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GNDU QUESTION PAPERS 2023
BA/BSc 6
th
SEMESTER
ECONOMICS
(Quantave Methods for Economists)
Time Allowed: 3 Hours Maximum Marks: 100
Note: Aempt Five quesons in all, selecng at least One queson from each secon. The
Fih queson may be aempted from any secon. All quesons carry equal marks.
SECTION – A
1.(a) Give the denions, types and examples of:
(i) Set
(ii) Relaons
(iii) Funcons
(b) Explain the maxima and minima condions for single variable funcon.
(c) Evaluate:
2.(a) Find
if:
(i)
(ii)
(b) Find the extreme values of the funcon:
SECTION – B
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3.(a) Calculate the mean and standard deviaon from the following data:
Value
20–29
30–39
40–49
50–59
Frequency
5
14
24
22
Value
60-69
70-79
80-89
Frequency
16
6
3
(b) Find the missing frequency from the following distribuon of daily sales of shops, given
that the median sale of shops is Rs. 2,400:
Sales (in ₹100)
0–10
10–20
20–30
30–40
40–50
No. of Shops
5
25
?
18
7
SECTION – C
4.(a) The arithmec mean and the standard deviaon of a set of 9 items are 43 and 5
respecvely.
If an item of value 63 is added to the set, nd:
(i) the new mean
(ii) the new standard deviaon of 10 items
(b) Find the Geometric Mean from the following data:
Marks
0–10
10–20
20–30
30–40
40–50
No. of Students
5
7
15
25
8
5.Calculate Karl Pearson’s coecient of skewness from the following data:
Size
1
2
3
4
5
6
7
Frequency
10
18
30
25
12
3
2
6.From the data given below, nd two regression equaons:
X
23
26
33
30
29
34
27
36
32
30
Y
41
44
47
39
34
30
29
28
31
37
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SECTION – D
7.Construct the Fisher’s Ideal Index for the following and show that it sases both the
factor reversal test and the me reversal test:
Commodity
Base Year
Price
Base Year
Quanty
Current Year
Price
Current Year
Quanty
A
6.5
500
10.8
560
B
2.8
124
8.2
148
C
4.7
69
13.4
78
D
10.9
38
10.8
25
E
8.6
49
13.4
20
8.From the following table, interpolate the missing values:
Year
1
2
3
4
5
6
Producon (‘000 tonnes)
200
220
260
?
350
430
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GNDU ANSWER PAPERS 2023
BA/BSc 6
th
SEMESTER
ECONOMICS
(Quantave Methods for Economists)
Time Allowed: 3 Hours Maximum Marks: 100
Note: Aempt Five quesons in all, selecng at least One queson from each secon. The
Fih queson may be aempted from any secon. All quesons carry equal marks.
SECTION – A
1.(a) Give the denions, types and examples of:
(i) Set
(ii) Relaons
(iii) Funcons
(b) Explain the maxima and minima condions for single variable funcon.
(c) Evaluate:
Ans: 1.(a) Definitions, Types, and Examples
Mathematics often studies collections of objects and the relationships between them.
Three very important ideas in this area are:
• Sets
• Relations
• Functions
Think of them as three levels of connection:
A set is just a group of items.
A relation shows how items from sets are connected.
A function is a special kind of relation with a strict rule.
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Let’s understand each in a simple way.
(i) Set
Definition of Set
A set is a well-defined collection of distinct objects.
“Well-defined” means:
We should be able to clearly tell whether an object belongs to the set or not.
For example:
• Students in your class → a set
• Days of the week → a set
• Even numbers less than 10 → a set
We usually write sets inside curly brackets { }.
Example:
• A = {1, 2, 3, 4}
• B = {Monday, Tuesday, Wednesday}
Types of Sets
Let’s understand some common types with easy examples.
Empty Set
A set with no elements.
Example:
• Set of months with 32 days
No such month exists
So the set =
Finite Set
A set with a limited number of elements.
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Example:
• Vowels in English = {a, e, i, o, u}
Infinite Set
A set with unlimited elements.
Example:
• Natural numbers = {1, 2, 3, 4, …}
Equal Sets
Two sets having exactly the same elements.
Example:
• A = {1,2,3}
• B = {3,2,1}
A = B (order doesn’t matter)
Subset
If every element of A is in B, then A is a subset of B.
Example:
• A = {1,2}
• B = {1,2,3,4}
Then A B
Real-Life Example of Set
Think of your phone contact list.
Each contact = element
All contacts together = set
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(ii) Relations
Now imagine two sets:
A = {1,2,3}
B = {4,5,6}
A relation shows how elements of A connect to elements of B.
Definition of Relation
A relation from set A to set B is a set of ordered pairs (a,b), where a A and b B.
Example:
A = {1,2,3}
B = {2,4,6}
Relation: “is double of”
So pairs:
• (1,2)
• (2,4)
• (3,6)
This is a relation from A to B.
Types of Relations
Let’s see the most common ones in simple words.
Reflexive Relation
Every element relates to itself.
Example:
A = {1,2,3}
Relation:
(1,1), (2,2), (3,3)
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Symmetric Relation
If (a,b) exists, then (b,a) also exists.
Example:
If 1 is friend of 2, then 2 is friend of 1.
Transitive Relation
If (a,b) and (b,c) exist, then (a,c) must exist.
Example:
If A is older than B
and B is older than C
Then A is older than C
Equivalence Relation
A relation that is:
✔ Reflexive
✔ Symmetric
✔ Transitive
Example:
“Having same age”
Real-Life Example of Relation
Think of Instagram followers:
• You follow someone → relation
• They follow you → relation
So relationships exist between users (sets).
(iii) Functions
A function is a very special kind of relation.
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Definition of Function
A function is a relation in which each element of set A has exactly one image in set B.
Important rule:
One input → only one output
Example of Function
A = {1,2,3}
B = {2,4,6}
Function: f(x) = 2x
Pairs:
• (1,2)
• (2,4)
• (3,6)
Each input has only one output → valid function ✔
Not a Function Example
If:
(1,2) and (1,3)
Then input 1 has two outputs → not function
Types of Functions
One-One Function
Different inputs → different outputs.
Example:
f(x)=2x
Many-One Function
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Different inputs → same output allowed.
Example:
f(x)=x²
(-2)² = 4
2² = 4
Onto Function
Every element in codomain has a preimage.
Example:
A={1,2,3}
B={2,4,6}
f(x)=2x
All elements in B used ✔
Into Function
Some elements in B unused.
Real-Life Example of Function
Think of:
Student → Roll number
Each student has one roll number
But roll number belongs to only one student
So this is a function.
1.(b) Maxima and Minima of Single Variable Function
Now let’s move to calculus.
Imagine you are climbing a hill.
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• The highest point → maximum
• The lowest valley → minimum
This is exactly what maxima and minima mean.
Definition
For function y = f(x):
• Maximum point → highest value nearby
• Minimum point → lowest value nearby
Condition for Maxima and Minima
The most important rule:
At maxima or minima:
f′(x) = 0
This means slope = 0
Graph is flat (horizontal tangent)
Second Derivative Test
After finding f′(x)=0, check f″(x).
Case 1: f″(x) < 0
Graph bends downward
Maximum
Case 2: f″(x) > 0
Graph bends upward
Minimum
Example
Find maxima/minima of:
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f(x) = x²
Step 1: First derivative
f′(x) = 2x
Set = 0:
2x = 0
x = 0
Step 2: Second derivative
f″(x) = 2
Since 2 > 0
Minimum at x=0
Value:
f(0)=0
So minimum point = (0,0)
Real-Life Example
Think of profit:
• Highest profit → maximum
• Lowest loss → minimum
Businesses use maxima/minima to:
• maximize profit
• minimize cost
1.(c) Evaluate Limit
We need to evaluate:
lim(x→3) (x² − 9)/(x − 3)
Step 1: Direct Substitution
Put x=3:
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Numerator:
3² − 9 = 9 − 9 = 0
Denominator:
3 − 3 = 0
So we get:
0/0
This is indeterminate form.
So we simplify.
Step 2: Factor Numerator
x² − 9 = (x−3)(x+3)
So expression becomes:
(x−3)(x+3)/(x−3)
Cancel (x−3):
= x + 3
Step 3: Take Limit
Now:
lim(x→3) (x+3)
Substitute x=3:
= 3+3
= 6
Final Answer:
lim(x→3) (x² − 9)/(x − 3) = 6
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2.(a) Find
if:
(i)
(ii)
(b) Find the extreme values of the funcon:
Ans: Part (a): Finding
This part is all about differentiation. Think of differentiation as a way of measuring how fast
something changes. If depends on , then
tells us how sensitive is when moves a little
bit.
(i)
Now, square roots can look scary, but they’re just powers in disguise. Remember:
So here:
When differentiating something like
, we use the chain rule. The chain rule is like
peeling an onion: you deal with the outer layer first, then the inner one.
• Outer layer: something
. Its derivative is
something
.
• Inner layer:
. Its derivative is just .
Put them together:
The “2” from the inner derivative cancels the “1/2” from the outer derivative:
And since negative exponents mean reciprocals:
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That’s it! Simple, right? The chain rule is like a relay race: the outer function hands the baton
to the inner one, and together they finish the race.
(ii)
This one looks bigger, but it’s the same idea. Again, chain rule.
• Outer layer:
something
. Its derivative is something
.
• Inner layer:
. Its derivative is .
So:
Multiply:
Done! Notice how the negative sign comes from the inner derivative. If you forget the chain
rule, you’ll miss that “-5,” and the whole answer changes. That’s why I always say: chain rule
is like checking both the outer wrapping and the inside gift.
Part (b): Extreme Values of
Now we move to the second part. This is about extreme values—basically, the peaks and
valleys of a curve. Imagine you’re hiking in the mountains: the highest points are maxima,
the lowest points are minima. In math, we find these by looking at where the slope
(derivative) becomes zero.
Step 1: Differentiate
Derivative:
This is the slope of the curve at any point.
Step 2: Find critical points
Critical points happen when slope = 0:
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This is a quadratic equation. Solve it:
So:
•
•
These are the critical points.
Step 3: Test for maxima or minima
We use the second derivative test. The second derivative tells us whether the curve is
bending upwards (like a bowl) or downwards (like a hill).
Now plug in the critical points:
• At :
Positive means the curve bends upwards → minimum.
• At
:
Negative means the curve bends downwards → maximum.
Step 4: Find the actual extreme values
We plug these -values back into the original function:
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• At :
So minimum value is at .
• At
:
Find common denominator (27):
So maximum value is
at
.
Why This Matters
You might wonder: “Okay, but why do I care about maxima and minima?” Well, in real life:
• Businesses use maxima/minima to maximize profit or minimize cost.
• Engineers use them to design safe structures (finding stress points).
• Even artists use them—curves in design often rely on smooth peaks and valleys.
So this isn’t just abstract math. It’s a universal tool for problem-solving.
Final Thought
Math often feels intimidating because we see symbols without the story. But once you see
differentiation as measuring change, and maxima/minima as finding highs and lows, the
whole picture becomes clear. It’s not about memorizing formulas—it’s about understanding
the journey of a curve.
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SECTION – B
3.(a) Calculate the mean and standard deviaon from the following data:
Value
20–29
30–39
40–49
50–59
Frequency
5
14
24
22
Value
60-69
70-79
80-89
Frequency
16
6
3
Ans:
Value (Class Interval)
Frequency
20–29
5
30–39
14
40–49
24
50–59
22
60–69
16
70–79
6
80–89
3
This means:
• 5 values lie between 20–29
• 14 values lie between 30–39
• 24 values lie between 40–49
• and so on…
So instead of raw numbers, we have grouped data.
To find mean and standard deviation from grouped data, we follow a systematic method.
Step 1: Find the Midpoint (Class Mark)
Since each class has a range (like 20–29), we take its middle value:
Midpoint
Lower limit + Upper limit
Let’s calculate:
Class
Midpoint (x)
20–29 → (20+29)/2 = 24.5
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30–39 → 34.5
40–49 → 44.5
50–59 → 54.5
60–69 → 64.5
70–79 → 74.5
80–89 → 84.5
Step 2: Make the Calculation Table
To find mean and standard deviation, we create a table with columns:
• Frequency (f)
• Midpoint (x)
• fx
• x²
• fx²
Let’s build it step by step.
Class
f
x
fx
x²
fx²
20–29
5
24.5
122.5
600.25
3001.25
30–39
14
34.5
483
1190.25
16663.5
40–49
24
44.5
1068
1980.25
47526
50–59
22
54.5
1199
2970.25
65345.5
60–69
16
64.5
1032
4160.25
66564
70–79
6
74.5
447
5550.25
33301.5
80–89
3
84.5
253.5
7140.25
21420.75
Step 3: Find Totals
Now add columns:
Step 4: Calculate Mean
Formula for grouped data mean:
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✔ Mean ≈ 51.17
What Does This Mean?
It means the average value of all observations lies around 51.
Since most frequencies are around 40–59, this makes sense.
Step 5: Calculate Standard Deviation
Formula for grouped data:
Now substitute values:
Mean square:
Now:
✔ Standard Deviation ≈ 14.21
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Final Answers
Mean = 51.17
Standard Deviation = 14.21
Simple Interpretation (Very Important for Exams)
Now imagine this like marks of students:
• Average marks ≈ 51
• Standard deviation ≈ 14
This tells us:
Most values lie between
So roughly between:
and
Which matches the data — most frequencies are in 40–69 range.
So the distribution is fairly spread around the mean.
Concept Understanding in Simple Words
Think of mean as:
“center point of data”
Think of standard deviation as:
“how far values spread from center”
Small SD → data close together
Large SD → data spread out
Here SD ≈ 14 → moderate spread.
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Easy Memory Trick for Exams
For grouped data:
Find midpoint
Multiply with frequency → fx
Mean = Σfx / Σf
SD formula:
Final
Mean:
Standard Deviation:
Conclusion
So, after carefully analyzing the grouped data, we found that the average value is about 51,
meaning the central tendency of the dataset lies around this point. The standard deviation
of about 14 shows that the values are neither too tightly packed nor too widely scattered —
they have a moderate spread around the mean. This matches the frequency distribution,
where most observations fall between 40 and 69.
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(b) Find the missing frequency from the following distribuon of daily sales of shops, given
that the median sale of shops is Rs. 2,400:
Sales (in ₹100)
0–10
10–20
20–30
30–40
40–50
No. of Shops
5
25
?
18
7
Ans: We’re given a frequency distribution of daily sales of shops. The sales are grouped into
intervals (like 0–10, 10–20, etc.), and for each interval we know how many shops fall into it.
But one frequency (the number of shops in the 20–30 interval) is missing. Our job is to find
that missing number, given that the median sale is ₹2,400.
Here’s the table again:
Sales (in ₹100)
0–10
10–20
20–30
30–40
40–50
No. of Shops
5
25
?
18
7
Step 1: Understanding the Median
The median is the middle value when all data is arranged in order. In grouped data, we
don’t list every single shop’s sales—we only know how many shops fall into each interval. So
we use a formula to estimate the median.
The formula for the median in a frequency distribution is:
Median
Where:
• = lower boundary of the median class
• = total frequency (total number of shops)
• = cumulative frequency before the median class
• = frequency of the median class
• = class width (size of the interval)
Step 2: Identifying the Median Class
We’re told the median sale is ₹2,400. Notice that sales are given in “₹100 units.” That
means:
• 0–10 means ₹0–₹1,000
• 10–20 means ₹1,000–₹2,000
• 20–30 means ₹2,000–₹3,000
• 30–40 means ₹3,000–₹4,000
• 40–50 means ₹4,000–₹5,000
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So ₹2,400 lies in the 20–30 interval. That’s our median class.
Step 3: Setting Up the Formula
Let’s assign values:
• (lower boundary of 20–30 interval, in rupees)
• (class width, since 20–30 means ₹2000–₹3000)
• (frequency of median class, the missing number we need to find)
• (total number of shops)
• (cumulative frequency before the median class)
Median is given as 2400. Plugging into the formula:
Step 4: Simplify the Equation
Subtract 2000 from both sides:
Divide both sides by 1000:
Step 5: Express in Terms of
Remember, . So:
Simplify numerator:
So:
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Step 6: Solve for
Simplify:
Multiply both sides by :
Bring terms together:
Step 7: Final Answer
The missing frequency is 25.
So the completed table looks like this:
Sales (in ₹100)
0–10
10–20
20–30
30–40
40–50
No. of Shops
5
25
25
18
7
Making It Relatable
Let’s pause and think about what we just did in everyday language.
Imagine you’re looking at the daily sales of shops in a marketplace. Some shops earn less
than ₹1,000, some between ₹1,000–₹2,000, and so on. We know how many shops fall into
each range, except one group (₹2,000–₹3,000). But we’re told the median sale—the middle
shop in the lineup—earns ₹2,400. That’s inside the ₹2,000–₹3,000 group. So we use the
median formula to figure out how many shops must be in that group to make ₹2,400 the
middle point.
It’s like solving a puzzle: the median acts as a clue, and the formula helps us balance the
distribution so that the middle shop lands exactly at ₹2,400.
Why This Matters
This isn’t just a classroom exercise. In real life:
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• Businesses use median sales to understand typical performance.
• If you only look at averages, one super-rich shop could distort the picture. The
median gives a fairer sense of what’s “normal.”
• By filling in missing data, analysts can reconstruct patterns and make better
decisions.
So this problem is really about data interpretation—using clues to complete the bigger
picture.
Final Thought
Math problems like this are less about crunching numbers and more about storytelling.
You’re piecing together a narrative: shops, sales, intervals, and the median acting as the
“middle character.” Once you see it that way, the formula feels less like a burden and more
like a tool to uncover the hidden frequency.
SECTION – C
4.(a) The arithmec mean and the standard deviaon of a set of 9 items are 43 and 5
respecvely.
If an item of value 63 is added to the set, nd:
(i) the new mean
(ii) the new standard deviaon of 10 items
Ans: Step 1 — Recall Important Formulas
Before solving, remember two key formulas:
Mean formula
Mean
Sum of observations
Number of observations
So,
Sum Mean
Standard Deviation formula (shortcut form)
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Where
• = SD
• = number of items
• = mean
•
= sum of squares of items
We will use this formula to reconstruct the data information.
Step 2 — Find the Sum of 9 Items
Given:
Mean = 43
Number = 9
So,
Sum of 9 items
So originally, total of all numbers = 387
Step 3 — Add the New Item
New item = 63
So new total sum:
Now number of items = 10
Step 4 — New Mean
New mean
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New mean = 45
So adding a large value (63) increased the average from 43 → 45.
This makes sense: a higher number pulls the average upward.
Step 5 — Find Sum of Squares of 9 Items
We use SD formula:
Given:
SD = 5 → variance =
Mean = 43
n = 9
So,
Now calculate:
So,
Add 1849 to both sides:
Multiply by 9:
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So sum of squares of 9 items = 16866
Step 6 — Add Square of New Item
New item = 63
New sum of squares:
So for 10 items:
Step 7 — New Standard Deviation
Use formula:
We know:
n = 10
mean = 45
sum of squares = 20835
So,
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New SD ≈ 7.65
Final Answers
(i) New Mean = 45
(ii) New Standard Deviation ≈ 7.65
Final Conclusion
Adding a large value to a dataset does two things:
• Pulls the mean toward itself
• Increases spread if it lies far from average
In this problem:
• Mean rose from 43 → 45
• SD rose from 5 → 7.65
So the dataset became both higher and more spread out.
(b) Find the Geometric Mean from the following data:
Marks
0–10
10–20
20–30
30–40
40–50
No. of Students
5
7
15
25
8
Ans: Step 1: Recall the Formula
For grouped data, the formula for the Geometric Mean is:
GM antilog
Where:
• = frequency (number of students in each interval)
• = mid-point of each class interval
• = total frequency (total number of students)
So the plan is:
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1. Find mid-points of each class.
2. Take logarithms of those mid-points.
3. Multiply each log by its frequency.
4. Add them all up.
5. Divide by total frequency.
6. Take the antilog to get the GM.
Step 2: Find Mid-Points
For each interval, the mid-point is simply the average of the lower and upper limits.
• 0–10 → midpoint = 5
• 10–20 → midpoint = 15
• 20–30 → midpoint = 25
• 30–40 → midpoint = 35
• 40–50 → midpoint = 45
So our midpoints are: 5, 15, 25, 35, 45.
Step 3: Build the Working Table
Let’s construct a table with all the necessary values:
Marks
Midpoint (x)
Frequency (f)
log(x) (approx)
f·log(x)
0–10
5
5
0.6990
3.495
10–20
15
7
1.1761
8.2327
20–30
25
15
1.3979
20.9685
30–40
35
25
1.5441
38.6025
40–50
45
8
1.6532
13.2256
Step 4: Add Up Frequencies and f·log(x)
• Total frequency
• Sum of
Step 5: Apply the Formula
Now take the antilog:
GM antilog
Since antilog means
:
GM
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Step 6: Final Answer
The Geometric Mean of the marks is approximately:
GM
5.Calculate Karl Pearson’s coecient of skewness from the following data:
Size
1
2
3
4
5
6
7
Frequency
10
18
30
25
12
3
2
Ans: Given Data
Size (x)
1
2
3
4
5
6
7
Frequency (f)
10
18
30
25
12
3
2
This is a discrete frequency distribution.
Formula to Use
Karl Pearson’s coefficient of skewness:
Sk
Mean Mode
Standard Deviation
So we need to calculate:
1. Mean
2. Mode
3. Standard deviation
Let’s do each slowly and clearly.
Step 1: Find Mean
Formula:
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Create a table:
x
f
fx
1
10
10
2
18
36
3
30
90
4
25
100
5
12
60
6
3
18
7
2
14
Now sum:
So:
Mean = 3.28
Step 2: Find Mode
Mode = value with highest frequency.
Look at frequencies:
10, 18, 30, 25, 12, 3, 2
Highest frequency = 30 at x = 3
Mode = 3
Step 3: Find Standard Deviation
Formula:
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We use mean = 3.28
Create calculation table:
x
f
x−3.28
(x−3.28)²
f(x−3.28)²
1
10
−2.28
5.1984
51.984
2
18
−1.28
1.6384
29.491
3
30
−0.28
0.0784
2.352
4
25
0.72
0.5184
12.96
5
12
1.72
2.9584
35.501
6
3
2.72
7.3984
22.195
7
2
3.72
13.8384
27.677
Now sum last column:
(approx)
Now divide by total frequency 100:
(approx)
Standard deviation = 1.35
Step 4: Karl Pearson’s Skewness
Formula:
Sk
Mean Mode
Substitute values:
Sk
(approx)
Final Answer
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Karl Pearson’s coefficient of skewness
Interpretation (Very Important in Exams)
Since skewness is positive:
Distribution is positively skewed
Tail is toward larger values
Mean > Mode
So the data slightly leans toward higher sizes (4–7 side).
Intuitive Understanding
Imagine 100 students choosing shoe sizes from 1 to 7.
Most students chose size 3 and 4.
Few chose larger sizes like 6 and 7.
Those few large values pull the mean slightly upward from 3 → 3.28.
6.From the data given below, nd two regression equaons:
X
23
26
33
30
29
34
27
36
32
30
Y
41
44
47
39
34
30
29
28
31
37
Ans: Step 1: Understanding the Task
Regression equations are basically lines that describe the relationship between two
variables. Here, we have (say marks or scores) and (say performance or results). We want
two equations:
1. Regression of on :
2. Regression of on :
Both equations show how one variable can be predicted from the other.
Step 2: Organize the Data
We’ll need sums of , ,
,
, and . Let’s compute them.
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X
Y
X²
Y²
XY
23
41
529
1681
943
26
44
676
1936
1144
33
47
1089
2209
1551
30
39
900
1521
1170
29
34
841
1156
986
34
30
1156
900
1020
27
29
729
841
783
36
28
1296
784
1008
32
31
1024
961
992
30
37
900
1369
1110
Now add them up:
•
•
•
•
•
Step 3: Means of X and Y
So the average values are , .
Step 4: Regression Coefficient
Formula:
Plugging values:
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Step 5: Regression Equation of Y on X
Equation form:
Simplify:
So the regression of on is:
Step 6: Regression Coefficient
Formula:
Step 7: Regression Equation of X on Y
Equation form:
Simplify:
So the regression of on is:
Step 8: Final Results
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The two regression equations are:
1. Regression of on :
2. Regression of on :
SECTION – D
7.Construct the Fisher’s Ideal Index for the following and show that it sases both the
factor reversal test and the me reversal test:
Commodity
Base Year
Price
Base Year
Quanty
Current Year
Price
Current Year
Quanty
A
6.5
500
10.8
560
B
2.8
124
8.2
148
C
4.7
69
13.4
78
D
10.9
38
10.8
25
E
8.6
49
13.4
20
Ans: What is Fisher’s Ideal Index?
When economists compare prices between two years, they want a measure that is fair and
balanced.
Two famous price indices exist:
• Laspeyres Price Index (L) → uses base-year quantities
• Paasche Price Index (P) → uses current-year quantities
But each has bias. So Irving Fisher suggested:
Take the geometric mean of Laspeyres and Paasche
Fisher Price Index
Because it combines both years, it is called the Ideal Index.
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Step 1: Write the Data Clearly
Commodity
p₀
q₀
p₁
q₁
A
6.5
500
10.8
560
B
2.8
124
8.2
148
C
4.7
69
13.4
78
D
10.9
38
10.8
25
E
8.6
49
13.4
20
Where
• p₀ = base price
• q₀ = base quantity
• p₁ = current price
• q₁ = current quantity
Step 2: Compute Required Columns
We need:
Calculations
A
• p₀q₀ = 6.5×500 = 3250
• p₁q₀ = 10.8×500 = 5400
• p₀q₁ = 6.5×560 = 3640
• p₁q₁ = 10.8×560 = 6048
B
• p₀q₀ = 2.8×124 = 347.2
• p₁q₀ = 8.2×124 = 1016.8
• p₀q₁ = 2.8×148 = 414.4
• p₁q₁ = 8.2×148 = 1213.6
C
• p₀q₀ = 4.7×69 = 324.3
• p₁q₀ = 13.4×69 = 924.6
• p₀q₁ = 4.7×78 = 366.6
• p₁q₁ = 13.4×78 = 1045.2
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D
• p₀q₀ = 10.9×38 = 414.2
• p₁q₀ = 10.8×38 = 410.4
• p₀q₁ = 10.9×25 = 272.5
• p₁q₁ = 10.8×25 = 270
E
• p₀q₀ = 8.6×49 = 421.4
• p₁q₀ = 13.4×49 = 656.6
• p₀q₁ = 8.6×20 = 172
• p₁q₁ = 13.4×20 = 268
Step 3: Take Totals
Step 4: Calculate Laspeyres and Paasche
Laspeyres Price Index
Paasche Price Index
Step 5: Fisher’s Ideal Index
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Fisher’s Ideal Price Index = 179.24
Meaning:
Prices increased about 79.24% from base year to current year.
Step 6: Time Reversal Test
A good index should give reciprocal when time is reversed:
Fisher satisfies this by theory, because:
Multiplying:
✔ Hence Time Reversal Test satisfied
Step 7: Factor Reversal Test
This test says:
Price Index Quantity Index Value Index
Value Index:
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Now Fisher Quantity Index:
Check Factor Reversal
✔ Factor Reversal Test satisfied
Final Answer
Fisher’s Ideal Price Index = 179.24
Since:
•
→ Time Reversal Test satisfied
•
→ Factor Reversal Test satisfied
Therefore, Fisher’s Ideal Index satisfies both tests and is called an Ideal Index.
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8.From the following table, interpolate the missing values:
Year
1
2
3
4
5
6
Producon (‘000 tonnes)
200
220
260
?
350
430
Ans: The Data
Year
1
2
3
4
5
6
Production (‘000 tonnes)
200
220
260
?
350
430
We need to find the missing production for Year 4.
Step 1: Understanding Interpolation
Interpolation is a method of estimating unknown values that fall within a given range of
data. Think of it like filling in a missing puzzle piece: we know the values before and after, so
we can guess the missing one by following the trend.
Here, production is steadily increasing over the years. We’ll use Newton’s Forward
Interpolation Formula because the data is sequential (Year 1, 2, 3, …).
Step 2: Newton’s Forward Formula
The formula is:
Where:
•
= first value (at Year 1)
•
• = the year we want (Year 4)
•
= starting year (Year 1)
• = interval between years (here, 1 year)
•
= differences
Step 3: Construct the Difference Table
Let’s build the difference table step by step.
Year
Production (y)
Δy
Δ²y
Δ³y
Δ⁴y
1
200
20
20
-10
70
2
220
40
10
60
3
260
?
?
4
?
?
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5
350
80
6
430
Now, let’s carefully compute:
• From Year 1 to Year 2: Δy = 220 - 200 = 20
• From Year 2 to Year 3: Δy = 260 - 220 = 40
• From Year 3 to Year 5 (since Year 4 is missing, we’ll use later values): Δy = 350 - 260 =
90
• From Year 5 to Year 6: Δy = 430 - 350 = 80
So the first differences are: 20, 40, 90, 80.
Now second differences:
• 40 - 20 = 20
• 90 - 40 = 50
• 80 - 90 = -10
So second differences: 20, 50, -10.
Third differences:
• 50 - 20 = 30
• -10 - 50 = -60
So third differences: 30, -60.
Fourth difference:
• -60 - 30 = -90
Step 4: Apply Formula
We want Year 4. So:
• ,
,
•
•
Now substitute:
Simplify step by step:
• First term: 200
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• Second term: 3 × 20 = 60
• Third term: (3 × 2 / 2) × 20 = 3 × 20 = 60
• Fourth term: (3 × 2 × 1 / 6) × 30 = 1 × 30 = 30
• Fifth term vanishes (because of ×0).
Add them up:
Step 5: Final Answer
The missing production for Year 4 is:
Production ‘
“This paper has been carefully prepared for educaonal purposes. If you noce any
mistakes or have suggesons, feel free to share your feedback.”
