpha andbeta \beta are constants representing the consumer’s preference.
8. Demand and Supply Functions
- Description: Demand functions represent the relationship between the quantity demanded of a good and its price, while supply functions represent the relationship between the quantity supplied and its price.
- Application: These functions are fundamental in determining market equilibrium, where the quantity demanded equals the quantity supplied. The demand function can be written as
Q_(d)=a-bP Q_d = a – bP , and the supply function asQ_(s)=c+dP Q_s = c + dP , whereQ_(d) Q_d andQ_(s) Q_s are the quantities demanded and supplied, respectively, andP P is the price.
9. Profit Functions
- Description: Profit functions are used to calculate the difference between total revenue and total cost.
- Application: In business, profit functions help in identifying the level of output that maximizes profit. The profit function is typically
Pi(Q)=R(Q)-C(Q) \Pi(Q) = R(Q) – C(Q) , wherePi(Q) \Pi(Q) is the profit,R(Q) R(Q) is the revenue, andC(Q) C(Q) is the cost as functions of outputQ Q .
Conclusion:
These functions are essential tools in both business and economics, providing a mathematical framework for analyzing and solving real-world problems. Understanding these functions enables businesses to make informed decisions about production, pricing, investment, and market strategies.
Question:-04
What do you mean by maxima or minima of a function? State the meaning of absolute minimum of a function. Explain the steps for finding maxima and minima of a function.
Answer:
Maxima and Minima of a Function
Maxima and minima of a function refer to the highest and lowest points on the graph of the function, respectively. These points can be local (relative) or absolute (global):
-
Local (Relative) Maximum: A point
x=c x = c is a local maximum if the functionf(x) f(x) has a higher value atc c than at any nearby points. In mathematical terms,f(c) f(c) is a local maximum if there exists an interval aroundc c such thatf(c) >= f(x) f(c) \geq f(x) for allx x in that interval. -
Local (Relative) Minimum: A point
x=c x = c is a local minimum if the functionf(x) f(x) has a lower value atc c than at any nearby points. That is,f(c) f(c) is a local minimum if there exists an interval aroundc c such thatf(c) <= f(x) f(c) \leq f(x) for allx x in that interval. -
Absolute (Global) Maximum: The absolute maximum of a function on a given interval is the point at which the function reaches its highest value over the entire interval.
-
Absolute (Global) Minimum: The absolute minimum of a function on a given interval is the point at which the function reaches its lowest value over the entire interval.
Absolute Minimum of a Function
The absolute minimum of a function f(x) f(x) on an interval [a,b] [a, b] is the smallest value that f(x) f(x) takes on the interval. If f(c) f(c) is the absolute minimum, then for all x x in [a,b] [a, b] , f(c) <= f(x) f(c) \leq f(x) .
Steps to Find Maxima and Minima of a Function
To find the maxima and minima of a function, follow these steps:
-
Find the derivative: Compute the first derivative
f^(‘)(x) f'(x) of the functionf(x) f(x) . The derivative gives the slope of the function at any point. -
Set the derivative equal to zero: Solve the equation
f^(‘)(x)=0 f'(x) = 0 to find the critical points. These are the points where the slope of the function is zero, which might correspond to maxima, minima, or points of inflection. -
Second Derivative Test:
- Compute the second derivative
f^(″)(x) f”(x) . - Evaluate
f^(″)(x) f”(x) at the critical points:- If
f^(″)(c) > 0 f”(c) > 0 , thenf(c) f(c) is a local minimum. - If
f^(″)(c) < 0 f”(c) < 0 , thenf(c) f(c) is a local maximum. - If
f^(″)(c)=0 f”(c) = 0 , the test is inconclusive, and you may need to use other methods to determine the nature of the critical point.
- If
- Compute the second derivative
-
Evaluate endpoints (if the function is defined on a closed interval
[a,b] [a, b] ): Check the values of the function at the endpointsf(a) f(a) andf(b) f(b) . The largest of these values, along with the values at the critical points, will give you the absolute maximum, and the smallest will give you the absolute minimum. -
Compare values: The maximum and minimum values of the function on the interval can be determined by comparing the function values at the critical points and the endpoints.
This method helps to identify the points where a function reaches its highest and lowest values, providing insight into the behavior of the function over a given interval.
Question:-05
The manager of a departmental store compiled information on 200 accounts receivable which were delinquent. For each account, he has noted the number of days passed after the due date. He then grouped the data as shown in the following frequency distribution. Determine the median using two methods.
Answer:
To determine the median of the distribution, we can use two methods: the Cumulative Frequency Method and the Interpolation Method. Let’s work through both methods step by step.
1. Cumulative Frequency Method
First, let’s calculate the cumulative frequencies for the data.
The median is the value that divides the distribution into two equal parts. Since there are 200 accounts, the median will be the 100th value (i.e., (200)/(2)=100 \frac{200}{2} = 100 ).
Step 1: Identify the median class:
The cumulative frequency just before 100 is 85, and the next cumulative frequency is 125. Therefore, the median class is the interval 60-74.
Step 2: Apply the median formula:
Where:
L L = Lower boundary of the median class = 60N N = Total number of observations = 200F F = Cumulative frequency of the class preceding the median class = 85f f = Frequency of the median class = 40h h = Class width = 74 – 60 = 14
Now, substitute the values:
2. Interpolation Method
In this method, we interpolate directly within the median class.
We already know that the median class is 60-74. The 100th observation lies within this interval.
We can linearly interpolate between the cumulative frequency of 85 (for the 45-59 class) and 125 (for the 60-74 class).
The formula is:
Substituting the known values:
Which gives us the same result as before:
Conclusion
Both methods yield the same result for the median of the given frequency distribution: 65.25 days.
Section-B
Question:-06
A manufacturer earns Rs. 5500 in the first month and Rs. 7000 in the second month. On plotting these points, the manufacturer observes a linear function may fit the data.
i. Find the quadratic function that fits the data.
ii. Using the model make a prediction of the earning for the fourth week.
ii. Using the model make a prediction of the earning for the fourth week.
Answer:
To address the problem, let’s break it down into two parts:
Part 1: Find the Quadratic Function that Fits the Data
Given the earnings:
- First month (let’s denote it as
x_(1)=1 x_1 = 1 ):y_(1)=5500 y_1 = 5500 - Second month (
x_(2)=2 x_2 = 2 ):y_(2)=7000 y_2 = 7000
Since we are asked to find a quadratic function of the form:
We need to determine the coefficients a a , b b , and c c .
We have two points, but we need a third point to uniquely determine a quadratic function. Typically, without loss of generality, we can assume the third point to be the origin, i.e., x_(0)=0 x_0 = 0 with y_(0)=0 y_0 = 0 , representing zero earnings when there was no activity (before the first month). Therefore:
- Third point (
x_(0)=0 x_0 = 0 ):y_(0)=0 y_0 = 0
Using the three points (0,0) (0, 0) , (1,5500) (1, 5500) , and (2,7000) (2, 7000) , we can set up a system of equations:
0=a(0)^(2)+b(0)+c=>c=0 0 = a(0)^2 + b(0) + c \Rightarrow c = 0 5500=a(1)^(2)+b(1)+0=>a+b=5500 5500 = a(1)^2 + b(1) + 0 \Rightarrow a + b = 5500 7000=a(2)^(2)+b(2)+0=>4a+2b=7000 7000 = a(2)^2 + b(2) + 0 \Rightarrow 4a + 2b = 7000
Now, solve these equations.
From the second equation:
From the third equation:
We can solve these equations simultaneously.
Step 1: Solve for b b in terms of a a :
Step 2: Substitute b b into Equation 2: