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GNDU Question Paper-2022

Bachelor of Business Administration

BBA 3

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

-6A

=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

(ii) Find Median wage from the following data:

Wages (Rs.)

No. of

workers

4. (i) Calculate Mean Deviation from Mean from the following data:

140 - 150

150 160

160 170

170 180

180 190

190 200

(ii) Calculate Standard Deviation for the following data:

SECTION-C

5.(i) Calculate Coefficient of Correlation between X and Y:

(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

6. (i) Fit a linear trend by principle of least squares for the following data:

Year

1997

1998

1999

2000

2001

2002

2003

Production

(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

(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

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

-6A

=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 wayflute three times, subtract

six violins, add seven pianos, and finally add two soft drumbeatsyoull 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 teachers mysterious instruction is exactly a polynomial in Aa linear combination

of I,A,A2,A3. When the combination evaluates to the zero matrix, weve 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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Well

prove this in two friendly ways:

1. by using the CayleyHamilton theorem (the elegant, theory-first route), and

2. by direct calculation (the hands-on, arithmetic route).

Both roads will meet at the same destinationsilence.

1) The elegant route: CayleyHamilton in action

The CayleyHamilton 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 lets 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 hasnt

vanished; its 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

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 students marks in a test. The marbles are already partitioned into six small

boxes labelled 010, 1020, , 5060. 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

boxs middle point. That gives her a quick, reliable estimate of the class average.

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This is exactly the method well use for the given grouped data.

Given data (classes and frequencies):

Classes (marks): 010, 1020, 2030, 3040, 4050, 5060

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 010 = (0 + 10) / 2 = 5

Midpoint of 1020 = (10 + 20) / 2 = 15

Midpoint of 2030 = (20 + 30) / 2 = 25

Midpoint of 3040 = (30 + 40) / 2 = 35

Midpoint of 4050 = (40 + 50) / 2 = 45

Midpoint of 5060 = (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 2030 marks interval (which had the highest

frequency: 34 students). So, picturing Ms. Raos 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 010 kg,

1020 kg, etc. He doesnt weigh every single grain; he simply assumes each sack in a bin is

the bins 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 bakers 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 2030

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.)

No. of

workers

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 earnsnot the richest one, not the poorest oneyou

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 lets 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

󹸱󹸲󹸰 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)

Lets 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 dont need to ask everyone. You

just need to look at the child standing at the middle positionthe one who separates the

line into two equal halves.

That special position is:

So, the 40th workers wage will be the median wage.

󼩕󼩖󼩗󼩘󼩙󼩚 Step 3: Prepare Cumulative Frequency (CF) Table

Now, lets arrange the data in cumulative form so we can easily locate the 40th worker.

Wages (Rs.)

No. of Workers

Cumulative Frequency (CF)

30 (12+18)

57 (30+27)

71 (57+14)

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, youll notice the 40th villager is standing inside this group of

₹15 earners. Thats why the median wage is ₹15.

So, even though some earn more and some earn less, the middle groundthe fair

representation of the typical workeris ₹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:

140 - 150

150 160

160 170

170 180

180 190

190 200

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 students height differs from the

class mean by about 10.56 units. In the same way, for our grouped data, the typical distance

of a classs 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:

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Ans: Imagine youre the manager of a small carnival game stall and youve 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: Whats

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). Lets 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:

Ans: A Different Beginning Lets 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:

"Dont 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."

Thats 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 Pearsons 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:

Y (F)

Now, lets make a table for calculations.

X²

Y²

135

225

128

256

196

169

121

144

100

Step 4: Summing Up

Now add them all:

ΣX = 45ΣX=45

ΣY = 108ΣY=108

ΣXY = 597ΣXY=597

ΣX

= 285ΣX2=285

ΣY

= 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 studentsgenerally, 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 1almost

perfect!

Thats 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

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 Xaviers marks is 100,

find the standard deviation of Yaras marks. Ill 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

(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 wasnt 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; its about stories hidden inside numbers. If you listen

carefully, numbers can tell you what happened in the past, whats happening now, and even

whisper about the future.

This clicked with Riya. She began to see numbers as storytellers. And today, were 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

Lets 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 theres

usually a general direction (upward, downward, or steady).

We use trend lines to capture this general movement.

2. Why Least Squares Method?

Now imagine youre trying to draw a straight line that passes as close as possible through

scattered points on a graph. You cant 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

1998

1999

2000

2001

2002

2003

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)

X²

1997

-3

-231

1998

-2

-176

1999

-1

-94

2000

2001

2002

196

2003

270

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 lets 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

1995

150

1996

165

1997

190

1998

210

1999

220

2000

230

2001

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 farmers market with her grandfather. She noticed that her grandfather

kept saying things like, Back in 1994, this rice was only ₹100. Now its ₹250!

Her grandfather was unknowingly using index numbers! He was comparing todays price

with a past reference year. Thats 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, Thats where trend lines help us, beta. They dont 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 todays weather with yesterdays.

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

(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 its 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 youll

find the last slice of pizza in the fridge when youre hungry at midnight.

In this answer, well not only solve the given problems but also explain them in such a

crystal-clear way that youll feel as if youre reading a short storybook rather than a math

solution.

Part 1: Tossing a Fair Coin 3 Times

Lets 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

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.

Lets 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, lets find them:

H H T

H T H

T H H

Thats 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

Lets 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

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(ii). Part 2: Selection of Candidates

Now lets 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, hadnt 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 isnt just numbers its 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 its 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

students 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.

Lets 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 doesnt 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, youll 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

populations 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 peoples 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.