Statistics

Frequency distribution

A frequency distribution systematically organizes raw statistical data by assigning frequencies to distinct values or class intervals, forming the bedrock of descriptive statistics. This topic covers discrete and continuous frequency distributions, cumulative frequencies, and class boundaries.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Grouped Frequency Distribution & Cumulative Aggregation

Let a set of $N$ observations be partitioned into $k$ mutually exclusive and exhaustive classes $[l_i, u_i)$ with respective frequencies $f_i$ such that $\sum_{i=1}^{k} f_i = N$. The cumulative frequency $F_i$ corresponding to the $i$-th class is defined as $F_i = \sum_{j=1}^{i} f_j$ for "less than" type and $F_i^* = \sum_{j=i}^{k} f_j$ for "more than" type distributions.

📐 Geometric Interpretation

Represented graphically via histograms, frequency polygons, and ogives (cumulative frequency curves), where the intersection of the 'less than' and 'more than' ogives precisely determines the median of the distribution.

⚙️ Algebraic & Structural Meaning

Facilitates compressed algebraic computation of central tendency (mean, median, mode) and dispersion (variance, standard deviation) without processing raw data arrays.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Exhaustiveness and Mutually Exclusivity	`N/A`	Strictly Mandatory	Overlapping or gapped class intervals lead to ambiguous frequency assignment and violate total count conservation ($\sum f_i \neq N$).
Class Interval Uniformity (Standard assumption)	`Constant width $h = u_i - l_i$`	undefined	Unequal class widths distort visual histograms and require width-correction factors for accurate frequency density calculations.
Continuity of Variate for Grouped Data	`Continuous real intervals`	Essential for interpolated median and mode formulas	Discrete integer data grouped into continuous intervals introduces grouping error (Sheppard's correction necessity).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Median Calculation via Ogive Intersection

$$ M_d = L + \left( \frac{\frac{N}{2} - C_f}{f} \right) \times h $$

Rule: Identify median class where cumulative frequency exceeds $N/2$, then apply direct proportional interpolation.

Application: Directly tested in UP PGT/TGT to find missing frequencies given the median.

Mode Calculation from Modal Class

$$ M_o = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h $$

Rule: Locate maximum frequency $f_1$, preceding $f_0$, succeeding $f_2$, and plug into the ratio.

Application: Frequent numerical questions in KVS and DSSSB PGT Mathematics.

Step-Deviation Method for Mean

$$ \bar{x} = A + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h, \quad u_i = \frac{x_i - A}{h} $$

Rule: Choose assumed mean $A$ at the mid-frequency class to simplify arithmetic coefficients to small integers.

Combined Mean of Frequency Distributions

$$ \bar{x}_{12} = \frac{N_1\bar{x}_1 + N_2\bar{x}_2}{N_1 + N_2} $$

Rule: Weighted average of subgroup means weighted by their respective total frequencies $N_1$ and $N_2$.

Empirical Relationship between Mean, Median, and Mode

$$ \text{Mode} = 3(\text{Median}) - 2(\text{Mean}) $$

Rule: For moderately skewed distributions, use this exact linear identity to find one missing central tendency measure when two are known.

Change of Origin and Scale for Variance

$$ \sigma_u^2 = \frac{1}{h^2} \sigma_x^2 \implies \sigma_x^2 = h^2 \sigma_u^2 $$

Rule: Variance is independent of change of origin ($A$) but scales quadratically with change of scale ($h$).

Application: Trick questions testing conceptual clarity in property-based MCQs.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Inclusive vs. Exclusive Class Intervals

Common Pitfall: Failing to convert inclusive series (e.g., $10-19, 20-29$) into exclusive series ($9.5-19.5, 19.5-29.5$) before calculating median, mode, or continuity-dependent statistics.

Counterexample: \text{Classes: } 10-19, 20-29 \implies \text{True limits: } 9.5-19.5, 19.5-29.5, \text{with } h = 10, \text{not } 9.

Why it fails: Using upper limit of inclusive class directly as boundary corrupts the lower-limit $L$ and class width $h$ in interpolation formulas.

🛡️ Exam Defense: Always check if the upper limit of one class matches the lower limit of the next. If not, apply continuity correction factor $\pm 0.5$.

⚠️ COMMON ERROR #02

Confusing Cumulative Frequency with Class Frequency

Common Pitfall: Using cumulative frequency values $C_f$ directly as $f_i$ in arithmetic mean or variance summation formulas.

Counterexample: \text{Given } C_f: 5, 12, 25 \implies \text{Actual } f: 5, 7, 13.

Why it fails: Summing cumulative frequencies double-counts previous observations, leading to grossly inflated mean values.

🛡️ Exam Defense: Check table headers. If terms are increasing monotonically up to $N$, they are cumulative frequencies; subtract consecutive terms to recover $f_i$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Arithmetic Mean	All frequencies defined, finite total $N$	$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$	Center of gravity of the mass distribution	Keywords: 'Average', 'Mean weight/marks'
Median	Data can be ordered or grouped	$M_d = L + \frac{\frac{N}{2} - C_f}{f} h$	Abscissa of the intersection of less-than and more-than ogives	Keywords: 'Middle observation', 'Ogive intersection'
Mode	Unimodal continuous distribution	$M_o = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} h$	Abscissa of the peak of the frequency polygon or histogram rectangle	Keywords: 'Most frequent', 'Maximum concentration'
Empirical Relation	Moderately asymmetric (skewed) distribution	$\text{Mode} = 3\text{Median} - 2\text{Mean}$	Positional balance point between mean, median, and mode in skew curves	Given two central tendencies, find the third

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

In a frequency distribution, if the mean is $26.8$ and the median is $27.9$, find the approximate mode using the empirical relation.

(A) $30.1$(B) $28.5$(C) $25.6$(D) $29.2$

📚 Method A: Formal Textbook Derivation

Step 1: Recall empirical formula $\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$. Step 2: Substitute values: $\text{Mode} = 3(27.9) - 2(26.8)$. Step 3: Compute $83.7 - 53.6 = 30.1$.

⚡ Method B: 25-Second Exam Shortcut

Mental calculation: $3 \times 28 \approx 84$, minus $53.6 \approx 30.1$. Match option (A).

💡 Tip: Direct formula substitution question; do not waste time deriving.

CANONICAL PROBLEM 02 Dual-Method Breakdown

For a continuous frequency distribution, the lower limit of the median class is $20$, frequency of median class is $10$, cumulative frequency of preceding class is $12$, and total frequency $N = 50$, with class width $h = 10$. Find the median.

(A) $23$(B) $25$(C) $27$(D) $29$

📚 Method A: Formal Textbook Derivation

Step 1: Identify parameters: $L = 20, N/2 = 25, C_f = 12, f = 10, h = 10$. Step 2: Apply formula $M_d = L + \frac{\frac{N}{2} - C_f}{f} h$. Step 3: $M_d = 20 + \frac{25 - 12}{10} \times 10 = 20 + 13 = 33$. Wait, let's re-verify: $N=50 \implies N/2 = 25$. If $C_f = 12$, $25-12 = 13$, $13/10 \times 10 = 13$, $20+13 = 33$. Let's adjust options: (A) 31, (B) 33, (C) 35, (D) 37. Correct option (B) 33.

⚡ Method B: 25-Second Exam Shortcut

Cancel $h$ and $f$ directly when class width equals frequency denominator magnitude.

💡 Tip: Ensure $N/2$ is computed correctly from total frequency sum.

Practice questions

Choose an answer to check it and see the solution:

If the sum of frequencies $\sum f_i = 50$ and $\sum f_i x_i = 1250$, the mean $\bar{x}$ is:

A $20$

B $25$

C $30$

D $35$

Derivation & Analysis:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1250}{50} = 25$. Option (B) is correct.

The curve obtained by plotting cumulative frequencies against upper class boundaries is called:

A Histogram

B Frequency polygon

C Ogive

D Frequency curve

Derivation & Analysis:

By definition, the cumulative frequency curve is termed an Ogive.

In a moderately skewed distribution, if Mean = $20$ and Mode = $26$, then Median is:

A $21$

B $22$

C $23$

D $24$

Derivation & Analysis:

Using $\text{Mode} = 3\text{Median} - 2\text{Mean} \implies 26 = 3M_d - 2(20) \implies 26 + 40 = 3M_d \implies 66 = 3M_d \implies M_d = 22$.

If each observation in a frequency distribution is multiplied by $3$, the variance of the new distribution:

A Remains unchanged

B Gets multiplied by $3$

C Gets multiplied by $9$

D Gets divided by $3$

Derivation & Analysis:

Variance scales quadratically with scale change: $\sigma_{new}^2 = 3^2 \sigma_{old}^2 = 9 \sigma_{old}^2$.

Class marks of a frequency distribution are given as $15, 25, 35, 45$. The class interval corresponding to class mark $25$ with uniform width $10$ is:

A $20 - 30$

B $15 - 25$

C $25 - 35$

D $10 - 20$

Derivation & Analysis:

Class interval is $[x_i - h/2, x_i + h/2] = [25 - 5, 25 + 5] = [20, 30]$.

The sum of deviations of all values $x_i$ from their arithmetic mean $\bar{x}$, weighted by frequencies $f_i$, is always:

A One

B Positive

C Zero

D Negative

Derivation & Analysis:

By algebraic property of mean, $\sum f_i(x_i - \bar{x}) = 0$.

For a symmetrical frequency distribution, which of the following holds true?

A $\text{Mean} < \text{Median} < \text{Mode}$

B $\text{Mean} > \text{Median} > \text{Mode}$

C $\text{Mean} = \text{Median} = \text{Mode}$

D $\text{Median} = \frac{\text{Mean} + \text{Mode}}{2}$

Derivation & Analysis:

In a perfectly symmetrical distribution, all three measures of central tendency coincide at the center.

If the total frequency $N = 100$, the median corresponds to the cumulative frequency value of:

A $25$

B $50$

C $75$

D $100$

Derivation & Analysis:

Median divides the total frequency into two equal halves, corresponding to $N/2 = 50$.

Which graphical representation is best suited for determining the Mode visually?

A Less than ogive

B More than ogive

C Histogram

D Frequency polygon

Derivation & Analysis:

The mode can be located graphically from the highest rectangle in a histogram using cross-diagonal lines.

Q10

If the coefficient of variation is $50$ and mean is $20$, the standard deviation is:

A $5$

B $10$

C $15$

D $20$

Derivation & Analysis:

$\text{CV} = \frac{\sigma}{\bar{x}} \times 100 \implies 50 = \frac{\sigma}{20} \times 100 \implies \sigma = 10$.

Common questions

Inclusive series have gaps between the upper limit of one class and the lower limit of the next (e.g., $10-19$ and $20-29$). Converting them to exclusive series ($9.5-19.5$ and $19.5-29.5$) ensures proper class boundaries and accurate median/mode interpolation.

Candidates often forget that variance scales with the square of the class width ($h^2$) and is unaffected by the choice of assumed mean (change of origin).

Memorize the exact identity $\text{Mode} = 3\text{Median} - 2\text{Mean}$ and apply mental arithmetic directly to the given options without writing down intermediate steps.

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