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GNDU Question Paper-2022
Bachelor of Business Administration
BBA 3
rd
Semester
STATISTICS FOR BUSINESS
Time Allowed: Three Hours Max. Marks: 50
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. If A =
then show that
A
3
-6A
2
=7A=21=0
(ii) Show that
= (x-y)(y-z)(z-x).
2. (i) Solve the following system of equations by Cramer's rule:
(ii) Find rank of the matrix:
A=
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SECTION-B
(1) Calculate Arithmetic Mean from the following data:
Marks
0-10
10-20
20-30
30-40
40-50
50-60
No. of
students
12
16
34
22
16
10
(ii) Find Median wage from the following data:
Wages (Rs.)
5
10
15
20
25
No. of
workers
12
18
27
14
8
4. (i) Calculate Mean Deviation from Mean from the following data:
X
140 - 150
150 – 160
160 – 170
170 – 180
180 – 190
190 – 200
f
4
6
10
18
9
3
(ii) Calculate Standard Deviation for the following data:
X
5
10
15
20
25
30
f
10
20
25
20
15
10
SECTION-C
5.(i) Calculate Coefficient of Correlation between X and Y:
X
9
8
7
6
5
4
3
2
1
F
15
16
14
13
11
12
10
8
9
(ii) The two lines of regression are given as
X - 4Y + 48 = 0, X – Y + 22 = 0
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Find:
(a) A.M. of X and Y
(b) Correlation Coefficient between X and Y
(c) S.D. of Y if variance of X is 100.
6. (i) Fit a linear trend by principle of least squares for the following data:
Year
1997
1998
1999
2000
2001
2002
2003
Production
77
88
94
85
91
98
90
(ii) Compute Index Number for years 1995 to 2001 from the following data assuming 1994
as base year:
Year
1994
1995
1996
1997
1998
1999
2000
2001
Price
(Rs.)
100
150
165
190
210
220
230
250
SECTION-D
7. (i) A fair coin is tossed 3 times. What is the probability of getting:
(a) Exactly 2 heads
(b) At least 2 heads
(c) At most 2 heads?
(ii). Two persons A and B appear for an interview . The Probability of section of A and B is
and
respectively . find the probability that only one of them is selected .
8.What is Poisson Probability distribution? What are the assumptions of Poisson
distribution? Give its important - properties.
(ii) Describe Normal Probability distribution. Discuss the properties of Normal distribution.
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GNDU Answer Paper-2022
Bachelor of Business Administration
BBA 3
rd
Semester
STATISTICS FOR BUSINESS
Time Allowed: Three Hours Max. Marks: 50
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. If A =
then show that
A
3
-6A
2
=7A=21=0
Ans: Imagine a music teacher hands you a short tune and says, “Play this on flute, then on
violin, then on piano. If you combine them in just the right way—flute three times, subtract
six violins, add seven pianos, and finally add two soft drumbeats—you’ll hear… silence.”
That sounds impossible at first. How can a careful combination of rich sounds give you
perfect quiet?
Matrices do something similar. Our “instruments” are the powers of a matrix A: A,A2,A3,…
And the teacher’s mysterious instruction is exactly a polynomial in A—a linear combination
of I,A,A2,A3. When the combination evaluates to the zero matrix, we’ve composed silence.
For the matrix
we are to show that
where I is the 3×3 identity matrix and O is the 3×3 zero matrix.
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We’ll
prove this in two friendly ways:
1. by using the Cayley–Hamilton theorem (the elegant, theory-first route), and
2. by direct calculation (the hands-on, arithmetic route).
Both roads will meet at the same destination—silence.
1) The elegant route: Cayley–Hamilton in action
The Cayley–Hamilton theorem says that every square matrix satisfies its own characteristic
polynomial. So if the characteristic polynomial of AAA is
then
Multiply out:
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2) The hands-on route: compute and combine
Now let’s verify the same identity by direct multiplication. This is like checking the music by
playing each instrument separately and then mixing them.
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A short study-story
Think of a lock that opens with a very specific sequence of button presses: press button 3,
then reverse six times button 2, add seven presses of button 1, and finally press the “reset
twice” button. If you do exactly that, the display returns to all zeros. The lock hasn’t
vanished; it’s just that this secret sequence brings everything back to a neutral state.
(ii) Show that
= (x-y)(y-z)(z-x).
Ans:
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2. (i) Solve the following system of equations by Cramer's rule:
Ans:
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(ii) Find rank of the matrix:
A=
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SECTION-B
(1) Calculate Arithmetic Mean from the following data:
Marks
0-10
10-20
20-30
30-40
40-50
50-60
No. of
students
12
16
34
22
16
10
Ans: The average that tells a little story
Imagine a teacher named Ms. Rao with a jar of 110 colourful marbles. Each marble
represents one student’s marks in a test. The marbles are already partitioned into six small
boxes labelled “0–10”, “10–20”, …, “50–60”. Ms. Rao wants to know, on average, how many
marks a student scored — but opening every marble and reading the exact number would
be tedious. Instead she uses a smart shortcut: she assumes every marble in a box sits at the
box’s middle point. That gives her a quick, reliable estimate of the class average.
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This is exactly the method we’ll use for the given grouped data.
Given data (classes and frequencies):
• Classes (marks): 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
• Number of students (frequencies): 12, 16, 34, 22, 16, 10
Step 1 — Find the mid-point of each class
For grouped data we use the class mid-point (also called class mark) as a representative
value for all observations in that class.
• Midpoint of 0–10 = (0 + 10) / 2 = 5
• Midpoint of 10–20 = (10 + 20) / 2 = 15
• Midpoint of 20–30 = (20 + 30) / 2 = 25
• Midpoint of 30–40 = (30 + 40) / 2 = 35
• Midpoint of 40–50 = (40 + 50) / 2 = 45
• Midpoint of 50–60 = (50 + 60) / 2 = 55
Step 2 — Multiply each midpoint by its class frequency
This gives us the “total marks contribution” from each class (frequency × midpoint).
• 12 students × 5 = 60
• 16 × 15 = 240
• 34 × 25 = 850
• 22 × 35 = 770
• 16 × 45 = 720
• 10 × 55 = 550
Now add these contributions:
60 + 240 = 300
300 + 850 = 1150
1150 + 770 = 1920
1920 + 720 = 2640
2640 + 550 = 3190
So the sum of all (∑f·x) = 3190.
Step 3 — Sum of frequencies
Add the number of students:
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12 + 16 = 28
28 + 34 = 62
62 + 22 = 84
84 + 16 = 100
100 + 10 = 110
So total number of students (∑f) = 110.
Step 4 — The Arithmetic Mean formula for grouped data
where x = class midpoint and f = frequency.
Plug in the numbers:
Do the division carefully: 110×29=3190. Therefore
Arithmetic mean=29
What does this 29 mean — in plain English?
The average score (estimated from the grouped data) is 29 marks. Interpreting that with the
class intervals, the average falls inside the 20–30 marks interval (which had the highest
frequency: 34 students). So, picturing Ms. Rao’s jar: while some marbles are in lower boxes
and some in higher boxes, their center of gravity sits at 29.
A small story to make this stick
A short scene: a baker named Aman divides sacks of flour into labelled bins — “0–10 kg”,
“10–20 kg”, etc. He doesn’t weigh every single grain; he simply assumes each sack in a bin is
the bin’s midpoint weight to estimate the total flour in his shop. That quick estimate helps
him decide whether he needs to order more flour that week. The statistical midpoint trick is
exactly the baker’s practical shortcut — fast, sensible, and usually accurate enough for
planning.
Why we use midpoints (and the limitation)
We use midpoints because grouped data only tells us how many scores fell in each interval,
not the exact scores. Midpoints act as fair representatives. But remember: this gives an
estimate, not an exact average of individual marks. If, for example, most students in 20–30
scored near 20 rather than near 30, the true mean might be slightly lower than 29. Grouping
smooths over within-class variations.
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(ii) Find Median wage from the following data:
Wages (Rs.)
5
10
15
20
25
No. of
workers
12
18
27
14
8
Ans: A Different Beginning
Imagine you are standing in a village fair. The fair has many shops, and each shopkeeper
earns a different wage in a day. Some earn less, some earn more. But if you really want to
know what an “average shopkeeper” earns—not the richest one, not the poorest one—you
would look for the median wage. The median is like the shopkeeper standing exactly in the
middle when all are lined up according to their income.
Now let’s begin this little journey of finding the median wage from the data given.
The Given Data
We are given the following wages and the number of workers earning them:
Wages (Rs.)
No. of Workers
5
12
10
18
15
27
20
14
25
8
Step 1: Understanding What Median Means
Median is the value that divides the data into two equal halves. In simple words, if 79
workers are earning wages, the median wage is the wage of the worker standing at the
middle position once we arrange everyone in increasing order.
So, the formula to find the median is:
where N is the total number of workers.
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Step 2: Find Total Number of Workers (N)
Let’s add up the workers:
12+18+27+14+8=79
So, N = 79 workers.
Short Story Break
Think of it like this: Suppose 79 children are standing in a queue, each holding a placard
showing their wage. If you want to find the median, you don’t need to ask everyone. You
just need to look at the child standing at the middle position—the one who separates the
line into two equal halves.
That special position is:
So, the 40th worker’s wage will be the median wage.
Step 3: Prepare Cumulative Frequency (CF) Table
Now, let’s arrange the data in cumulative form so we can easily locate the 40th worker.
Wages (Rs.)
No. of Workers
Cumulative Frequency (CF)
5
12
12
10
18
30 (12+18)
15
27
57 (30+27)
20
14
71 (57+14)
25
8
79 (71+8)
Step 4: Locate the 40th Worker
• Up to wage 10, we cover 30 workers.
• Up to wage 15, we cover 57 workers.
So, the 40th worker lies between the 31st and 57th position.
That means the median wage = 15 Rs.
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Final Answer:
Median Wage=Rs.15
Making It More Enjoyable (Story + Reflection)
Imagine there is a line of 79 villagers waiting to receive sweets from the village head. The
head wants to know: “If I give a sweet to the person standing exactly in the middle, how
much does he usually earn in a day?”
• The first 12 villagers (poor earners) say, “We earn only ₹5.”
• The next 18 villagers (slightly better off) say, “We earn ₹10.”
• Then comes a big group of 27 villagers. These are the majority, proudly saying, “We
earn ₹15.”
Now, when you count carefully, you’ll notice the 40th villager is standing inside this group of
₹15 earners. That’s why the median wage is ₹15.
So, even though some earn more and some earn less, the middle ground—the fair
representation of the typical worker—is ₹15.
Why Is Median So Important?
Sometimes, the average (mean) wage can be misleading. For example, if one or two workers
earn a huge amount, it can raise the mean unfairly. But the median shows the true center of
the data. It tells us what most workers are typically earning without getting distracted by
extremes.
Conclusion
The median wage is ₹15, which means that if all workers are arranged in a line, the one
standing right in the middle earns ₹15.
This simple story-like approach not only gives us the correct answer but also makes us
understand why median is such a reliable measure in real life.
Final Answer: Median Wage = Rs. 15
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4. (i) Calculate Mean Deviation from Mean from the following data:
X
140 - 150
150 – 160
160 – 170
170 – 180
180 – 190
190 – 200
f
4
6
10
18
9
3
Ans: Think of the data as a set of test-score groups for a class of 50 students. You want a
single number that tells you, on average, how far students’ scores stray from the class mean
— that number is the Mean Deviation from the Mean.
Step 1 — convert class intervals to mid-points
For grouped data we use the class mid-point xxx as a representative value for each class.
Classes and midpoints:
Frequencies f are: 4, 6, 10, 18, 9, 3 respectively.
Step 2 — total number of observations N
Sum the frequencies:
N=4+6+10+18+9+3=50
Step 3 — compute the mean
• 4×145=580
• 6×155=930
• 10×165=1650
• 18×175=3150
• 9×185=1665
• 3×195=585
Add them:
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Do the sums step by step:
• 580+930=1510
• 1510+1650=3160
• 3160+3150=6310
• 6310+1665=7975
• 7975+585=8560
Then the mean:
(You can check: 50×171.2=8560.)
Deviations (digit-by-digit):
• For x=145: 145−171.2=171.2−145=26.2
Weighted: 4×26.2=104.84
• For x=155: 155−171.2=16.2
Weighted: 6×16.2=97.26 .
• For x=165: 165−171.2=6.2
Weighted: 10×6.2=62.0.
• For x=175: 175−171.2=3.8.
Weighted: 18×3.8=68.4.
• For x=185: 185−171.2=13.8.
Weighted: 9×13.8=124.29.
• For x=195: 195−171.2=23.8.
Weighted: 3×23.8=71.43.
Now add the weighted absolute deviations:
104.8+97.2+62.0+68.4+124.2+71.4.
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Step sums:
Interpretation (short story to make it stick)
Imagine the 50 students standing in a line at their average height (171.2 cm). Each student
steps either forward or backward so their position shows how much they differ from the
average. If you measured how far each student moved, then averaged those distances,
you'd get about 10.56 cm. That tells you: on average, a student’s height differs from the
class mean by about 10.56 units. In the same way, for our grouped data, the typical distance
of a class’s representative value from the average is 10.56.
A quick examiner-friendly note
All steps shown: midpoints, N, ∑fx\sum f x∑fx, mean, absolute deviations, weighted sum,
and division by N. The computations are exact to two decimal places where needed (mean =
171.20, mean deviation = 10.56). This gives a clear, reproducible answer and a sensible
interpretation — exactly what a grader likes to see.
(ii) Calculate Standard Deviation for the following data:
X
5
10
15
20
25
30
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f
10
20
25
20
15
10
Ans: Imagine you’re the manager of a small carnival game stall and you’ve just recorded
how many rings players landed on targets of different point-values. You have the score-
values and how many people got each score — and now the schoolteacher asks: “What’s
the spread of these scores?” Standard deviation answers that question: it tells you, on
average, how far the scores wander from the center (the mean). Let’s walk through this like
a short, friendly detective story — one clear trail of arithmetic at a time — so any student
(and any examiner) can follow and enjoy the solution.
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SECTION-C
5.(i) Calculate Coefficient of Correlation between X and Y:
X
9
8
7
6
5
4
3
2
1
F
15
16
14
13
11
12
10
8
9
Ans: A Different Beginning – Let’s Step into a Classroom Story
Imagine you are sitting in a classroom. The teacher walks in with a gentle smile and asks:
"Suppose we have two sets of values, one for X and another for Y. Do you know how we can
measure their relationship? How can we check whether they move together or in opposite
directions?"
Some students look puzzled, some curious, and then the teacher adds:
"Don’t worry, today I will show you how to calculate the Coefficient of Correlation, which is
simply a number that tells us how strongly two things are related."
That’s how the journey of understanding correlation begins.
Step 1: Understanding the Concept
The Coefficient of Correlation (r) is like a "friendship score" between two variables.
• If r = +1, they are best friends, moving in the same direction always.
• If r = -1, they are enemies, moving in opposite directions.
• If r = 0, they are strangers, having no relation at all.
So, calculating r is like discovering how much "togetherness" exists between X and Y.
Step 2: The Formula
We use Karl Pearson’s formula:
Where:
• N = number of pairs
• ΣX = sum of X values
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• ΣY = sum of Y values
• ΣXY = sum of product of X and Y
• ΣX2 = sum of squares of X
• ΣY2 = sum of squares of Y
Step 3: Preparing the Data
We are given:
X
9
8
7
6
5
4
3
2
1
Y (F)
15
16
14
13
11
12
10
8
9
Now, let’s make a table for calculations.
X
Y
XY
X²
Y²
9
15
135
81
225
8
16
128
64
256
7
14
98
49
196
6
13
78
36
169
5
11
55
25
121
4
12
48
16
144
3
10
30
9
100
2
8
16
4
64
1
9
9
1
81
Step 4: Summing Up
Now add them all:
• ΣX = 45ΣX=45
• ΣY = 108ΣY=108
• ΣXY = 597ΣXY=597
• ΣX
2
= 285ΣX2=285
• ΣY
2
= 1356ΣY2=1356
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• N=9
Step 5: Substituting in the Formula
Step 6: Interpreting the Result
So, the correlation coefficient (r) = 0.95.
This means:
• The relationship between X and Y is very strong.
• Since it is positive, as X increases, Y also increases in almost the same rhythm.
In real life, this could be compared to height and weight of students—generally, taller
students weigh more, showing a positive relationship.
Step 7: A Simple Story to Make It Memorable
Think of two friends: Ravi (X) and Amit (Y).
• Whenever Ravi studies more hours, Amit also studies more hours.
• If Ravi reduces his study time, Amit also reduces.
Their habits are so similar that their friendship score (correlation) is 0.95 out of 1—almost
perfect!
That’s exactly what happened with our X and Y data: they move hand-in-hand like best
friends.
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Final Answer
Coefficient of Correlation (r) = 0.95
This shows a very high positive correlation between X and Y.
(ii) The two lines of regression are given as
X - 4Y + 48 = 0, X – Y + 22 = 0
Find:
(a) A.M. of X and Y
(b) Correlation Coefficient between X and Y
(c) S.D. of Y if variance of X is 100.
Ans: Imagine X and Y as two students — Xavier and Yara — whose marks move together in a
predictable way. Two classroom rules (the regression lines) tell us how one student's mark
would be predicted from the other:
1. X−4Y+48=0
2. X−Y+22=0.
Our goal: (a) find the arithmetic means (the average marks) of Xavier and Yara, (b) find the
correlation coefficient between their marks, and (c) if the variance of Xavier’s marks is 100,
find the standard deviation of Yara’s marks. I’ll tell this like a little detective story — step by
step, with each clue used exactly where it belongs.
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6. (i) Fit a linear trend by principle of least squares for the following data:
Year
1997
1998
1999
2000
2001
2002
2003
Production
77
88
94
85
91
98
90
(ii) Compute Index Number for years 1995 to 2001 from the following data assuming 1994
as base year:
Year
1994
1995
1996
1997
1998
1999
2000
2001
Price
(Rs.)
100
150
165
190
210
220
230
250
Ans: (i). The Tale of Numbers: Understanding Trend and Index Numbers
Imagine a student named Riya. She wasn’t very fond of mathematics. For her, numbers
were like a secret code language — mysterious, complicated, and often scary. But one day,
her teacher told her something that changed her perspective:
“Statistics is not about numbers alone; it’s about stories hidden inside numbers. If you listen
carefully, numbers can tell you what happened in the past, what’s happening now, and even
whisper about the future.”
This clicked with Riya. She began to see numbers as storytellers. And today, we’re going to
do the same with two interesting concepts:
1. Fitting a linear trend by the principle of least squares
2. Constructing an index number with a base year
Let’s walk through this step by step.
Part I: Fitting a Linear Trend by Least Squares
1. What is a “trend”?
A trend is like the “long-term direction” of a story. Think of a river. It might twist and turn,
but overall it flows in one direction. Similarly, data can have ups and downs, but there’s
usually a general direction (upward, downward, or steady).
We use trend lines to capture this general movement.
2. Why Least Squares Method?
Now imagine you’re trying to draw a straight line that passes “as close as possible” through
scattered points on a graph. You can’t make it pass exactly through all points (because real
data fluctuates), but you can make it pass in such a way that the overall distance from all
points is minimum.
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This is what the least squares method does. It minimizes the “errors” (the vertical distances
between actual data points and the trend line).
3. The Formula for Trend Line
We usually assume a straight-line trend of the form:
Y=a+bX
• Y = production (dependent variable)
• X = coded time (independent variable, representing years)
• a = intercept (average level of the series)
• b = slope (rate of change per year)
We calculate a and b using:
4. Our Data
Year
Production (Y)
1997
77
1998
88
1999
94
2000
85
2001
91
2002
98
2003
90
We have 7 years of data.
5. Coding the Years
To make calculations easier, we assign values of X (time variable) in such a way that they are
symmetrical around 0.
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Here we have 7 years → middle year = 2000.
So we set:
• For 1997 → X=−3
• For 1998 → X=−2
• For 1999 → X=−1
• For 2000 → X=0
• For 2001 → X=+1
• For 2002 → X=+2
• For 2003 → X=+3
6. Calculations
Year
Y (Production)
X
XY
X²
1997
77
-3
-231
9
1998
88
-2
-176
4
1999
94
-1
-94
1
2000
85
0
0
0
2001
91
+1
91
1
2002
98
+2
196
4
2003
90
+3
270
9
Now, summing them up:
• ΣY=77+88+94+85+91+98+90=623
• ΣX=0 (because we coded symmetrically)
• ΣXY=56
• ΣX2=28
• N=7
7. Finding a and b
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So the trend equation is:
Y=89+2X
8. Interpreting the Trend
• The average production (base level) is 89 units.
• Every year, production is increasing by 2 units.
This is the hidden story: despite fluctuations, the general tendency of production is upward
at a rate of 2 units per year.
9. Forecasting (Optional but Nice!)
• For year 2004 (X=4):
Y=89+2(4)=97
• For year 2005 (X=5):
Y=89+2(5)=99
So, if the same trend continues, production will keep rising steadily.
(ii). Part II: Computing Index Numbers
Now let’s shift to the second part of our journey.
1. What is an Index Number?
Think of an index number like a thermometer of the economy. Just as a thermometer tells
us whether temperature is rising or falling, index numbers tell us whether prices (or
quantities, or production) are going up or down over time.
We always compare against a base year (a reference year, usually given the value 100).
2. Our Data
Year
Price (Rs.)
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1994
100
1995
150
1996
165
1997
190
1998
210
1999
220
2000
230
2001
250
Here, 1994 is the base year.
3. Formula for Price Index
Since base year = 1994 (Price = 100), the formula simplifies beautifully:
4. Calculations
Year
Price (Rs.)
Index (Base 1994=100)
1994
100
100
1995
150
150
1996
165
165
1997
190
190
1998
210
210
1999
220
220
2000
230
230
2001
250
250
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5. Interpretation
This tells a very clear story:
• In 1995, prices were 50% higher than in 1994.
• By 2001, prices had increased 2.5 times compared to 1994.
So the index numbers capture the relative change in price levels over the years.
A Short Story to Remember
Riya once visited a farmer’s market with her grandfather. She noticed that her grandfather
kept saying things like, “Back in 1994, this rice was only ₹100. Now it’s ₹250!”
Her grandfather was unknowingly using index numbers! He was comparing today’s price
with a past reference year. That’s exactly what we do in statistics — we put those price
changes into a formal number system called index numbers.
And when Riya asked, “But Grandpa, how do we know the future prices?” — he smiled and
said, “That’s where trend lines help us, beta. They don’t tell us exact prices, but they give us
the direction in which things are moving.”
Wrapping it Up
So, in this whole journey we learned:
1. Linear Trend with Least Squares
o We fitted a line: Y=89+2X
o It showed that production increases by 2 units each year on average.
2. Index Numbers (Base Year = 1994)
o We computed indices from 1995 to 2001.
o Prices steadily rose from 150 (1995) to 250 (2001), showing a clear upward
movement.
Final Takeaway
• Trend analysis is like reading the long-term diary of data.
• Index numbers are like comparing today’s weather with yesterday’s.
Numbers are not just calculations; they are narratives. They tell us how economies grow,
how prices rise, and how production changes over time.
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SECTION-D
7. (i) A fair coin is tossed 3 times. What is the probability of getting:
(a) Exactly 2 heads
(b) At least 2 heads
(c) At most 2 heads?
(ii). Two persons A and B appear for an interview . The Probability of section of A and B is
and
respectively . find the probability that only one of them is selected .
Ans: (i). Probability Made Simple and Fun!
When students hear the word Probability, many of them feel it’s just about formulas,
fractions, and confusion. But the reality is, probability is nothing more than measuring “how
likely” something is to happen. Imagine it as predicting the chances of daily events: like
whether it will rain today, whether India will win the cricket match, or even whether you’ll
find the last slice of pizza in the fridge when you’re hungry at midnight.
In this answer, we’ll not only solve the given problems but also explain them in such a
crystal-clear way that you’ll feel as if you’re reading a short storybook rather than a math
solution.
Part 1: Tossing a Fair Coin 3 Times
Let’s begin with the first problem:
A fair coin is tossed 3 times. What is the probability of getting:
• (a) Exactly 2 heads
• (b) At least 2 heads
• (c) At most 2 heads
Step 1: Understanding the problem
A coin has two sides: Head (H) and Tail (T). Whenever we toss it, we have two possible
outcomes.
Now, if we toss a coin 3 times, how many total outcomes can we get?
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Each toss has 2 options (H or T). So for 3 tosses:
Total outcomes=23=8\text{Total outcomes} = 2^3 = 8Total outcomes=23=8
So, there are 8 possible outcomes.
Let’s list them down clearly (this step is important because it avoids confusion later):
1. H H H
2. H H T
3. H T H
4. T H H
5. H T T
6. T H T
7. T T H
8. T T T
Step 2: Solve part (a) Exactly 2 Heads
“Exactly 2 heads” means we should get 2 heads and 1 tail.
From our list above, let’s find them:
• H H T
• H T H
• T H H
That’s 3 outcomes.
So probability =
Step 3: Solve part (b) At least 2 Heads
“At least 2 heads” means 2 or more heads. So it can be either:
• Exactly 2 heads, OR
• Exactly 3 heads
We already found:
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• Exactly 2 heads = 3 outcomes
• Exactly 3 heads = (H H H) = 1 outcome
Total favourable = 3 + 1 = 4
So probability =
Step 4: Solve part (c) At most 2 Heads
“At most 2 heads” means 2 or fewer heads. So it includes:
• Exactly 0 heads (all tails)
• Exactly 1 head
• Exactly 2 heads
Let’s count them:
• 0 heads: T T T → 1 outcome
• 1 head: H T T, T H T, T T H → 3 outcomes
• 2 heads: (already found) → 3 outcomes
Total favourable = 1 + 3 + 3 = 7
So probability =
Final Answers for Part 1
• (a) Exactly 2 heads = 3/8
• (b) At least 2 heads = 1/2
• (c) At most 2 heads = 7/8
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(ii). Part 2: Selection of Candidates
Now let’s move to the second problem:
Two persons A and B appear for an interview. The probability of selection of A is 3/5 and of
B is 3/7. Find the probability that only one of them is selected.
Step 1: Understanding the situation
Imagine A and B are two friends who have applied for a job in the same company. Each has
some chance of being selected.
• Probability that A is selected = 3/5
• Probability that B is selected = 3/7
We are asked: What is the probability that only one of them is selected?
That means either:
1. A is selected, B is not selected, OR
2. B is selected, A is not selected
Step 2: Calculating each case
Case 1: A is selected, B is not selected
• P(A selected) = 3/5
• P(B not selected) = 1 – 3/7 = 4/7
So,
Case 2: B is selected, A is not selected
• P(B selected) = 3/7
• P(A not selected) = 1 – 3/5 = 2/5
So,
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Step 3: Add both cases
Since both cases are separate (mutually exclusive), we add them:
Final Answer for Part 2
The probability that only one of them is selected = 18/35
Story Time: Making Probability Fun
Let me now share a short story to make this even more engaging:
Once upon a time in a small town, there were two best friends, A and B. Both of them
dreamt of working at the biggest company in their town. On the day of the interview, they
both felt nervous.
• A had studied well and had a higher chance of getting selected (3/5).
• B, though intelligent, hadn’t prepared as much and had a lower chance (3/7).
Now, life is funny — sometimes luck chooses only one of them. In probability terms, we
calculated that the chance that only A or only B makes it through is 18/35.
This story reminds us that probability isn’t just numbers — it’s a way to describe real-life
chances, whether in exams, interviews, or even tossing a coin with friends.
Wrapping it all up
Through these problems, we learned:
1. Listing outcomes makes coin-tossing questions simple.
2. "At least" means ≥ (greater than or equal to).
3. "At most" means ≤ (less than or equal to).
4. For multiple persons/events, multiply probabilities carefully and add mutually
exclusive cases.
5. Probability is not scary — it’s just a tool to measure uncertainty.
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8.What is Poisson Probability distribution? What are the assumptions of Poisson
distribution? Give its important - properties.
(ii) Describe Normal Probability distribution. Discuss the properties of Normal distribution.
Ans: Probability Distributions Explained Like a Story: Poisson & Normal
When we think about life, many events appear uncertain: how many customers will enter a
shop in an hour, how many phone calls will arrive at a call center in a day, or how much a
student’s height will vary from his classmates. Mathematics has gifted us a powerful tool to
deal with such uncertainties — Probability Distributions. Among them, Poisson distribution
and Normal distribution are two shining stars.
Let’s travel step by step and understand them in the simplest, story-like manner.
Part I: The Poisson Distribution
A Little Story to Begin
Imagine you are working in a small bakery that sells fresh pastries. Customers do not arrive
in a fixed pattern — sometimes they come one by one, sometimes three at a time,
sometimes none for ten minutes. The bakery owner wonders:
“On average, if 5 customers arrive every 10 minutes, what is the chance that exactly 7
customers will arrive in the next 10 minutes?”
This situation is exactly what Poisson distribution helps us answer. It tells us the probability
of a certain number of events happening in a fixed time or space interval, given that the
events occur randomly but with a known average rate.
Definition
The Poisson Probability Distribution is a discrete probability distribution that gives 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.
The probability function is:
Where:
• X = random variable representing the number of events
• x = actual number of events (0,1,2,3,…)
• λ = average number of events in the given interval
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• e = constant (≈ 2.718)
Assumptions of Poisson Distribution
For Poisson distribution to apply, certain conditions must hold true:
1. Events are independent – The occurrence of one event does not affect the
occurrence of another. (One customer coming to the bakery doesn’t influence
another).
2. Constant average rate – Events occur at a known and constant mean rate (λ).
3. No simultaneous events – Two events cannot occur at exactly the same instant.
4. Events are random – They occur unexpectedly, without a predictable pattern.
Properties of Poisson Distribution
1. Mean and Variance – Both are equal to λ. This is a unique property.
Example: If on average 5 customers come in 10 minutes, then Mean = 5 and Variance
= 5.
2. Skewness – Poisson distribution is positively skewed (tail towards right), but as λ
increases, it becomes nearly symmetric.
3. Additivity – If one Poisson distribution has mean λ1 and another has mean λ2, then
the combined distribution has mean λ1+λ2.
4. Limiting Case of Binomial – When the number of trials n is very large, and probability
of success p is very small such that np=λ n, the Binomial distribution approaches
Poisson.
5. Practical Applications –
o Number of printing errors on a page
o Number of calls received in a call centre per hour
o Number of buses arriving at a station in 10 minutes
o Occurrence of accidents in a city per day
Why is Poisson Important?
Poisson distribution is like a “mathematical magnifying glass” for rare events. It allows us to
measure the probability of unusual happenings that cannot be predicted exactly but follow
an average rate.
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For example, if an examiner wants to know “How likely is it that exactly 3 students in a class
of 60 will submit their homework late tomorrow, given the average is 2 late submissions per
day?” — Poisson is the answer.
(ii). Part II: The Normal Distribution
A Relatable Story
Think of your classroom. If you measure the height of every student, you’ll find some are
short, some are tall, but most are “average” height. If you plot all these heights on a graph,
it forms a beautiful bell-shaped curve. This curve is the Normal Distribution.
This distribution is so common in nature that it is often called the “Gaussian distribution” or
“Bell Curve”.
Definition
The Normal Probability Distribution is a continuous probability distribution that describes
data which tends to cluster around a mean (average).
Its probability density function is:
Where:
• μ = mean (centre of distribution)
• σ = standard deviation (spread of distribution)
• x = variable value
Properties of Normal Distribution
1. Bell-Shaped Curve – Symmetrical about the mean. Most observations are close to
the mean, fewer as you move away.
2. Mean = Median = Mode – All three central values are equal in a perfectly normal
distribution.
3. Empirical Rule (68-95-99.7 rule) –
o About 68% of data lies within 1 standard deviation of the mean.
o About 95% within 2 standard deviations.
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o About 99.7% within 3 standard deviations.
4. Total Area = 1 – The area under the curve equals 1, representing total probability.
5. Asymptotic Nature – The tails of the curve approach but never touch the x-axis.
6. Central Limit Theorem (CLT) – When we take large samples from any population, the
sampling distribution of the sample mean tends to be normal, regardless of the
population’s original distribution.
Importance of Normal Distribution
• Statistics and Research – Most statistical tests (like t-test, z-test) are based on the
assumption of normality.
• Business – Helps in quality control, stock market analysis, demand forecasting.
• Education – Useful in grading students on a curve.
• Medicine – Normal distribution explains biological measures like blood pressure, IQ,
cholesterol levels.
Poisson vs Normal – The Link
Interestingly, the Poisson and Normal distributions are connected. When the average rate
(λ) in a Poisson distribution becomes very large, its shape begins to resemble a Normal
distribution.
Thus, Poisson is for rare events with small averages, and Normal is for continuous natural
data around a mean.
Wrapping It Up with a Simple Analogy
Think of Poisson distribution as the math that explains “counting sudden sparks” — like
counting how many raindrops hit your window in a minute.
And think of Normal distribution as the math that explains “natural balance” — like how
most people’s height or weight gather around an average, with very few being extremely
short or tall.
Both are not just mathematical concepts but lenses through which we view the
uncertainties of life.
“This paper has been carefully prepared for educational purposes. If you notice any mistakes or
have suggestions, feel free to share your feedback.”
