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🔷 TOPIC (v) — MEASURES OF CENTRAL TENDENCY

Syllabus topic (v) of 10 · runs until Topic (vi)

Covers: Construction of frequency distributions · inclusive vs exclusive method · Mean (individual, discrete, continuous; combined mean, missing values, correction, weighted mean) · Median · Mode · ➕ added: empirical relation, comparison of the three averages, GM & HM, partition values 🎯 Asked in the papers: 2024 · Q75 — arrange mean / median / mode / n in decreasing order

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1. Construction of a Frequency Distribution

Unarranged data can be distributed into classes and corresponding frequencies be assigned. Steps to follow

1. Sorting of Data.
2. Calculate the Range.
3. Decide number of classes.
4. Calculate class width.
5. Tally & count the observations.

e.g. 0, 1, 4, 3, 12, 13, 19, 15, 20, 23, 27, 29, 37, 31, 41, 46, 50, 48 $\rightarrow$ 18 observations.

Step 1 — Sorting the data $\underline{0}, 1, 3, 4, 12, 13, 15, 19, 20, 23, 27, 29, 31, 37, 41, 46, 48, \underline{50}$

Step 2 — calculate the range { 0 - 50 } = 50

Step 3 — Decide the number of classes. Let it be '5'

Step 4 — calculate class width. $\text{width} = \frac{\text{upper Range} - \text{Lowest Range}}{\text{No. of classes}}$

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$\text{width} = \frac{50 - 0}{5} = \frac{50}{5} = \text{==10==}$

Steps count observations & tally. No. of observations = frequency.

1.1 Inclusive vs Exclusive Method

• Inclusive Method / series (Both limits are included)

$\downarrow$ conversion to Exclusive Method distribution

$\text{adjustment } (h) = \frac{\text{Lower limit } (i_2) - \text{Upper limit } (i_1)}{2}$ $h = \frac{LL(i_2) - UL(i_1)}{2}$ $h = \frac{30 - 29}{2} = \text{==0.5==}$

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2. Mean

Definition — the average, or most common value, in a collection of numbers.

Two kinds:

• Simple Arithmetic Mean
• Weighted Mean

2.1 Simple Arithmetic Mean

Average of the collection of numbers. Formulas for calculating Simple Arithmetic Mean:

2.2 Individual Series

Q. Find the A. Mean of the given series: $10, 12, 15, 20, 30, 40$

Sol $\text{Mean}_{\text{Individual Series}} = \frac{\sum x}{N} = \frac{10 + 12 + 15 + 20 + 30 + 40}{6}$ $\bar{x} = \frac{127}{6} = \text{==21.16==}$

$\bar{x} = \frac{\sum x}{N} = \frac{75}{5} = \text{==15==}$

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2.3 Discrete Series

Note: N = $\sum f = 17$

Now $\bar{x} = \frac{\sum fx}{N} = \frac{88}{17} = 5.17$

2.4 Continuous Series

Now $\bar{x} = \frac{\sum fx}{\sum f} = \frac{780}{22} = \text{==35.45==}$

2.5 Combined Mean

$\bar{x}_{1,2,3...} = \frac{\bar{x}_1 f_1 + \bar{x}_2 f_2 + \bar{x}_3 f_3 + \dots}{f_1 + f_2 + f_3 + \dots}$

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$\bar{x}_{1,2,3,4} = \frac{\bar{x}_1 f_1 + \bar{x}_2 f_2 + \bar{x}_3 f_3 + \bar{x}_4 f_4}{f_1 + f_2 + f_3 + f_4}$ $= \frac{50 \times 30 + 60 \times 40 + 70 \times 20 + 100 \times 15}{30 + 40 + 20 + 15}$ $\Rightarrow \bar{x}_{1,2,3,4} = \frac{1500 + 2400 + 1400 + 1500}{105}$ $= \frac{6800}{105} = \text{==64.76==}$

2.6 Finding a Missing Value

Q. The mean temperature for four days noted is 120°c. If the temperature for day 1, day 2 & day 4 is 30, 35 and 40 respectively. Find the temperature of day 3?

Sol $\bar{x} = 120^\circ\text{c}$ $\Rightarrow \frac{\text{day } 1 + \text{day } 2 + \text{day } 3 + \text{day } 4}{4} = 120$ $\Rightarrow \frac{30 + 35 + \text{day } 3 + 40}{4} = 120$ $\Rightarrow \frac{105 + \text{day } 3}{4} = 120$ $\Rightarrow 105 + \text{day } 3 = 480$ $\Rightarrow \text{day } 3 = 480 - 105 = \text{==375}^\circ\text{==}$

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Q. Find the missing value if the mean value = 10.

$\bar{x} = \frac{\sum fx}{\sum f} = \frac{250 + 20x_1}{75}$ $\Rightarrow 10 = \frac{250 + 20x_1}{75}$ $\Rightarrow 750 = 250 + 20x_1$ $\Rightarrow 500 = 20x_1$ $\Rightarrow \frac{500}{20} = x_1$ $x_1 = \text{==25==}$

2.7 Correcting a Mean Value

Q. In a class, average marks of 30 students is 40. If the correct marks for one student is 46 which was misread as 42. calculate the new correct mean.

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Sol $\bar{x} = \frac{\sum x}{N}$ $40 = \frac{\sum x}{30} \Rightarrow \sum x = 40 \times 30 = 1200$

$\sum x =$ Sum of all the marks.

• Add correct marks
• Subtract wrong marks. $= \sum x + 46 - 42$ $= 1200 + 46 - 42 = \text{==1204==}$

New $\sum x = 1204$. $\therefore$ New correct mean $\bar{x} = \frac{\sum x}{N} = \frac{1204}{30} = \text{==40.13==}$

2.8 Weighted Arithmetic Mean

When the items of a series are not of equal importance / weightage. $\bar{x}_w = \frac{\sum wx}{\sum w}$ where $\bar{x}_w =$ weighted Mean. $\sum wx =$ sum of product of weights and items. $\sum w =$ sum of weights.

Q. student scores 20 marks in statistics, 35 in english, 40 in maths and 45 in Geography. calculate weighted mean, if the marks are weighted as 2, 1, 3, 4 respectively.

Sol

$\bar{x}_w = \frac{\sum wx}{\sum w} = \frac{375}{10} = \text{==37.5==}$

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3. Median

Definition — the middle value, which separates the higher half from the lower half of the data.

3.1 Individual Series

• Sort the data in ascending or descending order
• Calculate the median by the formula: $M = \text{value of } \left( \frac{n+1}{2} \right)^{\text{th}} \text{ item}$

e.g 1 $2, 7, 9, 3, 6, 8, 12$ Sol Sort: $2, 3, 6, 7, 8, 9, 12$ $\rightarrow$ Trick: (odd) Middle value = Median. here $n = 7 = \text{no. of observations / items}$ $\therefore M = \text{value of } \left( \frac{n+1}{2} \right)^{\text{th}} \text{ item}$ $= \text{value of } \left( \frac{7+1}{2} \right)^{\text{th}} \text{ item}$ $= \text{value of (4)}^{\text{th}} \text{ item}$ $\Rightarrow \text{==} M = 7 \text{==}$

e.g 2 $10, 15, 20, 30, 35, 40, 50, 60$ Sol Already sorted $\rightarrow$ Trick: $\frac{\text{Sum of two middle value}}{2} = M$ $M = \text{value of } \left( \frac{n+1}{2} \right)^{\text{th}} \text{ item}$ $= \text{value of } \left( \frac{8+1}{2} \right)^{\text{th}} \text{ item}$ $= \text{value of } (4.5)^{\text{th}} \text{ item}$ $= \text{value of } 4^{\text{th}} + 0.5(\text{difference between } 4^{\text{th}} \text{ \& } 5^{\text{th}} \text{ item})$ $= 30 + 0.5(5)$ $M = 30 + 2.5 = \text{==32.5==}$

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e.g 3

• sort values of X.

$M = \text{value of } \left( \frac{n+1}{2} \right)^{\text{th}} \text{ term}$ $= \text{v. of } \left( \frac{7+1}{2} \right)^{\text{th}} \text{ term}$ $= \text{v. of } 4^{\text{th}} \text{ term}$ $M = \text{==20==}$

3.2 Discrete Series

(X) (frequencies) *1. Sort values in Ascending or decending order. *2. Cummulate frequencies *3. $M = \text{value of } \left( \frac{n+1}{2} \right)^{\text{th}} \text{ item}$

e.g

$n = 63$

$M = \left( \frac{n+1}{2} \right)^{\text{th}} \text{ value}$ $= \left( \frac{63+1}{2} \right)^{\text{th}} \text{ value} = \text{==32==} \rightarrow$ check class and corresponding X-value $\therefore M = 50$

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3.3 Continuous Series

(1) Sort groups in ascending or decending order. (2) change inclusive series into exclusive series. (3) cummulate frequencies (4) Calculate $m = \text{value of } \left( \frac{n}{2} \right)^{\text{th}} \text{ item}$

$M = L_1 + \frac{i}{f} (m - c_f)$

e.g

$n = 27$ $m = \left( \frac{n}{2} \right)^{\text{th}} \text{ value} = \left( \frac{27}{2} \right)^{\text{th}} \text{ value} = \text{==13.5==} \rightarrow \text{Falls in 30-40 class.}$

Now $M = L_1 + \frac{i}{f} (m - c_f)$ $= 30 + \frac{10}{7} (13.5 - 10)$ $= 30 + \frac{10}{7} (3.5)$ $= 30 + \frac{10}{2}$ $= 30 + 5 = \text{==35==}$

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4. Mode (Z)

Definition — the value in the data set that appears most frequently; the value repeated the maximum number of times.

4.1 Individual Series

observation that is repeated maximum times e.g 1, 2, 3, 4, $\underline{3}$, 6, 5, 9, 10, 6, $\underline{3}$, 1 Mode = 3 (repeated maximum times)

e.g 2 1, 2, 3, 6, 7, 9 All Mode (no observation repeated).

e.g 3 4, $\underline{3}$, 4, 2, $\underline{3}$, 5, 6, 8. Mode = 3 & 4 (Bimodal).

• If there are more than two modes in a data set, it is called Multi-modal data.

4.2 Discrete Series

Observation that is corresponding to the maximum frequency.

$\therefore Z = \text{==20==}$

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4.3 Continuous Series

Sol

• check if for exclusiveness
• select the maximum/highest frequency
• its corresponding class becomes modal class. {40-50}

Now $Z = l + \left| \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right| \times h$

$l \rightarrow \text{lower limit of modal class}$ $f_1 \rightarrow \text{frequency of modal class}$ $f_0 \rightarrow \text{frequency pre-modal class}$ $f_2 \rightarrow \text{frequency post-modal class}$ $h = \text{upper - lower limit of modal class.}$

$Z = 40 + \left| \frac{12 - 10}{2 \times 12 - 10 - 6} \right| \times 10$ $= 40 + \left| \frac{2}{8} \right| \times 10 = 40 + \frac{1}{4} \times 10$ $= 40 + \frac{5}{2} = \text{==42.5==}$

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$X: 4, 6, 8, 10, 12$ If in the above example mean is increased by 2, what will happen to the individual observation if all are equally affected.

Sol $\bar{x} = \frac{\sum x}{N} = \frac{4+6+8+10+12}{5} = \frac{40}{5} = 8$

$\therefore \sum x = 40$ $\bar{x} = 8$ If mean is increased by 2 then New mean = $8 + 2 = 10$

$\therefore 10 = \frac{\sum x'}{5} \Rightarrow \sum x' = 50$

Now $\sum x' - \sum x = 50 - 40 = \text{==10==}$

This 10 needs to be equally distributed. each observation will get $\frac{10}{5} = 2$ increment.

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➕ Added — Relations Between the Averages, and Partition Values.

5. Relations Between the Averages

5.1 The Empirical Relation

$\textbf{Mode} = 3\,\textbf{Median} - 2\,\textbf{Mean}$

Rearranged forms you may need: $\text{Median} = \frac{\text{Mode} + 2\,\text{Mean}}{3} \qquad \text{Mean} = \frac{3\,\text{Median} - \text{Mode}}{2}$

• Also written Mean − Mode = 3 (Mean − Median)
• ⭐ For a SYMMETRIC distribution: Mean = Median = Mode
• Positively skewed: Mean > Median > Mode · Negatively skewed: Mean < Median < Mode

e.g. Mean = 25, Median = 24 → Mode = 3(24) − 2(25) = 72 − 50 = 22

5.2 Comparison of the Three Averages

⭐ Σ(x − x̄)² is MINIMUM when taken about the MEAN.

5.3 Types of Averages

Mathematical: Arithmetic Mean (AM) · Geometric Mean (GM) · Harmonic Mean (HM) Positional: Median · Mode

$GM = \sqrt[n]{x_1 \times x_2 \times \dots \times x_n} \qquad\qquad HM = \frac{n}{\sum \frac{1}{x}}$

⭐ AM ≥ GM ≥ HM (always; equal only when all values are identical) · ⭐ GM² = AM × HM Use GM for rates of growth / ratios / index numbers · Use HM for speeds and rates per unit.

6. Partition Values — Quartiles, Deciles, Percentiles

Values that divide an ordered data set into equal parts.

⭐ Also: Q₁ = P₂₅ · Q₃ = P₇₅ · D₁ = P₁₀

Individual & discrete series (N = number of items): $Q_i = \text{size of } \left(\frac{i(N+1)}{4}\right)^{th} \text{ item} \quad D_i = \left(\frac{i(N+1)}{10}\right)^{th} \quad P_i = \left(\frac{i(N+1)}{100}\right)^{th}$

Continuous series (interpolation, cf = cumulative frequency of the preceding class): $Q_i = l + \frac{\frac{iN}{4} - cf}{f} \times h$

Related: Quartile Deviation (Semi-Interquartile Range) $= \dfrac{Q_3 - Q_1}{2}$ · Interquartile Range $= Q_3 - Q_1$

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