Business Analytics (MBA-7302) hptu MBA DETAILED NOTES OF CURRENT SYLLABUS
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š INTRODUCTION TO BASIC STATISTICS
š¹ 1. MEASURES OF CENTRAL TENDENCY
Yeh wo values hoti hain jo data ko ek single representative value se summarize karti hain.
Common measures: Mean, Median, Mode
š§® (A) MEAN (Arithmetic Mean)
(i) Individual Series
Formula:
XĖ=N∑X
Example:
X = 10, 20, 30, 40, 50
XĖ=510+20+30+40+50=30
(ii) Discrete Series
Formula:
XĖ=∑f∑fX
Example:
| X | f |
|---|---|
| 10 | 2 |
| 20 | 3 |
| 30 | 5 |
XĖ=(2+3+5)(10×2+20×3+30×5)=10260=26
(iii) Continuous Series
Formula (using class intervals):
XĖ=A+∑f∑fd×i
Where:
A = Assumed mean,
d = (X - A)/i,
i = Class interval width.
Example:
| Class | f | Midpoint (X) | d = (X-A)/i | f*d |
|---|---|---|---|---|
| 0–10 | 4 | 5 | -2 | -8 |
| 10–20 | 6 | 15 | -1 | -6 |
| 20–30 | 10 | 25 | 0 | 0 |
| 30–40 | 8 | 35 | 1 | 8 |
| 40–50 | 2 | 45 | 2 | 4 |
XĖ=25+30(−2)×10=24.33
š (B) MEDIAN
Median = Middle value that divides data into two equal parts.
(i) Individual Series
Steps:
-
Arrange data in ascending order.
-
If n is odd → Median = 2n+1th item
If n is even → Median = Average of 2nth and 2n+1th items.
Example:
X = 10, 15, 20, 25, 30
Median = 3rd item = 20
(ii) Discrete Series
Steps:
-
Make cumulative frequency (c.f.)
-
Find 2N+1th item → see which class or value contains it.
Example:
| X | f | c.f |
|---|---|---|
| 10 | 2 | 2 |
| 20 | 3 | 5 |
| 30 | 5 | 10 |
N=10 → 210+1=5.5th item → lies in value 30
Median = 30
(iii) Continuous Series
Formula:
Median=L+(f2N−Cf)×i
Where:
L = Lower boundary of median class
Cf = cumulative frequency before median class
f = frequency of median class
i = class width
Example:
| Class | f | c.f |
|---|---|---|
| 0–10 | 4 | 4 |
| 10–20 | 6 | 10 |
| 20–30 | 10 | 20 |
| 30–40 | 8 | 28 |
N=28 → N/2 =14 → median class = 20–30
Median=20+10(14−10)×10=24
š (C) MODE
Mode = Most frequent or most repeated value.
(i) Individual Series
Value which occurs maximum times.
Example:
10, 20, 20, 25, 30
Mode = 20
(ii) Discrete Series
Value with highest frequency.
Example:
| X | f |
|---|---|
| 10 | 3 |
| 20 | 7 |
| 30 | 5 |
Mode = 20
(iii) Continuous Series
Formula:
Mode=L+(2f1−f0−f2f1−f0)×i
Where:
L = Lower boundary of modal class
f₁ = frequency of modal class
f₀ = frequency of class before modal class
f₂ = frequency of class after modal class
i = class width
Example:
| Class | f |
|---|---|
| 0–10 | 5 |
| 10–20 | 8 |
| 20–30 | 12 |
| 30–40 | 10 |
Modal class = 20–30
Mode=20+2(12)−8−10(12−8)×10=20+64×10=26.67
š MEASURES OF DISPERSION (Detailed Notes + Numericals)
Dispersion = data ke spread (kitna data mean ke aaspaas scattered hai).
Jitna zyada dispersion, utni kam consistency.
š¹ 1. RANGE
Meaning:
Range sabse simple measure hai jo maximum aur minimum value ka difference batata hai.
Formula:
R=L−SWhere,
L = Largest value,
S = Smallest value
Coefficient of Range:
C.R.=L+SL−SExample:
Data = 25, 28, 35, 40, 50, 60, 65
š Step 1:
Largest (L) = 65
Smallest (S) = 25
š Step 2:
Coefficient of Range=65+2565−25=9040=0.444✅ Final Answer:
Range = 40, Coefficient of Range = 0.444
š¹ 2. QUARTILE DEVIATION (Q.D.)
Meaning:
Quartile Deviation (Semi-Interquartile Range) middle 50% data ke spread ko batata hai.
Yeh 1st quartile (Q₁) aur 3rd quartile (Q₃) ke difference ka half hota hai.
Formulae:
Q.D.=2Q3−Q1Coefficient of Q.D.:
C.Q.D.=Q3+Q1Q3−Q1Interquartile Range (I.Q.R.):
I.Q.R.=Q3−Q1Example (Continuous Series):
| Class Interval | Frequency (f) | Cumulative Frequency (c.f.) |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 10 | 15 |
| 20–30 | 15 | 30 |
| 30–40 | 10 | 40 |
| 40–50 | 5 | 45 |
Total (N) = 45
Step 1: Find Q₁ and Q₃
Qn=L+(f4nN−Cf)×iFor Q₁ (N/4 = 11.25):
→ Lies in 10–20 class
L = 10, Cf = 5, f = 10, i = 10
For Q₃ (3N/4 = 33.75):
→ Lies in 30–40 class
L = 30, Cf = 30, f = 10, i = 10
Step 2: Find Q.D., I.Q.R., C.Q.D.
I.Q.R.=33.75−16.25=17.5 Q.D.=217.5=8.75 C.Q.D.=(33.75+16.25)17.5=5017.5=0.35✅ Final Answers:
-
Q₁ = 16.25
-
Q₃ = 33.75
-
I.Q.R. = 17.5
-
Q.D. = 8.75
-
Coefficient of Q.D. = 0.35
š¹ 3. MEAN DEVIATION (M.D.)
Meaning:
Mean Deviation measures average of absolute deviations (|X−A|) from a measure of central tendency (Mean, Median, or Mode).
Formulae:
M.D.=N∑∣X−A∣For discrete/continuous data:
M.D.=∑f∑f∣X−A∣Coefficient of M.D.:
C.M.D.=AM.D.(where A = Mean, Median or Mode used)
Example (Discrete Series):
| X | f |
|---|---|
| 10 | 2 |
| 20 | 3 |
| 30 | 5 |
Step 1: Find Mean
XĖ=∑f∑fX=10(10×2+20×3+30×5)=10260=26Step 2: Find |X−X̄| and f|X−X̄|
| X | f | |X−X̄| | f|X−X̄| |
|---|---|-----------|-----------|
| 10 | 2 | 16 | 32 |
| 20 | 3 | 6 | 18 |
| 30 | 5 | 4 | 20 |
Ī£f|X−X̄| = 70
Step 3:
M.D.=1070=7 C.M.D.=267=0.269✅ Final Answer:
M.D. = 7
Coefficient of M.D. = 0.269
š¹ 4. STANDARD DEVIATION (S.D.)
Meaning:
Standard deviation measures average deviation from mean in squared form, then square-rooted.
It is the most stable and accurate measure of dispersion.
Formulae:
Ļ=N∑(X−XĖ)2For discrete:
Ļ=∑f∑f(X−XĖ)2For continuous (Step-Deviation Method):
Ļ=∑f∑fd2−(∑f∑fd)2×iCoefficient of Variation (C.V.):
C.V.=XĖĻ×100Example (Continuous Series):
| Class | f | Mid (X) |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 10 | 15 |
| 20–30 | 15 | 25 |
| 30–40 | 10 | 35 |
| 40–50 | 5 | 45 |
Step 1: Assume Mean A = 25, i = 10
Find deviations: d = (X−A)/i
| X | f | d | f*d | f*d² |
|---|---|---|---|---|
| 5 | 5 | -2 | -10 | 20 |
| 15 | 10 | -1 | -10 | 10 |
| 25 | 15 | 0 | 0 | 0 |
| 35 | 10 | 1 | 10 | 10 |
| 45 | 5 | 2 | 10 | 20 |
Ī£f = 45, Ī£fd = 0, Ī£fd² = 60
Step 2: Formula
Ļ=4560−(450)2×10 Ļ=1.333×10=11.55Step 3: Mean = 25
C.V.=2511.55×100=46.2%✅ Final Answer:
-
Standard Deviation = 11.55
-
Coefficient of Variation = 46.2%
š FINAL REVISION TABLE
| Measure | Formula | Coefficient | Example Result |
|---|---|---|---|
| Range | L−S | (L−S)/(L+S) | R=40, C=0.44 |
| Quartile Deviation | (Q₃−Q₁)/2 | (Q₃−Q₁)/(Q₃+Q₁) | QD=8.75, C=0.35 |
| Interquartile Range | Q₃−Q₁ | — | 17.5 |
| Mean Deviation | Ī£f | X−A | /Ī£f |
| Standard Deviation | √(Ī£f(X−X̄)²/Ī£f) | (Ļ/X̄)×100 | SD=11.55, CV=46.2% |
š CHEBYSHEV’S THEOREM
š¹ (A) Hinglish Explanation
š Meaning:
Chebyshev’s Theorem ek universal rule hai jo batata hai ki koi bhi data distribution (chahe normal ho ya non-normal) me kitne percentage values mean ke certain distance ke andar rahengi.
Iska matlab:
Chahe data skewed ho ya random ho — Chebyshev’s theorem hamesha apply hoti hai.
š Formula:
P=1−k21Where:
-
P = proportion of data within k standard deviations from mean
-
k = number of standard deviations (k > 1)
š” Matlab Simple Bhasha Me:
Agar tum mean se k standard deviation duri tak values ko lo,
to kam se kam (1 - 1/k²) proportion of data us range ke andar hoga.
š Example Values:
| k | Minimum % of data within k SD |
|---|---|
| 2 | 1−41=0.75 → 75% |
| 3 | 1−91=0.888... → 88.9% |
| 4 | 1−161=0.9375 → 93.75% |
š¢ Example Question:
Ek company ke salaries ka mean ₹40,000 hai aur standard deviation ₹5,000 hai.
Find minimum proportion of employees having salary between ₹30,000 and ₹50,000.
Solution:
-
Mean = 40,000
-
SD = 5,000
-
Range = 30,000 to 50,000 → i.e., ±10,000 from mean
-
So, k=5,00010,000=2
✅ At least 75% employees ki salaries 30,000 se 50,000 ke beech hongi.
š§ Key Points:
-
Yeh rule sirf minimum guarantee deta hai (actual data zyada ho sakta hai).
-
Chebyshev’s Theorem applies to any type of distribution (not only normal).
-
Useful jab humein distribution shape nahi pata hota.
š¹ (B) Easy English Explanation
š Meaning:
Chebyshev’s Theorem states that for any data distribution, no matter how skewed or irregular it is,
a certain minimum percentage of data will always lie within a specific number of standard deviations from the mean.
Formula:
P=1−k21, where k>1Here,
-
P = proportion of observations within k standard deviations
-
k = number of standard deviations away from the mean
Interpretation:
-
At least 75% of values lie within 2 SDs of the mean.
-
At least 89% of values lie within 3 SDs.
-
At least 94% of values lie within 4 SDs.
Example:
If the mean weight of people = 60 kg and standard deviation = 5 kg,
find minimum % of people between 50 kg and 70 kg.
✅ So, at least 75% of people weigh between 50 and 70 kg.
Key Takeaways:
-
Chebyshev’s Theorem gives minimum limits — actual data may have higher % inside that range.
-
Works for any type of data — normal or not.
-
Helps understand data spread when distribution shape is unknown.
š§¾ Comparison with Normal Distribution (for reference)
| k (SD) | Chebyshev Minimum % | Normal Distribution (Actual) |
|---|---|---|
| 1 | 0% (Not defined) | 68.3% |
| 2 | 75% | 95.4% |
| 3 | 88.9% | 99.7% |
✅ In Short (Exam Line):
“According to Chebyshev’s theorem, for any distribution, at least (1 − 1/k²) of observations lie within k standard deviations from the mean, where k > 1.”
___________________________________________________________________
š UNIT II – PROBABILITY AND PROBABILITY DISTRIBUTIONS (DETAILED NOTES)
š¹ PART A: THEORY OF PROBABILITY
1. Concept of Probability
Probability ka matlab hota hai kisi event ke hone ki sambhavana (likelihood or chance).
Kisi event ke hone ki probability 0 aur 1 ke beech hoti hai.
0≤P(E)≤1
-
P(E) = 0 → Event impossible
-
P(E) = 1 → Event certain
-
P(E) = 0.5 → Event equally likely
2. Basic Terms
| Term | Definition | Example |
|---|---|---|
| Experiment | Random process with uncertain result | Tossing a coin |
| Trial | Each repetition of experiment | Throwing a die once |
| Sample Space (S) | All possible outcomes | For a die, S = {1,2,3,4,5,6} |
| Event (E) | Subset of sample space | Getting even number = {2,4,6} |
| Mutually Exclusive Events | Dono ek sath nahi ho sakte | Getting head & tail together |
| Exhaustive Events | Sab possible outcomes included | Die → 6 outcomes |
| Equally Likely Events | Har outcome ka chance same | Fair coin toss |
3. Classical Definition of Probability
P(E)=Total number of outcomesNumber of favourable outcomes
Example 1:
A coin is tossed → P(getting Head) = 1/2
Example 2:
A die is rolled → P(getting even number) = 3/6 = 1/2
4. Types of Probability
| Type | Meaning | Example |
|---|---|---|
| Theoretical (Classical) | Based on reasoning | Coin → 1/2 |
| Empirical | Based on past observation | Rainy days data |
| Subjective | Based on judgment or belief | Chance of exam success |
⚙️ 5. Laws of Probability
(A) Addition Law (OR Rule)
Used when we want either event A or event B to occur.
(i) For Mutually Exclusive Events
P(A or B)=P(A)+P(B)
Example:
Rolling a die:
P(2 or 4) = 1/6 + 1/6 = 1/3
(ii) For Non-Mutually Exclusive Events
P(A or B)=P(A)+P(B)−P(A and B)
Example:
Let P(A)=0.5, P(B)=0.4, P(A and B)=0.2
P(A or B)=0.5+0.4−0.2=0.7
(B) Multiplication Law (AND Rule)
Used when we want both events A and B to occur.
(i) Independent Events
P(A and B)=P(A)×P(B)
Example:
Two coins tossed → both heads?
P(A)=1/2,P(B)=1/2⇒P(A and B)=1/4
(ii) Dependent Events
P(A and B)=P(A)×P(B∣A)
Example:
Two cards drawn without replacement:
P(1stAce)=4/52, P(2ndAce∣1stAce)=3/51 P(both Ace)=4/52×3/51=1/221
š 6. Conditional Probability
It measures the probability of event A given that event B has already occurred.
P(A∣B)=P(B)P(A and B)
Example:
P(A∩B)=0.2, P(B)=0.4
P(A∣B)=0.2/0.4=0.5
š 7. Bayes’ Theorem
š¹ Concept:
Used to find reverse (posterior) probability — probability of a cause given an observed effect.
š¹ Formula:
P(Ai∣B)=∑j=1nP(Aj)P(B∣Aj)P(Ai)P(B∣Ai)
Where:
-
A1,A2,...An = mutually exclusive causes
-
B = observed event (effect)
š¹ Example:
A company has 3 machines A₁, A₂, A₃ producing 20%, 30%, 50% of items respectively.
Defective rate: 1%, 2%, 3%.
Find probability that a defective item came from A₃.
Solution:
P(A1)=0.2,P(A2)=0.3,P(A3)=0.5 P(D∣A1)=0.01,P(D∣A2)=0.02,P(D∣A3)=0.03
Now:
P(D)=(0.2×0.01)+(0.3×0.02)+(0.5×0.03)=0.023 P(A3∣D)=0.0230.5×0.03=0.652
✅ Probability defective item is from A₃ = 65.2%
š¹ PART B: PROBABILITY DISTRIBUTIONS
1. BINOMIAL DISTRIBUTION
š¹ Concept:
Used when an experiment has two possible outcomes:
Success (p) or Failure (q = 1−p), repeated n times.
š¹ Conditions:
-
Fixed number of trials (n)
-
Two outcomes per trial (Success/Failure)
-
Probability (p) remains constant
-
Trials are independent
š¹ Formula:
P(X=x)=(xn)pxqn−x (xn)=x!(n−x)!n!
Mean = np
Variance = npq
š¹ Example:
Find probability of exactly 3 heads in 5 coin tosses.
n=5, x=3, p=0.5, q=0.5
P(X=3)=(35)(0.5)3(0.5)2=10(0.03125)=0.3125
✅ Probability = 0.3125
2. POISSON DISTRIBUTION
š¹ Concept:
Used when events are rare but trials are large.
It’s a limiting form of Binomial when n→∞ and p→0.
š¹ Formula:
P(X=x)=x!e−mmx
Where:
-
m = mean = np
-
e = 2.71828
Mean = m
Variance = m
š¹ Example:
If 3 typing errors occur per page on average,
Find probability that a page has no error.
m=3,x=0 P(0)=0!e−330=e−3=0.0498
✅ Probability = 0.0498 (≈5%)
3. NORMAL DISTRIBUTION
š¹ Concept:
Normal distribution ek continuous, symmetrical, bell-shaped curve hoti hai.
Most real-life data (height, weight, IQ, marks) follow normal distribution.
š¹ Formula:
f(x)=Ļ2Ļ1e−2Ļ2(x−μ)2
Where:
μ = mean, Ļ = standard deviation
š¹ Properties:
-
Symmetrical about mean
-
Mean = Median = Mode
-
Total area under curve = 1
-
Area within ±1Ļ = 68.27%
Area within ±2Ļ = 95.45%
Area within ±3Ļ = 99.73%
š¹ Example:
Heights of students follow N(μ=170, Ļ=5).
Find probability of height between 165 and 175.
z1=5165−170=−1,z2=5175−170=1
Area between −1 and +1 = 0.6826
✅ Probability = 68.26%
4. COMPARISON TABLE
| Feature | Binomial | Poisson | Normal |
|---|---|---|---|
| Type | Discrete | Discrete | Continuous |
| Parameters | n, p | m | μ, Ļ |
| Mean | np | m | μ |
| Variance | npq | m | ϲ |
| Shape | Symmetrical/Skewed | Right skewed | Bell-shaped |
| Example | Coin toss | Rare accidents | Heights, IQ |
š§ KEY TAKEAWAYS
| Concept | Key Idea | Formula |
|---|---|---|
| Probability | Chance of event | Favourable/Total |
| Addition Law | “OR” | P(A)+P(B)−P(A∩B) |
| Multiplication Law | “AND” | P(A)×P(B |
| Conditional Prob. | Given event | P(A |
| Bayes Theorem | Reverse prob. | P(Ai |
| Binomial | Repeated trials | nCx pĖ£ qāæ⁻Ė£ |
| Poisson | Rare events | e⁻įµ mĖ£/x! |
| Normal | Continuous data | (1/Ļ√2Ļ)e⁻(x−μ)²/2ϲ |
š” Short Easy English Summary
-
Probability: Measures likelihood of an event (0–1).
-
Addition Law: For “either/or” events.
-
Multiplication Law: For “both/and” events.
-
Bayes Theorem: Finds reverse probability (cause after effect).
-
Binomial Distribution: Repeated trials (success/failure).
-
Poisson Distribution: Rare events.
-
Normal Distribution: Continuous, bell-shaped, natural phenomena.
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