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GNDU Question Paper-2024
BBA 3
rd
Semester
STATISTICS FOR BUSINESS
Time Allowed: Three Hours Max. Marks: 100
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks
SECTION-A
1.(a) Solve the equation for x, y, z and t if:
(b) if A =
C=
2. (a) If A=
-1
if it exists.
(b) Solve the following system of linear equations by matrix method:
2x + 3y + 3z = 5
x - 2y + z = - 4
3x - y - 2z = 3
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SECTION-B
(a) Calculate Arithmetic Mean from the following data:
Class Interval
Number of students
10-25
2
25-40
3
40-55
7
55-70
6
70-85
6
85-100
6
(b) Calculate Median from the following data:
Class interval
Frequency
135-140
4
140-145
7
145-150
18
150-155
11
155-160
6
160-165
5
4. Find Variance and Standard Deviation from the following data:
Class
Frequency
30-40
3
40-50
7
50-60
12
60-70
15
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70-80
8
80-90
3
90-100
2
SECTION-C
5.(a) Fit a straight line to the following data:
X
1
2
3
4
6
8
y
2.4
3
3.6
4
5
8
(b) What are important properties of Correlation Coefficient?
6.(a) Calculate Coefficient of Rank Correlation from the following data:
(b) Discuss Time Series and its various components.
Marks in Maths X
Marks in Science Y
20
80
30
70
40
40
50
50
60
10
70
20
80
30
SECTION-D
7. (a) A die is thrown twice. Write down all possible outcomes. What is the probability
that 5 will come up at least once?
3 (b) If a sample space S = AB, P(A) =
P(B)=
find P(AB).
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8.(a) Explain Poisson distribution and its important properties.
(b) What is Normal distribution? Give its important properties.
GNDU Answer Paper-2024
BBA 3
rd
Semester
STATISTICS FOR BUSINESS
Time Allowed: Three Hours Max. Marks: 100
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks
SECTION-A
1.(a) Solve the equation for x, y, z and t if:
(b) if A =
C=
Ans: Understanding the Problem
Imagine you are a detective in the world of mathematics. You’re faced with a mysterious
case involving matrices—yes, those rectangular arrays of numbers that can do so much
more than just sit on your notebook page. Today, your mission is to solve for unknowns in a
matrix equation and also perform matrix multiplication in a clever way.
We have two parts in our investigation:
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Part (a): Solving the Matrix Equation
Think of a matrix equation as a puzzle. Each entry in the matrix is a piece of that puzzle. To
solve it, we use the simple rule:
Two matrices are equal if and only if their corresponding entries are equal.
This means we can compare each position in the matrices like detectives comparing
fingerprints.
Step 1: Expand the left-hand side
The left-hand side of our equation is:
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2. (a) If A=
-1
if it exists.
Ans: Part (b): Finding A(BC)
Now, the second part is a matrix multiplication story. Remember: matrix multiplication is
like combining two groups of workers where the rows of the first team interact with the
columns of the second team to produce results.
We are asked to find:
Step 1: Multiply B and C first
Matrix multiplication rule:
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(b) Solve the following system of linear equations by matrix method:
2x + 3y + 3z = 5
x - 2y + z = - 4
3x - y - 2z = 3
Ans:
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SECTION-B
3. (a) Calculate Arithmetic Mean from the following data:
Class Interval
Number of students
10-25
2
25-40
3
40-55
7
55-70
6
70-85
6
85-100
6
(b) Calculate Median from the following data:
Class interval
Frequency
135-140
4
140-145
7
145-150
18
150-155
11
155-160
6
160-165
5
Ans: Understanding the Magic Behind Arithmetic Mean and Median
Imagine you are a teacher in a school. Every day, you see students bustling into your
classroom, each carrying their own stories, experiences, and of course, marks in the last
test. You want to get a sense of how your class performed overall. You could look at each
score individually, but it would take ages and be confusing. What if there was a way to
summarize all that data into a single number that tells you the “average” performance of
your students? That’s where Arithmetic Mean comes in.
And what if you wanted to know the “middle” student—the one whose performance splits
the class into two equal halves? That’s when Median becomes your friend. Let’s explore
both using the data you provided.
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Part (a): Calculating the Arithmetic Mean
Step 1: Understanding the Data
We are given a table showing class intervals and the number of students in each interval.
Here’s the data:
Class Interval
Number of Students (Frequency, f)
10-25
2
25-40
3
40-55
7
55-70
6
70-85
6
85-100
6
So, in total, we have 2 + 3 + 7 + 6 + 6 + 6 = 30 students.
Step 2: Finding the Midpoints
For the Arithmetic Mean of grouped data, we first calculate the midpoint (x) of each class
interval. The midpoint is the average of the lower and upper limits of the class:
Lower limit + Upper limit
Let’s calculate it:
• 10–25 →
• 25–40 →
• 40–55 →
• 55–70 →
• 70–85 →
• 85–100 →
Now our table looks like this:
Class Interval
Frequency (f)
Midpoint (x)
10-25
2
17.5
25-40
3
32.5
40-55
7
47.5
55-70
6
62.5
70-85
6
77.5
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85-100
6
92.5
Step 3: Multiply Frequency and Midpoint
To find the weighted contribution of each class, multiply frequency (f) by midpoint (x):
Class Interval
f
x
f·x
10-25
2
17.5
35
25-40
3
32.5
97.5
40-55
7
47.5
332.5
55-70
6
62.5
375
70-85
6
77.5
465
85-100
6
92.5
555
Now, sum up all the f·x values:
Step 4: Apply the Formula
The Arithmetic Mean (AM) formula for grouped data is:
AM
AM
So, the average marks of the students is 62.
This tells us that, in a sense, if all students had the same marks, each would have 62, which
represents the overall performance.
Visualizing the Arithmetic Mean
Think of it like balancing a seesaw. If each student’s score is a weight on the seesaw at their
midpoint, the arithmetic mean is the point where the seesaw balances perfectly. In our
case, 62 is the balancing point—right in the middle of the data’s "weight".
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Part (b): Calculating the Median
Now, let’s find the Median, which is the middle value dividing the data into two equal
halves.
We are given another set of data:
Class Interval
Frequency (f)
135-140
4
140-145
7
145-150
18
150-155
11
155-160
6
160-165
5
Step 1: Calculate Cumulative Frequency (CF)
To locate the middle, we first calculate Cumulative Frequency (CF):
Class Interval
f
CF
135-140
4
4
140-145
7
11
145-150
18
29
150-155
11
40
155-160
6
46
160-165
5
51
The total frequency .
Step 2: Find the Median Class
Median is located using the formula:
Median Class Class interval containing
observation
Look at the cumulative frequency column:
• CF before 145-150 = 11
• CF for 145-150 = 29
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So, the 25.5th observation lies in 145-150, making it the median class.
Step 3: Apply the Median Formula
Median
before
median
Where:
• → lower limit of median class
•
before
→ cumulative frequency before median class
•
median
→ frequency of median class
• → class width
Median
Step by step:
So, the Median = 149.03, meaning the middle student scored approximately 149 marks.
Visualizing the Median
Think of standing in a line of 51 students sorted by marks. The 26th student (middle
student) marks the median. The median doesn’t care about extreme highs or lows; it just
tells us the center.
If we draw a histogram, we can also see it:
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The median class (145–150) is the tallest bar, showing that most students scored around
this range.
Key Insights and Humanized Understanding
1. Arithmetic Mean gives the “overall average,” which balances all marks. In our first
example, 62 marks tell us the typical performance of the class.
2. Median gives the “middle point,” unaffected by extreme scores. In the second
example, 149 marks show the central tendency even if some students scored very
low or very high.
3. Both measures complement each other: Mean is sensitive to all values, while
Median gives a fair picture when data is skewed.
4. The process might seem technical, but it’s like telling the story of your students:
o AM = “If all students had equal marks, this is where we would be.”
o Median = “The student in the middle, whose score splits the class in half.”
Conclusion
In simple words, Arithmetic Mean and Median are like two guides helping a teacher
understand student performance. The mean is a balancing point; the median is the middle
mark. By calculating them carefully using the formulas for grouped data, we can summarize
and interpret huge sets of data in just a few numbers.
From our examples:
• Arithmetic Mean = 62 (Part a)
• Median = 149.03 (Part b)
Both are powerful tools in statistics, helping make sense of numbers, understand trends,
and even make decisions in education, business, and research.
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4. Find Variance and Standard Deviation from the following data:
Class
Frequency
30-40
3
40-50
7
50-60
12
60-70
15
70-80
8
80-90
3
90-100
2
Ans: Understanding Variance and Standard Deviation Through a Story
Imagine a classroom filled with students who have just taken a mathematics test. The
teacher, Mr. Sharma, is curious—not just about how many students scored in certain
ranges, but also about how the scores are spread out. Are most students scoring close to
the average, or are the marks scattered all over? To answer this, Mr. Sharma wants to
calculate two important statistical measures: Variance and Standard Deviation.
He looks at the scores grouped in intervals, something he finds easier than listing every
single mark:
Class Interval
Frequency
30–40
3
40–50
7
50–60
12
60–70
15
70–80
8
80–90
3
90–100
2
At first glance, this seems like a simple table of numbers, but each frequency tells a story
about the students. For example, only 3 students scored between 30–40, while 15 students
scored between 60–70. Clearly, most students are around the middle. But “around the
middle” is vague. Mr. Sharma wants to measure exactly how the scores are spread. This is
where variance and standard deviation come into play.
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Step 1: Understanding the Concept
Before crunching numbers, Mr. Sharma explains the idea to the students:
• Mean (Average): Think of it as the balance point of all the scores. If we imagine the
scores as weights on a number line, the mean is where the seesaw would perfectly
balance.
• Variance: This tells us, on average, how far each score is from the mean. A low
variance means most students scored close to the average. A high variance means
scores are widely spread out.
• Standard Deviation: This is just the square root of variance. While variance gives us
the spread in squared units (like square marks), standard deviation brings it back to
the original units of marks, which is easier to interpret.
Step 2: Calculating the Midpoints
Since the scores are given in intervals, Mr. Sharma decides to use midpoints of each class as
representative scores.
• The midpoint is simply the average of the lower and upper limits of each class.
• This makes sense because, for calculations, we assume all students in a class scored
around the midpoint.
Here’s how he calculates them:
Class Interval
Midpoint (x)
Frequency (f)
30–40
35
3
40–50
45
7
50–60
55
12
60–70
65
15
70–80
75
8
80–90
85
3
90–100
95
2
Now the table is ready for the calculations. Think of these midpoints as the “representative
marks” of each group.
Step 3: Calculating the Mean
The formula for the mean (average) of grouped data is:
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Where:
• = frequency of each class
• = midpoint of each class
First, Mr. Sharma multiplies frequency and midpoint for each class (f × x):
Class
Midpoint (x)
Frequency (f)
f × x
30–40
35
3
105
40–50
45
7
315
50–60
55
12
660
60–70
65
15
975
70–80
75
8
600
80–90
85
3
255
90–100
95
2
190
Now, he adds all f × x values:
The total frequency (sum of f) is:
So, the mean is:
The average score of the class is 62 marks. Mr. Sharma notes that the average falls in the
60–70 range, which matches our earlier observation that most students scored around this
middle interval.
Step 4: Calculating the Variance
Variance measures how much the scores deviate from the mean. The formula for variance
in grouped data is:
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1. Find the deviation:
2. Square the deviation:
3. Multiply by frequency:
Let’s go step by step.
Class
Midpoint (x)
f
x - mean
(x - mean)²
f × (x - mean)²
30–40
35
3
35 - 62 = -27
729
2187
40–50
45
7
45 - 62 = -17
289
2023
50–60
55
12
55 - 62 = -7
49
588
60–70
65
15
65 - 62 = 3
9
135
70–80
75
8
75 - 62 = 13
169
1352
80–90
85
3
85 - 62 = 23
529
1587
90–100
95
2
95 - 62 = 33
1089
2178
Next, sum all
values:
Finally, divide by total frequency (50):
So, the variance is 201.
Mr. Sharma smiles and explains: “This means that the squared deviations of scores from the
mean add up to 201 per student, on average.” Squared units are tricky, which is why we use
standard deviation.
Step 5: Calculating the Standard Deviation
The standard deviation is simply the square root of the variance:
This tells us that, on average, the students’ scores deviate about 14 marks from the mean of
62. So, most scores lie roughly between 62 - 14 = 48 and 62 + 14 = 76.
This aligns well with our observation from the frequency table—the largest number of
students scored in the 50–70 range.
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Step 6: Interpreting the Results
1. Variance = 201: The squared deviation from the mean gives us a sense of the spread
of the data. A variance of 201 indicates a moderate spread; scores are neither tightly
packed nor extremely scattered.
2. Standard Deviation ≈ 14.18: In simple terms, most students scored within 14 marks
of the mean. This is easier to visualize and interpret because it uses the same unit as
the scores.
3. Data Visualization Idea: Imagine a bell-shaped curve centered around 62, where
most students are near the mean, and fewer students are at the extremes. This
picture perfectly represents the distribution and confirms our calculations.
Optional Diagram for Revision
To make it even more intuitive, you can draw a simple histogram:
• X-axis: Class intervals (30–40, 40–50, …)
• Y-axis: Frequency
• Draw bars for each interval based on frequency.
• Draw a vertical line at the mean (62) and shade ±1 standard deviation (48–76) to
show where most scores fall.
This visual helps anyone quickly understand the spread and central tendency of the data.
Step 7: Why This Matters
Mr. Sharma tells the students a simple moral:
“Variance and standard deviation aren’t just numbers. They tell us the story of the data—
how consistent it is, how much it spreads, and where most students lie. In exams, business,
research, or even sports, knowing the spread is often as important as knowing the average.”
For instance:
• If the standard deviation were very low, it would mean almost all students scored
around 62.
• If it were very high, scores were very unpredictable.
This is why statisticians, engineers, and scientists rely on these measures to make decisions,
detect errors, or understand patterns.
Conclusion
To summarize:
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1. Mean = 62 marks
2. Variance = 201
3. Standard Deviation ≈ 14.18
By using midpoints, multiplying with frequencies, calculating deviations, and taking
averages, we could measure not just the “central tendency” but also the spread of marks in
a class. This approach turns a simple table into a story, showing how scores are distributed
and helping teachers, researchers, and students make sense of the data.
By combining arithmetic with visualization, we can easily see the patterns, predict
outcomes, and even compare one set of scores with another in a meaningful way.
Thus, variance and standard deviation do more than measure numbers—they tell the story
hidden in the data.
SECTION-C
5.(a) Fit a straight line to the following data:
X
1
2
3
4
6
8
y
2.4
3
3.6
4
5
8
(b) What are important properties of Correlation Coefficient?
Ans: Understanding the Relationship Between X and Y: A Story of Data
Imagine you are a scientist trying to understand how something changes with time—or
maybe you are a shopkeeper trying to see how sales change with the number of
advertisements. You have some observations, but they are scattered—sometimes high,
sometimes low. You wonder: Can I find a pattern? This is exactly what statistics helps us do.
And today, we are going to explore two powerful tools: fitting a straight line (linear
regression) and measuring the strength of the relationship with the correlation coefficient.
We are given the following data:
X
1
2
3
4
6
8
Y
2.4
3
3.6
4
5
8
We want to fit a straight line to this data and also understand the correlation between X
and Y.
(a) Fitting a Straight Line (Linear Regression)
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When we try to fit a straight line, we assume the relationship between X and Y can be
expressed as:
Where:
• Y is the dependent variable (what we want to predict, in this case y).
• X is the independent variable (what we know, in this case x).
• a is the intercept of the line (where the line crosses the Y-axis).
• b is the slope of the line (how steep the line is).
Step 1: Find the slope (b)
The formula for the slope is:
To do this, first calculate:
1. The mean of X (
)
2. The mean of Y (
)
Step 2: Calculate (X_i - X), (Y_i - Ȳ), and their product
X
Y
X-X
Y-Ȳ
(X-X)(Y-Ȳ)
(X-X)²
1
2.4
-3
-1.933
5.799
9
2
3
-2
-1.333
2.666
4
3
3.6
-1
-0.733
0.733
1
4
4
0
-0.333
0
0
6
5
2
0.667
1.334
4
8
8
4
3.667
14.668
16
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Now sum the columns:
Step 3: Calculate slope (b)
Step 4: Calculate intercept (a)
The intercept formula is:
Step 5: Write the regression line
This is our best-fit straight line. It tells us:
• When X = 0, Y ≈ 1.369
• For every unit increase in X, Y increases by 0.741 units.
Step 6: Plotting the Line (Optional Diagram)
Here, the stars roughly represent the observed values, and the line Y = 1.369 + 0.741X
would pass close to most of them.
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(b) Important Properties of Correlation Coefficient
Now, let’s talk about correlation coefficient, a measure that tells us how strongly X and Y
are related.
The correlation coefficient is usually denoted by r (for sample) and ranges from -1 to +1.
• r = +1 → Perfect positive correlation (as X increases, Y increases perfectly)
• r = -1 → Perfect negative correlation (as X increases, Y decreases perfectly)
• r = 0 → No linear correlation (X and Y are independent linearly)
Step 1: Formula for correlation coefficient (r)
We already have (X-X)(Y-Ȳ) and (X-X)². Let’s calculate (Y-Ȳ)²:
Y-Ȳ
(Y-Ȳ)²
-1.933
3.737
-1.333
1.777
-0.733
0.537
-0.333
0.111
0.667
0.445
3.667
13.444
Step 2: Calculate r
Step 3: Interpret r
• r ≈ 0.968, which is very close to 1.
• This means X and Y are strongly positively correlated. As X increases, Y almost
always increases in a predictable pattern.
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Step 4: Important Properties of Correlation Coefficient
1. Range: Always between -1 and +1.
2. Significance of Sign: Positive r → positive relationship, negative r → negative
relationship.
3. No Units: r is a pure number, independent of the units of X and Y.
4. Symmetry: r(X,Y) = r(Y,X)
5. Insensitive to Linear Transformation: Adding or multiplying X or Y by a constant
doesn’t change r.
6. Perfect Correlation: r = 1 or r = -1 only if all points lie exactly on a straight line.
7. Measure of Linear Relationship: r measures strength of linear association, not
causation.
8. Correlation vs Causation: High correlation does not mean one variable causes the
other; it only shows relationship.
A Story-Like Interpretation
Let’s put it in a real-world story:
Suppose you are a baker trying to predict how many cakes to bake based on the number of
customers visiting your shop. You notice a pattern in the past six days. By plotting the data
and fitting a straight line, you get a formula:
Cakes Sold Number of Customers
This means even if no customers come (X=0), you sell about 1 cake (maybe a default order).
And for every extra customer, cake sales increase by 0.741.
Checking the correlation coefficient r = 0.968, you can confidently say: “Wow! The more
customers come, the more cakes I sell, and this relationship is very strong!”
This is the beauty of regression and correlation—they turn scattered numbers into a story
that predicts the future and explains patterns.
Conclusion
1. Regression Line:
2. Correlation Coefficient: r ≈ 0.968 (strong positive relationship)
3. Interpretation: X and Y move together in a predictable linear manner.
4. Applications: Economics, business, social sciences, engineering, medicine—you
name it! Wherever relationships exist between variables, regression and correlation
help us understand and predict.
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6.(a) Calculate Coefficient of Rank Correlation from the following data:
(b) Discuss Time Series and its various components.
Marks in Maths X
Marks in Science Y
20
80
30
70
40
40
50
50
60
10
70
20
80
30
Ans: Statistics: Making Sense of Numbers Through Stories
Imagine you are a teacher in a small school in Punjab. You love numbers, but more than
that, you love understanding your students—how their abilities in different subjects are
connected, and how trends change over time. One day, you notice an interesting pattern:
some students who do well in Mathematics don’t always do as well in Science, and vice
versa. You wonder: Is there a relationship between their marks in these two subjects?
This is exactly where statistics comes in to help. Let’s embark on a journey to understand
this, step by step.
Part (a) Calculate Coefficient of Rank Correlation
Step 1: Understanding the Problem
You have marks of 7 students in Mathematics (X) and Science (Y):
Marks in Maths (X)
Marks in Science (Y)
20
80
30
70
40
40
50
50
60
10
70
20
80
30
You want to know: Do higher marks in Maths mean higher marks in Science, or is it the
opposite?
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For this, we use Spearman’s Rank Correlation Coefficient (ρ or rₛ). This coefficient measures
the strength and direction of a relationship between two variables when the data is in
ranks (or can be ranked).
Step 2: Rank the Data
We start by giving ranks to each set of marks. The lowest mark gets rank 1, the next higher
gets rank 2, and so on.
Ranking Maths (X):
Maths (X)
Rank of X
20
1
30
2
40
3
50
4
60
5
70
6
80
7
Ranking Science (Y):
Science marks from lowest to highest: 10, 20, 30, 40, 50, 70, 80.
Science (Y)
Rank of Y
80
7
70
6
40
4
50
5
10
1
20
2
30
3
Step 3: Calculate Differences in Ranks
For each student, subtract Rank of Maths (Rₓ) - Rank of Science (Rᵧ), and then square the
difference.
Maths Rank Rₓ
Science Rank Rᵧ
d = Rₓ - Rᵧ
d²
1
7
-6
36
2
6
-4
16
3
4
-1
1
4
5
-1
1
5
1
4
16
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6
2
4
16
7
3
4
16
Now, sum up the squared differences:
Step 4: Apply the Formula
Spearman’s Rank Correlation formula:
Where (number of students).
Step by step calculation:
1.
2.
3.
4.
Step 5: Interpretation
The rank correlation coefficient is approximately -0.82.
• A negative value indicates an inverse relationship: students who scored higher in
Maths scored lower in Science.
• The magnitude, 0.82, is close to 1, which means it is a strong negative correlation.
In simple words: in this small class, excelling in Maths did not mean doing well in Science—
almost the opposite!
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This calculation not only shows numbers but also tells a story: sometimes, strengths in one
subject may come at the expense of another, and as a teacher, this insight can help design
balanced learning strategies.
Part (b) Time Series and Its Components
Now, imagine another scenario. You are tracking the monthly attendance of students in
your school over several years. Every month, you note down the number of students
attending. At first, the numbers fluctuate, sometimes high, sometimes low. You wonder:
Can we make sense of this? Can we forecast future attendance?
This is exactly where Time Series Analysis comes in.
Step 1: Understanding Time Series
A Time Series is a sequence of observations recorded at regular intervals over time. It
could be hours, days, months, quarters, or years. The main idea is to study patterns over
time, detect trends, seasonal effects, and make predictions.
Examples:
• Daily temperature of a city
• Monthly sales of a shop
• Annual population growth
• Yearly rainfall in Punjab
Mathematically, we can write a time series as:
Where each component has a meaning.
Step 2: Components of Time Series
1. Trend (Tₜ)
o Trend represents the long-term movement of the series.
o It shows whether the data is generally increasing, decreasing, or constant
over time.
o For example, student attendance may gradually increase over the years due
to more enrollment.
o Trend can be linear (steady growth) or non-linear (curved growth).
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2. Seasonal Component (Sₜ)
o Seasonal variation is the short-term, repetitive pattern that occurs at fixed
intervals.
o Example: sales of umbrellas peak every monsoon season; attendance may
drop every summer vacation.
o Seasonal changes are predictable because they repeat every season or
month.
3. Cyclic Component (Cₜ)
o Cyclic variations are long-term fluctuations around the trend, lasting more
than a year.
o Unlike seasonal effects, cycles don’t have fixed lengths and depend on
economic or business conditions.
o Example: school attendance may rise during good economic years and fall
during crises.
4. Irregular Component (Iₜ)
o Also called residual or random component, this is unpredictable and caused
by unusual events.
o Example: a sudden flood closing schools, or an epidemic affecting student
attendance.
o This component is random and cannot be easily forecasted.
Step 3: Types of Time Series Models
1. Additive Model:
Used when the seasonal variations are roughly constant over time.
2. Multiplicative Model:
Used when seasonal variations increase with the trend (e.g., sales grow larger each year).
Step 4: Diagram of Time Series Components
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The overall upward path represents Trend (Tₜ)
• The waves above and below the trend show Seasonal (Sₜ) and Cyclic (Cₜ)
fluctuations
• Random scattered points above or below the line indicate Irregular (Iₜ) variation
Think of it like observing a river over a year:
• The river generally flows downstream → Trend
• Waves during rainy season → Seasonal
• Floods or droughts over several years → Cyclic
• Sudden rock falling → Irregular
Step 5: Importance of Time Series Analysis
1. Forecasting: Predicting future attendance, sales, or weather patterns.
2. Planning: Helps schools, companies, and governments plan resources.
3. Decision Making: Understanding trends and cycles aids policy and investment
choices.
4. Identifying Patterns: Recognizes underlying behavior that raw numbers alone
cannot show.
Conclusion
In summary:
• Coefficient of Rank Correlation: A simple yet powerful tool to detect relationships
between two variables, here showing a strong negative correlation between Maths
and Science marks.
• Time Series and Components: A tool to analyze data over time, with trend,
seasonal, cyclic, and irregular components, helping us understand patterns and
forecast the future.
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By studying these tools in a story-like context—seeing marks as students’ lives and time
series as the flow of events—we not only perform calculations but also bring data to life,
making learning engaging and meaningful.
SECTION-D
7. (a) A die is thrown twice. Write down all possible outcomes. What is the probability
that 5 will come up at least once?
(b) If a sample space S = AB, P(A) =
P(B)=
find P(AB).
Ans: Understanding Probability through a Story: Dice and Events
Imagine you are sitting in a quiet room on a rainy afternoon, sipping a cup of tea, and you
have a pair of dice in front of you. Not just any dice, but the standard six-faced dice with
numbers 1 to 6 on their faces. You pick them up and start thinking: “What are the chances
that a 5 will show up if I throw these dice twice?”
At first, it might seem like a simple question. But when you begin to explore all the possible
outcomes, it turns into a little adventure—a journey through the world of probability.
Part (a): A Die is Thrown Twice
Step 1: Understanding the Experiment
When we say a die is thrown twice, we are performing two independent events. This means
the result of the first throw does not affect the second throw. Think of it like tossing a coin
twice; whatever happens on the first toss doesn’t change the second toss.
For a single die, there are 6 possible outcomes: 1, 2, 3, 4, 5, or 6.
Since we throw it twice, the total number of outcomes is the product of the possibilities of
each throw:
So, there are 36 equally likely outcomes.
Step 2: Listing All Possible Outcomes
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Now, let’s list them in an organized way. We can think of the first throw as giving the row,
and the second throw as giving the column in a grid.
1
2
3
4
5
6
1
(1,1)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
2
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
3
(3,1)
(3,2)
(3,3)
(3,4)
(3,5)
(3,6)
4
(4,1)
(4,2)
(4,3)
(4,4)
(4,5)
(4,6)
5
(5,1)
(5,2)
(5,3)
(5,4)
(5,5)
(5,6)
6
(6,1)
(6,2)
(6,3)
(6,4)
(6,5)
(6,6)
Here, the first number in each pair represents the first throw, and the second number
represents the second throw. As we said, there are 36 outcomes, and all are equally likely.
Step 3: Identifying Favorable Outcomes
Now comes the main question: “What is the probability that 5 will come up at least once?”
“At least once” means we want all outcomes where a 5 appears in either the first throw,
the second throw, or both.
Let’s find them:
1. 5 on the first throw: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes
2. 5 on the second throw: (1,5), (2,5), (3,5), (4,5), (6,5) → 5 outcomes
Notice that (5,5) was already counted in the first set, so we don’t count it twice.
So the total favorable outcomes = 6 + 5 = 11.
Step 4: Calculating the Probability
Probability is defined as:
Number of favorable outcomes
Total number of outcomes
So here:
5 at least once
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And there we have it! The probability that a 5 will appear at least once is 11/36, which is
approximately 0.3056, or about 30.56%.
Step 5: Visual Representation
Sometimes, a diagram helps to see the concept clearly. Think of a grid of 6 × 6, where the
row number is the first throw and the column number is the second throw. The cells where
a 5 appears are highlighted:
(1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,5)
A visual representation makes it easy to spot patterns and avoid double counting.
Part (b): Probability of Intersection
Now let’s shift gears from dice to a slightly more abstract idea. Imagine you have two
events, A and B, in a sample space S.
We are given:
•
•
•
We are asked to find , the probability that both events occur together.
Step 1: Using the Union Formula
Here’s a story approach: Think of A and B as two overlapping circles in a Venn diagram.
• Circle A represents all outcomes of event A.
• Circle B represents all outcomes of event B.
• The overlapping area is (both happen).
• The total area covered by both circles is .
The formula for the probability of a union of two events is:
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Step 2: Substituting the Known Values
We know that the entire sample space is A B, which means .
So we substitute the values:
Step 3: Solving for
First, convert fractions to a common denominator:
Subtract 1 from both sides:
So the probability that both events A and B happen together is 1/4, or 25%.
Step 4: Diagrammatic Explanation
A simple Venn diagram helps:
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• Circle A: P(A) = 1/2
• Circle B: P(B) = 3/4
• Intersection (A∩B): 1/4
• Union (AB): 1 (entire space)
The overlapping portion clearly shows 1/4, which is exactly what we calculated.
Part (c): Connecting Both Stories
If you look closely, the two parts of this question—dice and set theory—are not very
different. Both are about counting favorable outcomes and understanding overlaps:
1. With dice, we counted how many outcomes had a 5 (overlap happens if 5 appears on
both throws).
2. With events A and B, we counted the probability of both happening (overlap in the
Venn diagram).
In both cases, the key is careful counting and avoiding double counting.
Think of probability as a story of chances—where every outcome has its own part to play,
and sometimes stories overlap. Understanding probability is just learning to read these
overlapping stories.
Step 5: Summary Table
Part
Scenario
Total
Outcomes
Favorable
Outcomes
Probability
(a)
Throwing die twice, at least
one 5
36
11
11/36 ≈
0.306
(b)
Events A and B in sample
space
1 (S)
A∩B = 1/4
1/4 = 0.25
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Step 6: Moral of the Story
Probability is not just about numbers. It’s about visualizing possibilities, understanding
overlaps, and thinking logically. Each throw of a die, each event in a sample space, is like a
tiny story with its own chance of happening.
By imagining outcomes as grids, diagrams, and overlapping circles, you not only get the
correct answer but also understand why it works. And that’s the secret to enjoying and
mastering probability.
8.(a) Explain Poisson distribution and its important properties.
(b) What is Normal distribution? Give its important properties.
Ans: Understanding Poisson and Normal Distributions: A Story of Randomness and
Patterns
Imagine you are sitting at a bus stop on a quiet morning. You notice that buses arrive at
random times. Some days, two buses come within 10 minutes, while other days, there
might be none for half an hour. You start thinking: “Is there any way to understand these
random events? Can we somehow predict the likelihood of buses arriving in a given time?”
This is exactly the world of probability distributions, which are tools statisticians use to
make sense of random events. Among the many distributions, two stand out in practical
applications: Poisson distribution and Normal distribution. Let’s explore them, step by step,
in a way that makes them both simple and relatable.
(a) Poisson Distribution
Imagine a small town where accidents at a particular road intersection are rare but
occasionally happen. One day, you are asked to calculate the probability of exactly 0, 1, 2, or
3 accidents happening in a day.
Here comes the Poisson distribution to the rescue. Named after the French mathematician
Simeon Denis Poisson, this distribution helps us describe the probability of a given number
of events happening in a fixed interval of time or space, when these events occur
independently, and the average rate is known.
Definition of Poisson Distribution
If the average number of events in a time interval is λ (lambda), the probability of observing
exactly x events is given by the formula:
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Where:
• X = the number of events (0, 1, 2, …)
• λ = average number of events per interval
• e = approximately 2.718 (base of natural logarithms)
• x! = factorial of x
Think of λ as the “usual rate” of events. For instance, if normally 3 accidents happen per day
on that road, λ = 3.
Important Properties of Poisson Distribution
1. Discrete Nature: Poisson is used for discrete events. You can’t have 2.5 accidents;
it’s always 0, 1, 2, …
2. Mean and Variance: Both the mean and variance of a Poisson distribution are equal
to λ.
Mean Variance
This unique property makes Poisson distribution easy to work with.
3. Rare Events: It’s suitable for events that are rare or occur randomly in time or space.
Examples:
o Number of phone calls to a call center in an hour
o Number of typographical errors in a page of a book
o Number of cars passing a checkpoint in a minute
4. Additivity Property: If you have two independent Poisson processes with means λ₁
and λ₂, the total process is also Poisson with mean λ₁ + λ₂.
5. Memoryless Approximation: Although technically the Poisson process is not
memoryless (that’s exponential for waiting times), it assumes independent events,
so past events do not influence future events.
Example of Poisson Distribution
Suppose a hospital receives an average of 2 emergency calls per hour (λ = 2). What is the
probability that exactly 3 calls will arrive in the next hour?
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So, there is an 18% chance that exactly 3 calls will come in the next hour. Simple, yet
powerful!
Diagram for Poisson Distribution
You can imagine a bar chart with X-axis = number of events (0,1,2,3…) and Y-axis =
probability. For small λ, the chart is skewed to the right, with the highest bar near λ, and
then it tapers down.
This visual helps students grasp how rare events (like accidents or calls) are distributed
around an average rate.
(b) Normal Distribution
Now, let’s move from rare events to something more universal. Imagine you go to your
college canteen and notice the heights of students. Some are tall, some are short, but most
are average height. If you plotted the heights on a graph, a special pattern would appear: a
bell-shaped curve, highest in the middle and tapering symmetrically on both sides. This is
the Normal distribution, also known as the Gaussian distribution.
Definition of Normal Distribution
A Normal distribution is a continuous probability distribution characterized by the famous
bell-shaped curve. It is defined by two parameters:
• μ (mu) = mean (center of the distribution)
• σ (sigma) = standard deviation (spread of the distribution)
The probability density function (PDF) is given by:
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Where:
• x = value of the variable
• μ = mean
• σ = standard deviation
• e = 2.718
• π = 3.1416
This formula might look intimidating, but the essence is simple: values close to the mean are
more likely, and values far from the mean are less likely.
Important Properties of Normal Distribution
1. Symmetry: The curve is perfectly symmetrical around the mean. Left side = Right
side.
2. Mean, Median, Mode are Equal: In a Normal distribution, μ = median = mode. This
is why it’s often called the “ideal distribution.”
3. Bell-Shaped Curve: Most data points are near the mean. Approximately:
o 68% of data lies within 1σ of the mean
o 95% within 2σ
o 99.7% within 3σ
4. Continuous Distribution: Unlike Poisson, Normal deals with continuous variables,
like heights, weights, test scores, or temperature.
5. Asymptotic: The tails of the curve never touch the X-axis. This means there is always
a tiny chance of extreme values.
6. Standard Normal Distribution: If we standardize a Normal variable using Z = (X-μ)/σ,
it becomes a Standard Normal Distribution, with μ = 0 and σ = 1.
Example of Normal Distribution
Suppose the heights of college students are normally distributed with μ = 170 cm and σ = 10
cm. What is the probability of finding a student with height between 160 cm and 180 cm?
Here, you would standardize the values using Z-scores:
From Z-tables, the probability of Z being between -1 and 1 ≈ 68%, meaning most students
are of average height.
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Diagram for Normal Distribution
Imagine a smooth bell curve:
The highest point is the mean, and the curve falls off symmetrically on both sides. This visual
makes the concept intuitive and helps in exams.
Comparing Poisson and Normal Distributions
To make it relatable, let’s compare them like two different friends:
Feature
Poisson Distribution
Normal Distribution
Nature
Discrete (number of events)
Continuous (measurements)
Shape
Skewed for small λ, symmetric for
large λ
Perfectly bell-shaped
Parameters
λ (average rate)
μ (mean), σ (standard
deviation)
Mean =
Variance?
Yes
No
Typical Example
Accidents, emails, calls
Heights, weights, exam scores
Interestingly, when λ in Poisson is large, the Poisson distribution approximates a Normal
distribution. This connection shows how the two distributions are related in practice.
Real-Life Applications
Poisson:
• Counting the number of typing errors per page
• Number of customers entering a shop per hour
• Rare disease occurrence in a population
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Normal:
• Heights and weights of people
• IQ scores
• Measurement errors in scientific experiments
Conclusion
Both Poisson and Normal distributions help us make sense of randomness. Poisson is like
counting rare sparks in a night sky, discrete and sporadic. Normal is like watching the waves
of the sea, continuous and flowing smoothly. Together, they form the backbone of
probability theory and statistics, enabling us to model real-world events and make informed
predictions.
Understanding these distributions is not just about memorizing formulas—it’s about seeing
patterns in chaos, predicting the unpredictable, and using mathematics as a tool to
understand life around us.
“This paper has been carefully prepared for educational purposes. If you notice any mistakes or
have suggestions, feel free to share your feedback.”
