Business Mathematics and Statistics व्यावसायिक गणित व सांख्यिकी

Quick summary

Ratio, proportion and percentage are the basic tools of business arithmetic. They are used in sharing profits among partners, mixing goods, calculating profit and loss, discounts, commission and changes in prices. This topic covers the types of ratios, the properties of proportion, and the main percentage applications asked in university exams.

Key points

• Ratio (गुणोत्तर) compares two quantities of the same kind: a : b = a/b. It has no unit.
• A ratio stays the same if both terms are multiplied or divided by the same non-zero number.
• Types: duplicate (a² : b²), triplicate (a³ : b³), sub-duplicate (√a : √b), inverse (b : a), compound (product of ratios).
• Proportion (प्रमाण): equality of two ratios, a : b = c : d, i.e. a × d = b × c (product of extremes = product of means).
• Continued proportion: a : b = b : c, so b² = ac; b is the mean proportional.
• Direct variation: one quantity rises in the same ratio as the other; inverse variation: one rises as the other falls (more workers, fewer days).
• Percentage means "per hundred"; x% = x/100.
• Successive changes are multiplied, not added: +20% then −20% gives a net 4% fall.
• Profit and loss are always calculated on cost price (CP) unless stated otherwise.
• Trade discount is on list (marked) price; cash discount is on the price after trade discount.

Important terms

• Antecedent / consequent – first and second terms of a ratio.
• Fourth proportional – x in a : b = c : x.
• Third proportional – c in a : b = b : c.
• Commission – percentage of sales paid to an agent.
• Brokerage – commission paid to a broker on deals.

Formats and formulas

• Dividing an amount S in ratio a : b : c → shares S·a/(a+b+c), S·b/(a+b+c), S·c/(a+b+c)
• Fourth proportional of a, b, c: x = bc/a
• Mean proportional of a and c: b = √(ac)
• Percentage change = (Change ÷ Original) × 100
• Profit % = (Profit ÷ CP) × 100; SP = CP × (100 + profit%)/100
• Net effect of two successive changes x% and y% = x + y + xy/100

Worked examples:

5 important questions

Q1. Find the fourth proportional to 3, 5 and 12.

3 : 5 = 12 : x, so 3x = 5 × 12 = 60 and x = 20.

Q2. A salesman gets 5% commission on sales up to ₹50,000 and 8% on sales above that. Find his commission on sales of ₹80,000.

On ₹50,000: 5% = ₹2,500. On the remaining ₹30,000: 8% = ₹2,400. Total commission = ₹4,900.

Q3. Divide ₹1,20,000 among P, Q and R so that P : Q = 2 : 3 and Q : R = 4 : 5.

Make Q common: P : Q = 8 : 12 and Q : R = 12 : 15, so P : Q : R = 8 : 12 : 15 (35 parts). Each part = 1,20,000 ÷ 35 ≈ ₹3,428.57. Shares: P ≈ ₹27,428.57, Q ≈ ₹41,142.86, R ≈ ₹51,428.57 (total ₹1,20,000).

Q4. If 12 workers finish a job in 15 days, how many days will 20 workers take?

Inverse variation: workers × days is constant. 12 × 15 = 180 worker-days. Days = 180 ÷ 20 = 9 days.

Q5. A trader marks goods 25% above cost and allows a 10% discount. Find his profit percent.

Let CP = ₹100. Marked price = ₹125. SP = 125 × 0.90 = ₹112.50. Profit = ₹12.50 on ₹100, so profit = 12.5%.

Exam tip

Show the formula, the substitution and the final answer with a unit (₹, days, %). Most marks are lost by skipping the working.

Quick summary

Interest is the price paid for using borrowed money. Under simple interest it is charged only on the original principal; under compound interest it is charged on principal plus the interest already added. This topic covers both methods, conversion periods, the effective rate of interest and compound depreciation, with worked numericals.

Key points

• Principal (P) is the sum borrowed or invested; amount (A) = principal + interest.
• Simple interest (सरळ व्याज) is the same every year.
• Compound interest (चक्रवाढ व्याज) grows every period because interest earns interest.
• Conversion period: the interval after which interest is added to principal – yearly, half-yearly, quarterly, monthly.
• For half-yearly compounding: rate is halved and periods are doubled; for quarterly: rate ÷ 4, periods × 4.
• Nominal rate is the stated annual rate; effective rate is the actual annual rate after compounding.
• For the same P, R and time over 1 year, CI is greater than SI; for exactly 1 year with yearly compounding they are equal.
• Depreciation on reducing balance follows the compound formula with a minus sign.

Important terms

• Rate of interest (R) – interest on ₹100 for one year.
• Period (N or n) – time in years (or number of conversion periods).
• Effective rate – rate that gives the same interest in one year with yearly compounding.
• Compound depreciation – fall in value at a fixed percentage of the reduced value every year.

Formats and formulas

• SI = (P × N × R) ÷ 100; A = P + SI
• CI amount: A = P(1 + r/100)ⁿ; CI = A − P
• Compounded k times a year: A = P(1 + r/(100k))^(nk)
• Effective rate = [(1 + r/(100k))^k − 1] × 100
• CI − SI for 2 years = P(R/100)²
• Depreciated value: V = P(1 − r/100)ⁿ

Worked examples (P = ₹10,000):

5 important questions

Q1. Find the SI on ₹15,000 at 9% p.a. for 2 years 6 months.

N = 2.5 years. SI = 15,000 × 2.5 × 9 ÷ 100 = ₹3,375. Amount = ₹18,375.

Q2. At what rate will a sum double itself in 8 years at simple interest?

Interest = P when it doubles. P = P × 8 × R ÷ 100, so R = 100 ÷ 8 = 12.5% p.a.

Q3. Find the compound interest on ₹20,000 at 12% p.a. for 1 year, compounded quarterly.

Rate per quarter = 3%, n = 4. A = 20,000 × (1.03)⁴ = 20,000 × 1.12550881 = ₹22,510.18. CI = ₹2,510.18.

Q4. A machine costs ₹50,000 and depreciates at 10% p.a. on the reducing balance. Find its value after 2 years and the total depreciation.

V = 50,000 × (0.90)² = 50,000 × 0.81 = ₹40,500. Total depreciation = 50,000 − 40,500 = ₹9,500.

Q5. The difference between CI and SI on a sum for 2 years at 5% p.a. is ₹50. Find the sum.

CI − SI = P(R/100)² → 50 = P × (0.05)² = 0.0025P, so P = 50 ÷ 0.0025 = ₹20,000.

Exam tip

Before using the CI formula, convert the rate and the number of periods for the compounding frequency. Write (1.05)⁴, not 1.05 × 4.

Quick summary

An annuity is a series of equal payments made at equal intervals, like monthly rent, insurance premiums, SIP instalments or loan EMIs. This topic explains the types of annuities and how to find their future (accumulated) value and present value. It ends with the EMI formula used by banks for home, vehicle and personal loans.

Key points

• Annuity (वार्षिकी/हप्ता): equal periodic payments for a fixed number of periods.
• Ordinary annuity (immediate): payment at the end of each period (most loans, EMIs).
• Annuity due: payment at the beginning of each period (rent, insurance premium, SIP on the 1st).
• Annuity certain: fixed number of payments; contingent annuity: depends on an event (life insurance pension); perpetuity: payments go on forever.
• Future value (amount) answers "how much will my instalments grow to?"
• Present value answers "what lump sum today equals all future instalments?"
• An annuity due earns one more period of interest than an ordinary annuity, so its value = ordinary value × (1 + i).
• Sinking fund: equal amounts set aside regularly to repay a debt or replace an asset later; it uses the future-value formula.
• EMI (Equated Monthly Instalment) is an ordinary annuity whose present value equals the loan amount; each EMI has an interest part (large at first) and a principal part (grows over time).

Important terms

• i – rate of interest per period (annual rate ÷ periods per year, as a decimal).
• n – total number of payments.
• P (or a) – periodic payment.
• Amortisation – repaying a loan through instalments; the schedule shows interest and principal in each EMI.
• Perpetuity – annuity with no end.

Formats and formulas

• Future value of ordinary annuity: A = P × [(1 + i)ⁿ − 1] ÷ i
• Present value of ordinary annuity: V = P × [1 − (1 + i)⁻ⁿ] ÷ i
• Annuity due: multiply the ordinary result by (1 + i)
• Present value of perpetuity: V = P ÷ i
• EMI = L × r × (1 + r)ⁿ ÷ [(1 + r)ⁿ − 1], where L = loan, r = monthly rate, n = number of months
• Sinking fund instalment: P = A × i ÷ [(1 + i)ⁿ − 1]

Worked examples:

EMI example: loan ₹1,00,000 at 12% p.a. for 12 months. r = 12 ÷ 12 = 1% = 0.01, n = 12, (1.01)¹² = 1.126825.
EMI = 1,00,000 × 0.01 × 1.126825 ÷ 0.126825 = ₹8,884.88 (approx).
Total paid ≈ 12 × 8,884.88 = ₹1,06,618.56, so total interest ≈ ₹6,618.56.
First month: interest = 1% of 1,00,000 = ₹1,000; principal repaid = 8,884.88 − 1,000 = ₹7,884.88.

5 important questions

Q1. Distinguish between ordinary annuity and annuity due.

In an ordinary annuity each payment is made at the end of the period (loan EMI). In an annuity due it is made at the beginning (rent paid in advance). For the same P, i and n, an annuity due has a higher value by a factor (1 + i).

Q2. What is a sinking fund?

A fund built by setting aside equal amounts at regular intervals, invested at compound interest, to collect a known sum at a future date, e.g. to repay debentures or replace a machine.

Q3. Find the amount of an annuity of ₹2,000 paid at the end of every 6 months for 2 years at 8% p.a. compounded half-yearly.

i = 4% = 0.04, n = 4. A = 2,000 × [(1.04)⁴ − 1] ÷ 0.04 = 2,000 × (1.16985856 − 1) ÷ 0.04 = 2,000 × 4.246464 = ₹8,492.93.

Q4. Find the EMI for a loan of ₹1,00,000 at 12% p.a. repayable in 12 monthly instalments. Also find the total interest.

Steps as in the worked example: r = 0.01, n = 12 → EMI ≈ ₹8,884.88 → total paid ≈ ₹1,06,618.56 → interest ≈ ₹6,618.56. Mention that interest falls and principal rises every month.

Q5. A company must repay ₹5,00,000 after 5 years. How much should it set aside each year at 10% p.a. compound interest?

Sinking fund: P = 5,00,000 × 0.1 ÷ [(1.1)⁵ − 1] = 50,000 ÷ 0.61051 = ₹81,898.74 (approx) at the end of each year.

Exam tip

First decide: end-of-period or beginning, future value or present value. Then pick the formula. Write i per period, not per year.

Quick summary

A matrix is a rectangular arrangement of numbers in rows and columns. Businesses use matrices to store and combine data such as quantities sold, prices and costs across branches. A determinant is a single number linked with a square matrix; it is used to find the inverse of a matrix and to solve simultaneous equations by Cramer's rule.

Key points

• A matrix with m rows and n columns has order m × n.
• Types: row, column, square, zero (null), diagonal, scalar, identity (unit) matrix I, upper/lower triangular, symmetric (A' = A).
• Addition and subtraction are possible only for matrices of the same order (add corresponding elements).
• Multiplication AB is possible only if columns of A = rows of B; the result has order (rows of A) × (columns of B).
• In general AB ≠ BA (matrix multiplication is not commutative).
• Transpose A' (or Aᵀ): rows become columns.
• A determinant (निर्धारक) exists only for a square matrix.
• A matrix is singular if |A| = 0 (no inverse) and non-singular if |A| ≠ 0.
• Cramer's rule solves n equations in n unknowns using determinants, provided D ≠ 0.

Important terms

• Element – each number in a matrix, written aᵢⱼ (row i, column j).
• Minor – determinant left after deleting the row and column of an element.
• Cofactor – minor with sign (−1)^(i+j).
• Adjoint (adj A) – transpose of the cofactor matrix.
• Inverse A⁻¹ – matrix such that A × A⁻¹ = I.

Formats and formulas

• For A = [a b; c d] (rows separated by ;): |A| = ad − bc
• Inverse of 2 × 2: A⁻¹ = (1/|A|) × [d −b; −c a]
• General inverse: A⁻¹ = adj A ÷ |A|
• 3 × 3 determinant (expansion along row 1): |A| = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁)
• Cramer's rule (2 unknowns): x = Dx ÷ D, y = Dy ÷ D

Worked examples with A = [2 3; 1 4] and B = [1 0; 2 5]:

Check: A × A⁻¹ row 1 = (2×0.8 + 3×(−0.2), 2×(−0.6) + 3×0.4) = (1, 0). Correct.

5 important questions

Q1. Find the determinant of [1 2 3; 0 1 4; 5 6 0].

|A| = 1(1×0 − 4×6) − 2(0×0 − 4×5) + 3(0×6 − 1×5) = 1(−24) − 2(−20) + 3(−5) = −24 + 40 − 15 = 1.

Q2. When is a matrix singular? Can it have an inverse?

A square matrix is singular when its determinant is zero. A singular matrix has no inverse, because A⁻¹ = adj A ÷ |A| needs division by |A|.

Q3. Solve by Cramer's rule: 2x + 3y = 13 and x + 4y = 14.

D = |2 3; 1 4| = 8 − 3 = 5. Dx = |13 3; 14 4| = 52 − 42 = 10. Dy = |2 13; 1 14| = 28 − 13 = 15. x = 10 ÷ 5 = 2, y = 15 ÷ 5 = 3. Check: 2(2) + 3(3) = 13 and 2 + 4(3) = 14.

Q4. Two shops sell pens and notebooks. Shop A sells 10 pens and 5 notebooks; Shop B sells 8 pens and 12 notebooks. A pen costs ₹10 and a notebook ₹40. Find each shop's revenue using matrices.

Quantity matrix Q = [10 5; 8 12], price matrix P = [10; 40] (2 × 1). Revenue = QP = [10×10 + 5×40; 8×10 + 12×40] = [300; 560]. Shop A earns ₹300 and Shop B earns ₹560.

Q5. Find the inverse of A = [3 2; 4 3] and verify.

|A| = 9 − 8 = 1. A⁻¹ = (1/1) × [3 −2; −4 3] = [3 −2; −4 3]. Verify: row 1 of A × A⁻¹ = (3×3 + 2×(−4), 3×(−2) + 2×3) = (1, 0); row 2 = (4×3 + 3×(−4), 4×(−2) + 3×3) = (0, 1). So A × A⁻¹ = I.

Exam tip

Before multiplying, write the orders (2 × 2 · 2 × 1 = 2 × 1). It tells you instantly whether the product exists and what shape the answer has.

Quick summary

Statistics begins with data. This topic explains how data are collected (primary and secondary sources, census and sample methods), how raw data are classified into a frequency distribution, and how they are presented through tables, diagrams and graphs so that patterns can be seen at a glance.

Key points

• Statistics (सांख्यिकी) in the plural means numerical facts; in the singular, the science of collecting, organising, presenting, analysing and interpreting data.
• Primary data: collected first-hand for the purpose (original). Methods: direct personal interview, indirect oral investigation, questionnaire by post/e-mail/online form, schedules filled by enumerators, information through local correspondents.
• Secondary data: already collected by others (published reports, Census of India, RBI bulletins, journals, websites). Check reliability, suitability and adequacy before use.
• Census method: every item of the population is studied. Sample method: only a representative part is studied; it saves time and cost.
• Sampling methods: random (lottery, random number tables), stratified, systematic, cluster, and non-random (judgement, quota, convenience).
• Classification: arranging data into groups by common characteristics – geographical, chronological, qualitative, quantitative.
• A frequency distribution groups values into class intervals with their frequencies.
• Exclusive classes (0–10, 10–20: upper limit excluded) vs inclusive classes (0–9, 10–19).
• Tabulation: presenting data in rows and columns. Parts of a table: table number, title, caption (column heads), stub (row heads), body, head note, footnote, source.
• Diagrams: simple, multiple, subdivided and percentage bar diagrams; pie diagram. Graphs: histogram, frequency polygon, frequency curve, ogives (less-than and more-than cumulative frequency curves).

Important terms

• Population (universe) – all items under study.
• Sample – a part of the population selected for study.
• Class interval – the range of a class (upper limit − lower limit).
• Class mark (mid-value) – (lower limit + upper limit) ÷ 2.
• Cumulative frequency – running total of frequencies.
• Ogive – graph of cumulative frequencies; used to find median graphically.

Formats and formulas

• Class width = Upper limit − Lower limit
• Mid-value = (Lower limit + Upper limit) ÷ 2
• Pie diagram angle for a part = (Part ÷ Total) × 360°

Worked example. Marks of 20 students: 12, 25, 37, 8, 45, 22, 31, 18, 27, 39, 15, 28, 33, 5, 41, 24, 29, 19, 35, 26. Prepare a frequency distribution with classes 0–10, 10–20, ... (exclusive).

Pie example: a family spends ₹6,000 of ₹24,000 on rent. Angle = (6,000 ÷ 24,000) × 360° = 90°.

5 important questions

Q1. Distinguish between primary and secondary data.

Primary data are collected for the first time by the investigator for a specific purpose; they are original but costly and slow. Secondary data were collected earlier by someone else and are used again; they are cheap and quick but may not exactly suit the purpose.

Q2. What is a sample? State two merits of the sample method.

A sample is a small representative part of the population selected for study. Merits: it saves time, money and labour, and it is the only possible method when testing destroys the item (e.g. testing bulbs or medicines).

Q3. Explain the methods of collecting primary data.

Outline: direct personal investigation → indirect oral investigation → information from local agents/correspondents → mailed questionnaire → schedules through enumerators → online surveys → merits, demerits and suitability of each → qualities of a good questionnaire.

Q4. What is tabulation? Explain the parts of a good statistical table.

Outline: meaning and objectives → parts (number, title, head note, captions, stubs, body, footnote, source) with a small specimen table → rules of tabulation (clear title, units mentioned, totals, comparisons side by side).

Q5. Prepare a frequency distribution and draw a histogram and less-than ogive (numerical).

Steps: find range → decide classes (5–10) → tally marks → frequencies → cumulative frequencies → histogram: class limits on x-axis, frequency on y-axis, adjacent bars → ogive: plot upper limits against less-than cumulative frequencies and join smoothly (see the worked example).

Exam tip

Always write the title, units and source with any table or diagram. These small items carry marks and are often forgotten.

Quick summary

A measure of central tendency (average) is a single value that represents the whole data set. The main averages are the arithmetic mean, median and mode; geometric and harmonic means are used for special cases like growth rates and speeds. This topic gives their formulas for individual, discrete and continuous series, with a fully worked grouped-data example.

Key points

• An average (सरासरी / मध्यवर्ती प्रवृत्ती) should be rigidly defined, easy to understand, based on all values, not unduly affected by extreme values, and capable of further algebraic treatment.
• Arithmetic mean: sum of values ÷ number of values; based on all items but affected by extreme values.
• Median: the middle value when data are arranged in order; not affected by extreme values; useful for income, wages and qualitative data.
• Mode: the value that occurs most often; useful for sizes of shoes and clothes, and the most popular item.
• Weighted mean: used when items have different importance (index numbers, average marks with different credits).
• Geometric mean: average growth rate; harmonic mean: average speed or rate.
• Quartiles (Q1, Q3), deciles and percentiles divide data into 4, 10 and 100 parts.
• Empirical relation for moderately skewed data: Mode ≈ 3 Median − 2 Mean.
• Median can be found graphically from the intersection of the less-than and more-than ogives; mode from a histogram.

Important terms

• Median class – class where the cumulative frequency first reaches N/2.
• Modal class – class with the highest frequency.
• Step deviation – (x − A) ÷ h, used to simplify mean calculation.
• Bimodal – data with two modes.

Formats and formulas

• Mean (individual): x̄ = Σx ÷ n; (discrete/continuous): x̄ = Σfx ÷ N, where x = mid-value
• Step deviation: x̄ = A + (Σfd' ÷ N) × h, where d' = (x − A)/h
• Median (continuous): Md = L + [(N/2 − cf) ÷ f] × h
• Mode (continuous): Mo = L + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h
• L = lower limit of the class; cf = cumulative frequency before the class; f = frequency of the class; f₁ = modal class frequency, f₀ and f₂ = frequencies of the classes before and after; h = class width
• GM = (x₁ × x₂ × ... × xₙ)^(1/n); HM = n ÷ Σ(1/x)

Worked example (N = 40):

• Mean = 1,020 ÷ 40 = 25.5
• Median: N/2 = 20 → median class 20–30 (L = 20, cf = 13, f = 12, h = 10). Md = 20 + [(20 − 13) ÷ 12] × 10 = 20 + 5.83 = 25.83
• Mode: modal class 20–30 (f₁ = 12, f₀ = 8, f₂ = 10). Mo = 20 + [(12 − 8) ÷ (24 − 8 − 10)] × 10 = 20 + (4 ÷ 6) × 10 = 26.67

Check: 3 × 25.83 − 2 × 25.5 = 26.5, close to 26.67, as expected for nearly symmetric data.

5 important questions

Q1. What are the qualities of a good average?

It should be rigidly defined, simple to understand and calculate, based on all observations, least affected by extreme values and sampling fluctuations, and suitable for further algebraic treatment. The arithmetic mean satisfies most of these.

Q2. Find the median of 7, 3, 9, 12, 5, 10.

Arrange: 3, 5, 7, 9, 10, 12 (n = 6, even). Median = average of the 3rd and 4th items = (7 + 9) ÷ 2 = 8.

Q3. Calculate the mean, median and mode of a grouped frequency distribution (numerical).

Steps: mid-values → fx → mean → cumulative frequency → median class and formula → modal class and formula → interpret (see the worked example).

Q4. Compare mean, median and mode: merits, demerits and uses.

Outline: definitions → merits and demerits of each → uses (mean for marks and output; median for income, open-end classes; mode for fashion sizes and popular items) → empirical relation.

Q5. A vehicle travels 60 km at 30 km/h and returns at 60 km/h. Find the average speed.

Equal distances, so use the harmonic mean: HM = 2 ÷ (1/30 + 1/60) = 2 ÷ (3/60) = 40 km/h. (Check: total time = 2 h + 1 h = 3 h for 120 km = 40 km/h.)

Exam tip

Write the values of L, cf, f, f₀, f₂ and h separately before putting them in the formula. It avoids slips and earns step marks.

Quick summary

Two data sets can have the same average but very different spread. Dispersion (variation) measures how far the values are scattered around the average. This topic covers absolute measures (range, quartile deviation, mean deviation, standard deviation) and relative measures (their coefficients and the coefficient of variation), which are used to compare the consistency of two or more series.

Key points

• Dispersion (अपस्करण / विचलन) shows the reliability of an average: less dispersion means the average represents the data better.
• Absolute measures are in the original units (₹, kg); relative measures are pure numbers (ratios or percentages) used for comparison.
• Range = largest − smallest value; simplest but depends only on two extreme items.
• Quartile deviation (semi-interquartile range) = (Q3 − Q1)/2; ignores the top and bottom 25%, so it is not affected by extreme values; useful for open-end classes.
• Mean deviation: average of absolute deviations from mean or median (signs ignored).
• Standard deviation (σ) (प्रमाण विचलन): square root of the mean of squared deviations from the arithmetic mean; the best and most widely used measure (Karl Pearson, 1893).
• Variance = σ².
• Coefficient of variation (CV) = (σ ÷ mean) × 100. A series with lower CV is more consistent, uniform or stable.
• SD is not changed by adding a constant to every value (change of origin), but it is changed by multiplying (change of scale).
• Skewness measures lack of symmetry; kurtosis measures peakedness (introductory).

Important terms

• Consistency – less variation; lower CV.
• Q1, Q3 – values below which 25% and 75% of observations lie.
• Absolute deviation – deviation with the sign ignored, |x − x̄|.
• Karl Pearson's coefficient of skewness = (Mean − Mode) ÷ σ.

Formats and formulas

• Range = H − L; Coefficient of range = (H − L) ÷ (H + L)
• QD = (Q3 − Q1) ÷ 2; Coefficient of QD = (Q3 − Q1) ÷ (Q3 + Q1)
• Mean deviation = Σ|x − x̄| ÷ n (or Σf|x − x̄| ÷ N)
• σ = √[Σ(x − x̄)² ÷ n]; grouped: σ = √[Σf(x − x̄)² ÷ N]
• Shortcut: σ = √[Σfd² ÷ N − (Σfd ÷ N)²], d = x − A
• CV = (σ ÷ x̄) × 100

Worked example: 2, 4, 4, 4, 5, 5, 7, 9 (n = 8, mean = 40 ÷ 8 = 5).

• Range = 9 − 2 = 7; coefficient = 7 ÷ 11 = 0.64
• Mean deviation = 12 ÷ 8 = 1.5
• σ = √(32 ÷ 8) = √4 = 2; variance = 4
• CV = (2 ÷ 5) × 100 = 40%
• Q1 = value of the (n + 1)/4 = 2.25th item = 4; Q3 = 6.75th item = 5 + 0.75 × (7 − 5) = 6.5; QD = (6.5 − 4) ÷ 2 = 1.25

5 important questions

Q1. What is the coefficient of variation? Why is it used?

CV = (σ ÷ mean) × 100. It is a relative measure of dispersion used to compare the variability or consistency of two or more series, even when their units or averages differ. The series with the lower CV is more consistent.

Q2. Two workers' daily output: A has mean 50 units, σ 5; B has mean 40 units, σ 6. Who is more consistent?

CV of A = (5 ÷ 50) × 100 = 10%. CV of B = (6 ÷ 40) × 100 = 15%. A has the lower CV, so A is more consistent.

Q3. Calculate the standard deviation of a grouped frequency distribution (numerical).

Steps: mid-values → mean → deviations → squared deviations × f → σ = √[Σf(x − x̄)² ÷ N]. Example: classes 0–10, 10–20, 20–30, 30–40, 40–50 with f = 5, 8, 12, 10, 5 give mean 25.5, Σf(x − x̄)² = 5,790, variance 144.75, σ ≈ 12.03 and CV ≈ 47.2%.

Q4. Compare the various measures of dispersion.

Outline: range, QD, MD, SD with definition, formula, merits and demerits → why SD is the best (based on all values, algebraic treatment, used in sampling and correlation) → when QD is preferred (open-end classes).

Q5. Explain the meaning and objectives of measuring dispersion.

Outline: meaning → objectives (test reliability of average, compare consistency, control quality, basis for further analysis like correlation and regression) → properties of a good measure → absolute vs relative measures.

Exam tip

For comparison questions, never compare σ alone when the means differ; always compute the CV and state which series is more consistent.

Quick summary

Correlation measures the degree and direction of the relationship between two variables, such as advertising and sales or price and demand. Regression goes one step further: it gives an equation to estimate the value of one variable from a known value of the other. This topic covers Karl Pearson's coefficient, Spearman's rank correlation and the two regression lines.

Key points

• Correlation (सहसंबंध): positive (both move in the same direction: income and spending), negative (opposite directions: price and demand), or zero (no relation).
• Linear vs non-linear; simple (two variables) vs multiple and partial correlation.
• Methods: scatter diagram, Karl Pearson's coefficient (r), Spearman's rank correlation (ρ), concurrent deviation method.
• r lies between −1 and +1: +1 perfect positive, −1 perfect negative, 0 no linear correlation.
• r is independent of change of origin and scale.
• Correlation does not prove cause and effect; two variables can be correlated by chance or through a third factor.
• Spearman's rank method is used for qualitative data (beauty, honesty) or when only ranks are given.
• Regression (प्रतीपगमन): there are two regression lines – Y on X (estimate Y from X) and X on Y (estimate X from Y). Both pass through the point (x̄, ȳ).
• Regression coefficients: byx and bxy; r = ±√(byx × bxy), with the same sign as the coefficients. Both coefficients have the same sign, and both cannot be greater than 1.
• If r = ±1, the two regression lines coincide; if r = 0, they are perpendicular.

Important terms

• Scatter diagram – plot of paired values to see the pattern of relationship.
• Covariance – average of the products of deviations: Σ(x − x̄)(y − ȳ) ÷ n.
• Probable error – PE = 0.6745 × (1 − r²) ÷ √n; if r > 6 PE, correlation is significant.
• Regression coefficient (byx) – change in Y for one unit change in X.

Formats and formulas

• r = Σxy ÷ √(Σx² × Σy²), where x = X − X̄ and y = Y − Ȳ (deviations from means)
• Spearman: ρ = 1 − [6Σd² ÷ n(n² − 1)], d = difference in ranks
• byx = Σxy ÷ Σx²; bxy = Σxy ÷ Σy²; also byx = r × σy/σx
• Regression of Y on X: Y − Ȳ = byx (X − X̄)
• Regression of X on Y: X − X̄ = bxy (Y − Ȳ)

Worked example: X = 1, 2, 3, 4, 5 (advertising, ₹ thousand); Y = 2, 4, 5, 4, 5 (sales, ₹ lakh). X̄ = 3, Ȳ = 4.

• r = 6 ÷ √(10 × 6) = 6 ÷ 7.746 = 0.77 (high positive correlation)
• byx = 6 ÷ 10 = 0.6 → Y − 4 = 0.6(X − 3) → Y = 0.6X + 2.2
• bxy = 6 ÷ 6 = 1 → X − 3 = 1(Y − 4) → X = Y − 1
• Check: √(0.6 × 1) = 0.7746 = r
• Estimate: if advertising X = 6, sales Y = 0.6 × 6 + 2.2 = 5.8

5 important questions

Q1. What is the meaning of r = −0.9?

It shows a high degree of negative correlation: when one variable increases, the other decreases strongly and almost linearly, e.g. price and quantity demanded.

Q2. Find Spearman's rank correlation if two judges rank 5 candidates as (1, 2, 3, 4, 5) and (2, 1, 4, 3, 5).

d = −1, 1, −1, 1, 0; Σd² = 4. ρ = 1 − (6 × 4) ÷ [5 × (25 − 1)] = 1 − 24/120 = 0.8. There is a high positive agreement between the judges.

Q3. Calculate Karl Pearson's coefficient of correlation (numerical).

Steps: find means → deviations x and y → x², y², xy → totals → apply formula → interpret (see worked example). For large values use assumed means with the formula r = [nΣdxdy − ΣdxΣdy] ÷ √{[nΣdx² − (Σdx)²][nΣdy² − (Σdy)²]}.

Q4. Obtain the two regression equations and estimate Y when X = 6 (numerical).

Steps: means → byx and bxy → write both equations through (x̄, ȳ) → simplify → substitute X = 6 in the Y-on-X line → check that r² = byx × bxy (see worked example: Y = 0.6X + 2.2, X = Y − 1, estimate 5.8).

Q5. Distinguish between correlation and regression.

Outline: correlation measures degree and direction; regression gives a functional relation for estimation → correlation is symmetric (rxy = ryx); regression has two different lines → correlation does not show cause and effect; regression treats one variable as dependent → r is a pure number; regression coefficients have units → uses in business forecasting.

Exam tip

Always use the Y-on-X line to estimate Y and the X-on-Y line to estimate X. Using the wrong line is the commonest error in regression problems.