Basic Concepts of Statistics

Q: What is the difference between mean, median and mode?

The mean is the total of all observations divided by their number. The median is the middle value when the data is arranged in order. The mode is the value that occurs most often. The mean uses every value, so extreme values affect it; the median and mode are positional and resist extreme values.

Q: How do you find the median of grouped data?

Make a cumulative frequency column and find N/2. The median class is the first class whose cumulative frequency reaches N/2. Then median = l + ((N/2 - cf)/f) x h, where cf is the cumulative frequency of the class before the median class, f its frequency and h its width.

Q: What is the formula for the mode of grouped data?

Mode = l + ((f1 - f0)/(2f1 - f0 - f2)) x h. Here l is the lower limit of the modal class, f1 its frequency, f0 and f2 the frequencies of the classes just before and after it, and h the class width. All classes must have equal width.

Q: Why is the sum of deviations from the mean always zero?

Because the sum of (xi - mean) equals the sum of xi minus n times the mean, and n times the mean is exactly the total of the xi. The positive deviations of values above the mean cancel the negative deviations below it, which is why the mean is called the balance point of the data.

Q: When should the median be used instead of the mean?

Use the median when the data has extreme values or is skewed, such as incomes, house prices or waiting times. One very large value can pull the mean far from the typical value, but the median depends only on the middle position and hardly changes. The median also works for data that can only be ranked.

Q: What happens to mean, median and mode when every observation is changed to ax + b?

All three averages follow the same rule: the new mean is a times the old mean plus b, the new median is a times the old median plus b, and the new mode is a times the old mode plus b. Adding b shifts every average by b, and multiplying by a scales every average by a.

Q: How are mean, median and mode asked in JEE Main?

JEE Main questions usually test combined and corrected means, missing frequencies for a given mean, means of series such as xi + 2i, the median after some observations change, and the effect of adding or multiplying every value by a constant. Most can be solved in under a minute using totals and the shift and scale rules.

Q: Is statistics part of the JEE Advanced syllabus?

Yes. JEE Advanced includes measures of central tendency and dispersion: mean, median and mode of grouped and ungrouped data, along with mean deviation, variance and standard deviation. Advanced questions often combine these with series sums, binomial coefficients or algebra, as in the weighted mean and binomial frequency examples on this page.

MathsStatisticsFor JEE aspirants

Mean, median and mode are the three measures of central tendency: single values that represent a whole set of data. The mean shares the total equally among all observations, the median is the middle value of the arranged data, and the mode is the value that occurs most often. This page covers mean, median and mode of ungrouped and grouped data, shortcut methods, weighted and combined means, key properties and solved examples. The topic is part of the JEE Main and JEE Advanced syllabus.

Key Formulas - Quick Reference

Mean of ungrouped data: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
Mean of a frequency distribution: $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{N}$ , where $N = \sum f_{i}$
Step-deviation method: $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{N}$ , where $u_{i} = \frac{x _{i} - a}{h}$
Weighted mean: $\overset{x}{ˉ}_{w} = \frac{\sum w _{i} x _{i}}{\sum w _{i}}$
Combined mean: $\overset{x}{ˉ} = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2}}{n _{1} + n _{2}}$
$\sum (x_{i} - \overset{x}{ˉ}) = 0$ , and the mean of $a x_{i} + b$ is $a \overset{x}{ˉ} + b$
Median of $n$ arranged values: $(\frac{n + 1}{2})^{th}$ value if $n$ is odd; mean of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ values if $n$ is even
Median of grouped data: $l + (\frac{N /2 - c f}{f}) \times h$
Mode of grouped data: $l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h$

1. Measures of Central Tendency

A large set of observations is hard to read at a glance. A measure of central tendency (an average) is one value that describes the whole distribution. The three averages in the syllabus are:

Arithmetic mean: a calculated average that uses every observation.
Median: a positional average, the middle value of the arranged data.
Mode: the value (or class) with the highest frequency.

The method of calculation depends on the form in which the data is given.

Form of data	What is given	Example
Ungrouped (individual) data	A list of $n$ observations $x_{1}, x_{2}, \dots, x_{n}$	3, 1, 4, 0, 2, 1
Discrete frequency distribution	Values $x_{i}$ with frequencies $f_{i}$	$x$ : 5, 8, 11 and $f$ : 4, 5, 6
Grouped (continuous) frequency distribution	Class intervals with frequencies	0-10: 8 students, 10-20: 30 students

Class mark of a class interval $= \frac{lower limit + upper limit}{2}$ . The class width $h$ is the difference between the upper and lower limits. For grouped data, every observation in a class is represented by its class mark.

2. Arithmetic Mean

2.1 Mean of Ungrouped Data

The arithmetic mean is the sum of all observations divided by the number of observations.

\overset{x}{ˉ} = \frac{x _{1} + x _{2} + \dots + x _{n}}{n} = \frac{1}{n} i = 1 \sum n x_{i}

When the values are large, choose a convenient assumed mean $a$ and work with the deviations $d_{i} = x_{i} - a$ :

\overset{x}{ˉ} = a + \frac{\sum d _{i}}{n}

2.2 Mean of a Frequency Distribution

If the values $x_{1}, x_{2}, \dots, x_{n}$ occur with frequencies $f_{1}, f_{2}, \dots, f_{n}$ , each value is counted as many times as it occurs:

\overset{x}{ˉ} = \frac{f _{1} x _{1} + f _{2} x _{2} + \dots + f _{n} x _{n}}{f _{1} + f _{2} + \dots + f _{n}} = \frac{\sum f _{i} x _{i}}{N}, N = \sum f_{i}

For grouped data, $x_{i}$ is the class mark. This assumes the observations in each class are spread evenly around its mid-point.

2.3 Shortcut Methods for Frequency Distributions

All three methods give the same mean. The shortcuts only reduce the arithmetic.

Method	Working variable	Formula
Direct method	$x_{i}$	$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{N}$
Assumed-mean (deviation) method	$d_{i} = x_{i} - a$	$\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{N}$
Step-deviation method	$u_{i} = \frac{x _{i} - a}{h}$	$\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{N}$

$a$ and $h$ can be any convenient numbers. Take $a$ as a class mark near the middle of the table and $h$ as the common class width, so that the $u_{i}$ become small integers such as $- 2, - 1, 0, 1, 2$ .

2.4 Weighted Arithmetic Mean

When observations have different importance, each value $x_{i}$ is given a weight $w_{i}$ :

\overset{x}{ˉ}_{w} = \frac{w _{1} x _{1} + w _{2} x _{2} + \dots + w _{n} x _{n}}{w _{1} + w _{2} + \dots + w _{n}} = \frac{\sum w _{i} x _{i}}{\sum w _{i}}

The frequency formula is a weighted mean in which the weights are the frequencies. If all weights are equal, the weighted mean is the ordinary mean.

2.5 Combined Mean of Two or More Groups

If two groups have $n_{1}$ and $n_{2}$ observations with means $\overset{x}{ˉ}_{1}$ and $\overset{x}{ˉ}_{2}$ , their totals are $n_{1} \overset{x}{ˉ}_{1}$ and $n_{2} \overset{x}{ˉ}_{2}$ . So the mean of the combined group is

\overset{x}{ˉ} = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2}}{n _{1} + n _{2}}, \overset{x}{ˉ} = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2} + n _{3} x ˉ _{3} + \dots}{n _{1} + n _{2} + n _{3} + \dots}

The combined mean always lies between the smallest and the largest group means.

2.6 Properties of the Arithmetic Mean

Sum of deviations is zero: $\sum (x_{i} - \overset{x}{ˉ}) = 0$ for ungrouped data and $\sum f_{i} (x_{i} - \overset{x}{ˉ}) = 0$ for a frequency distribution. The mean is the balance point of the data (Figure 1).
Least sum of squares: $\sum (x_{i} - a)^{2}$ is smallest when $a = \overset{x}{ˉ}$ .
Change of origin: if every observation is increased by $k$ , the mean increases by $k$ .
Change of scale: if every observation is multiplied by $k$ , the mean is multiplied by $k$ . Together, the mean of $a x_{i} + b$ is $a \overset{x}{ˉ} + b$ .
Uses every value: one extreme value can shift the mean a lot (Section 5).

Figure 1: The mean

\overset{x}{ˉ} = 6

balances the data 2, 3, 7, 8, 10. The negative deviations (

- 7

) cancel the positive deviations (

+ 7

), so

\sum (x_{i} - \overset{x}{ˉ}) = 0

JEE Advanced

Why the sum of squared deviations is least about the mean. Let $S (a) = \sum (x_{i} - a)^{2} = \sum x_{i}^{2} - 2 a \sum x_{i} + n a^{2}$ . This is a quadratic in $a$ with positive leading coefficient, so it is least at $a = \frac{\sum x _{i}}{n} = \overset{x}{ˉ}$ . The minimum value is $S (\overset{x}{ˉ}) = n σ^{2}$ , which links the mean to the variance.

Exam Trick

Corrected mean: if a value $w$ was wrongly recorded in place of the correct value $c$ , then new mean $=$ old mean $+ \frac{c - w}{n}$ . When observations are removed or added, work with totals: new total divided by new count.

3. Median

The median is the middle value of the data when the observations are arranged in ascending or descending order. It divides the arranged data into two equal halves: as many observations lie below it as above it.

3.1 Median of Ungrouped Data

Arrange the $n$ observations in order first.

$n$ odd: median $= (\frac{n + 1}{2})^{th}$ observation.
$n$ even: median $=$ mean of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.

Example: for 13, 14, 16, 18, 20, 22 we have $n = 6$ , so the median is the mean of the 3rd and 4th values: $\frac{16 + 18}{2} = 17$ .

3.2 Median of a Discrete Frequency Distribution

Arrange the values in ascending order and write the cumulative frequency ( $c f$ ) column.
Find $N = \sum f_{i}$ .
If $N$ is odd, the median is the $(\frac{N + 1}{2})^{th}$ observation. If $N$ is even, it is the mean of the $(\frac{N}{2})^{th}$ and $(\frac{N}{2} + 1)^{th}$ observations.
Use the $c f$ column: the observation at a given position is the first value whose $c f$ is greater than or equal to that position number.

3.3 Median of Grouped Data

Write the $c f$ column and find $\frac{N}{2}$ .
The median class is the first class whose $c f$ is greater than or equal to $\frac{N}{2}$ .
Apply the formula below.

Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h

Symbol	Meaning
$l$	lower limit of the median class
$N$	total frequency $\sum f_{i}$
$c f$	cumulative frequency of the class before the median class
$f$	frequency of the median class
$h$	width of the median class

The formula assumes the observations inside the median class are spread evenly. The same idea can be seen on a graph of cumulative frequency, called an ogive.

Figure 2: The less than and more than ogives cross where the cumulative frequency reaches half the total,

N /2 = 50

. Reading down from that point gives the median, 23 marks, the same value the grouped median formula gives.

A less than ogive plots each upper class limit against the cumulative frequency up to it. A more than ogive plots each lower class limit against the number of observations from that limit onwards. The two ogives cross at the median.

3.4 Properties of the Median

Resists extreme values: changing the largest or smallest observations (while they stay on the same side of the middle) does not change the median.
Change of origin and scale: the median of $a x_{i} + b$ is $a M + b$ , where $M$ is the original median.
Least sum of absolute deviations: $\sum ∣ x_{i} - A ∣$ is smallest when $A$ is the median. This result is used in mean deviation.
Works for ranked data: the median can be found for qualities that can only be put in order, such as ranks in a merit list.

4. Mode

The mode is the value that occurs most frequently. In a frequency distribution it is the value with the maximum frequency. A data set can have one mode, more than one mode (bimodal, multimodal), or no mode when all values occur equally often.

4.1 Mode of Ungrouped and Discrete Data

Arrange the values with their frequencies and pick the value with the highest frequency. For 5, 7, 2, 7, 7, 4, 7, 7, 6, 7 the value 7 occurs six times, so the mode is 7.

4.2 Mode of Grouped Data

The class with the highest frequency is the modal class. The mode inside it is found with

Mode = l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h

Symbol	Meaning
$l$	lower limit of the modal class
$f_{1}$	frequency of the modal class
$f_{0}$	frequency of the class just before the modal class
$f_{2}$	frequency of the class just after the modal class
$h$	width of the modal class

Figure 3: Join each top corner of the modal bar to the facing top corner of the neighbouring bar. The lines cross above

x \approx 22.6

, exactly the value of

l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h

The classes must have equal width. If the modal class is the first class, take $f_{0} = 0$ ; if it is the last class, take $f_{2} = 0$ . Some books write $f$ for the modal class frequency and $f_{1}, f_{2}$ for its neighbours: the formula is the same, only the labels change.

5. Mean, Median and Mode Compared

Feature	Mean	Median	Mode
Based on	every observation	middle position	highest frequency
Effect of extreme values	strongly affected	hardly affected	not affected
Every value increased by $k$	increases by $k$	increases by $k$	increases by $k$
Every value multiplied by $k$	multiplied by $k$	multiplied by $k$	multiplied by $k$
Best suited to	symmetric numerical data	skewed data such as incomes	most common size or category

For the marks of 100 students used in Figures 2 and 3, the mean is 23.6, the median is 23 and the mode is about 22.6. The three averages are close because the data is nearly symmetric. With an extreme value the picture changes:

Figure 4: One extreme value (95) drags the mean from 25 to about 36.7, while the median moves only from 25 to 26.5. For skewed data such as incomes, the median is the better average.

Exam Trick

Average of rates = total ÷ total. For an average speed over equal distances, or an average price when equal amounts of money are spent, do not take the simple mean of the rates. Divide the total distance by the total time, or the total money by the total quantity bought.

6. Solved Examples

Solved Example 1

A group of 10 items has mean 6. If the mean of 4 of these items is 7.5, then the mean of the remaining items is
(A) 6.5
(B) 5.5
(C) 4.5
(D) 5.0

Solution:

Sum of all 10 items $= 10 \times 6 = 60$ .
Sum of the 4 items $= 4 \times 7.5 = 30$ .
Sum of the remaining 6 items $= 60 - 30 = 30$ , so their mean $= \frac{30}{6} = 5$ .

Answer: (D) 5.0

Solved Example 2

The mean of a set of observations is

\overset{x}{ˉ}

. If each observation is divided by

α (α \neq = 0)

and then increased by 10, the mean of the new set is

(A) $\frac{x ˉ}{α}$

(B) $\frac{x ˉ + 10}{α}$

(D) $α \overset{x}{ˉ} + 10$

Solution:

Each new observation is $y_{i} = \frac{x _{i}}{α} + 10$ .
By the change of scale and origin property, $\overset{y}{ˉ} = \frac{x ˉ}{α} + 10$ .
Writing over a common denominator: $\overset{y}{ˉ} = \frac{x ˉ + 10 α}{α}$ .

Answer: (C) $\frac{x ˉ + 10 α}{α}$

Solved Example 3

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}

, the arithmetic mean of

C_{0}, 2 C_{1}, 3 C_{2}, \dots, (n + 1) C_{n}

(A) $\frac{2 ^{n - 1} ( n + 2 )}{n + 1}$

(B) $\frac{2 ^{n} ( n + 2 )}{n + 1}$

(D) $\frac{2 ^{n} ( n + 1 )}{n + 2}$

Solution:

Sum $S = \sum_{r = 0}^{n} (r + 1) C_{r} = \sum r C_{r} + \sum C_{r}$ .
Use $\sum C_{r} = 2^{n}$ and $\sum r C_{r} = n \cdot 2^{n - 1}$ (differentiate $(1 + x)^{n}$ and put $x = 1$ ).
$S = n \cdot 2^{n - 1} + 2^{n} = 2^{n - 1} (n + 2)$ .
There are $n + 1$ terms, so the mean $= \frac{2 ^{n - 1} ( n + 2 )}{n + 1}$ .

Answer: (A) $\frac{2 ^{n - 1} ( n + 2 )}{n + 1}$

Solved Example 4

A variable takes the values

0, 1, 2, \dots, n

with frequencies

^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}

. The arithmetic mean is

(A) $n$

(B) $\frac{2 ^{n}}{n}$

(D) $\frac{n}{2}$

Solution:

$N = \sum^{n} C_{r} = 2^{n}$ .
$\sum f_{i} x_{i} = \sum r^{n} C_{r} = n \cdot 2^{n - 1}$ .
$\overset{x}{ˉ} = \frac{n \cdot 2 ^{n - 1}}{2 ^{n}} = \frac{n}{2}$ .

Answer: (D) $\frac{n}{2}$

Solved Example 5

The mean of

x_{1}, x_{2}, \dots, x_{n}

\overset{x}{ˉ}

. The mean of the numbers

x_{i} + 2 i

i = 1, 2, \dots, n

, is
(A)

\overset{x}{ˉ} + 2 n

(B)

\overset{x}{ˉ} + n + 1

(C)

\overset{x}{ˉ} + 2

(D)

\overset{x}{ˉ} + n

Solution:

New sum $= \sum x_{i} + 2 (1 + 2 + \dots + n) = n \overset{x}{ˉ} + n (n + 1)$ .
New mean $= \frac{n x ˉ + n ( n + 1 )}{n} = \overset{x}{ˉ} + n + 1$ .

Answer: (B) $\overset{x}{ˉ} + n + 1$

Solved Example 6

A variable takes the values

α + 4, α - \frac{7}{2}, α - \frac{5}{2}, α - 3, α - 2, α + \frac{1}{2}, α - \frac{1}{2}, α + 5 (α > 0)

. The median is

(A) $α - \frac{5}{4}$

(B) $α - \frac{1}{2}$

(D) $α + \frac{5}{4}$

Solution:

Arrange: $α - \frac{7}{2}, α - 3, α - \frac{5}{2}, α - 2, α - \frac{1}{2}, α + \frac{1}{2}, α + 4, α + 5$ .
$n = 8$ is even, so the median is the mean of the 4th and 5th values.
Median $= \frac{( α - 2 ) + ( α - \frac{1}{2} )}{2} = α - \frac{5}{4}$ .

Answer: (A) $α - \frac{5}{4}$

Solved Example 7

The mean weight of 29 male workers in a factory is 71 kg and that of 31 female workers is 48 kg. Find the combined mean weight of all 60 workers.

Solution:

Total weight $= 29 \times 71 + 31 \times 48 = 2059 + 1488 = 3547$ kg.
Combined mean $= \frac{3547}{60} \approx 59.1$ kg.

Answer: about 59.1 kg

Solved Example 8

The mean annual salary paid to 1000 employees of a company is ₹3400. The mean annual salaries paid to male and female employees are ₹200 and ₹4200. The percentages of male and female employees are
(A) 50, 50
(B) 40, 60
(C) 70, 30
(D) 20, 80

Solution:

Let the number of male employees be $m$ , so there are $1000 - m$ female employees.
Combined mean: $\frac{200 m + 4200 ( 1000 - m )}{1000} = 3400$ .
$4200000 - 4000 m = 3400000$ , so $m = 200$ and females $= 800$ .
Males $= 20%$ , females $= 80%$ .

Answer: (D) 20, 80

Solved Example 9

The average weight of 25 boys was calculated as 78.4 kg. It was later found that one weight was misread as 69 kg instead of 96 kg. The correct average is
(A) 79 kg
(B) 79.48 kg
(C) 81.32 kg
(D) none of these

Solution:

Incorrect total $= 25 \times 78.4 = 1960$ kg.
Correct total $= 1960 - 69 + 96 = 1987$ kg.
Correct average $= \frac{1987}{25} = 79.48$ kg. (Shortcut: $78.4 + \frac{96 - 69}{25} = 79.48$ .)

Answer: (B) 79.48 kg

Solved Example 10

A car owner buys petrol at ₹7.50, ₹8.00 and ₹8.50 per litre in three successive years. If he spends ₹4000 each year, the average cost per litre is
(A) ₹8
(B) ₹8.25
(C) ₹7.98
(D) none of these

Solution:

Litres bought: $\frac{4000}{7.5} \approx 533.33$ , $\frac{4000}{8} = 500$ and $\frac{4000}{8.5} \approx 470.59$ ; total $\approx 1503.92$ litres.
Total money spent $=$ ₹12000.
Average cost $= \frac{12000}{1503.92} \approx$ ₹7.98 per litre. The simple mean ₹8 is wrong because more litres were bought when petrol was cheaper.

Answer: (C) ₹7.98

Solved Example 11

The mean of the following frequency distribution is 50 and the total frequency is 120. Find the missing frequencies

f_{1}

and

f_{2}

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	$f_{1}$	32	$f_{2}$	19	120

(A) 28, 24
(B) 24, 36
(C) 36, 28
(D) none of these

Solution:

Class marks: 10, 30, 50, 70, 90. So $\sum f_{i} x_{i} = 170 + 30 f_{1} + 1600 + 70 f_{2} + 1710 = 3480 + 30 f_{1} + 70 f_{2}$ .
Mean condition: $\frac{3480 + 30 f _{1} + 70 f _{2}}{120} = 50$ , which gives $3 f_{1} + 7 f_{2} = 252$ .
Total frequency: $17 + f_{1} + 32 + f_{2} + 19 = 120$ , which gives $f_{1} + f_{2} = 52$ .
Solving: $3 (52 - f_{2}) + 7 f_{2} = 252$ , so $f_{2} = 24$ and $f_{1} = 28$ .

Answer: (A) $f_{1} = 28$ , $f_{2} = 24$

Solved Example 12

Find the arithmetic mean of the following frequency distribution.

$x$	5	8	11	14	17
$f$	4	5	6	10	20

Solution:

$N = 4 + 5 + 6 + 10 + 20 = 45$ .
$\sum f_{i} x_{i} = 5 (4) + 8 (5) + 11 (6) + 14 (10) + 17 (20) = 20 + 40 + 66 + 140 + 340 = 606$ .
$\overset{x}{ˉ} = \frac{606}{45} \approx 13.47$ .

Answer: 13.47 (approx.)

Solved Example 13

Find the mean of the following frequency distribution using an assumed mean.

$x$	5	15	25	35	45	55
$f$	12	18	27	20	17	6

Solution:

$N = 100$ . Take $a = 35$ , so $d_{i} = x_{i} - 35 = - 30, - 20, - 10, 0, 10, 20$ .
$\sum f_{i} d_{i} = 12 (- 30) + 18 (- 20) + 27 (- 10) + 20 (0) + 17 (10) + 6 (20) = - 700$ .
$\overset{x}{ˉ} = 35 + \frac{- 700}{100} = 28$ .

Answer: 28

Solved Example 14

Find the weighted mean of the first

n

natural numbers when their weights are equal to their squares.

Solution:

Weighted mean $= \frac{\sum k \cdot k ^{2}}{\sum k ^{2}} = \frac{\sum k ^{3}}{\sum k ^{2}}$ .
$\sum k^{3} = \frac{n ^{2} ( n + 1 ) ^{2}}{4}$ and $\sum k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ .
Weighted mean $= \frac{n ^{2} ( n + 1 ) ^{2}}{4} \times \frac{6}{n ( n + 1 ) ( 2 n + 1 )} = \frac{3 n ( n + 1 )}{2 ( 2 n + 1 )}$ .

Answer: $\frac{3 n ( n + 1 )}{2 ( 2 n + 1 )}$

Solved Example 15

The mean income of one group of persons is ₹400 and that of another group is ₹480. The mean income of both groups together is ₹430. Find the ratio of the numbers of persons in the two groups.

Solution:

$430 = \frac{400 n _{1} + 480 n _{2}}{n _{1} + n _{2}}$ .
$430 n_{1} + 430 n_{2} = 400 n_{1} + 480 n_{2}$ , so $30 n_{1} = 50 n_{2}$ .
$n_{1} : n_{2} = 5 : 3$ .

Answer: $5 : 3$

Solved Example 16

Find the median of the following marks obtained by 100 students.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	8	30	40	12	10

Solution:

Class	Frequency $f$	Cumulative frequency $c f$
0-10	8	8
10-20	30	38
20-30	40	78
30-40	12	90
40-50	10	100

$N = 100$ , so $\frac{N}{2} = 50$ .
The first $c f$ that reaches 50 is 78, so the median class is 20-30.
$l = 20$ , $c f = 38$ , $f = 40$ , $h = 10$ .
Median $= 20 + \frac{50 - 38}{40} \times 10 = 20 + 3 = 23$ .

Answer: 23 marks (see Figure 2)

Solved Example 17

In an examination the weights of Physics, Chemistry, English and Mathematics are 2, 1, 1 and 2. A student scores 60 in Physics, 70 in Chemistry, 70 in English and 80 in Mathematics. The weighted mean is
(A) 60
(B) 70
(C) 80
(D) 85

Solution:

$\sum w_{i} x_{i} = 2 (60) + 1 (70) + 1 (70) + 2 (80) = 420$ and $\sum w_{i} = 6$ .
Weighted mean $= \frac{420}{6} = 70$ .

Answer: (B) 70

Solved Example 18

Find the mean marks of the 100 students in Solved Example 16 by the step-deviation method.

Solution:

Class	Class mark $x_{i}$	$f_{i}$	$u_{i} = \frac{x _{i} - 25}{10}$	$f_{i} u_{i}$
0-10	5	8	$- 2$	$- 16$
10-20	15	30	$- 1$	$- 30$
20-30	25	40	0	0
30-40	35	12	1	12
40-50	45	10	2	20
Total		100		$- 14$

Take $a = 25$ and $h = 10$ .
$\overset{x}{ˉ} = 25 + 10 \times \frac{- 14}{100} = 25 - 1.4 = 23.6$ .

Answer: 23.6 marks

Solved Example 19

Find the mode of the marks of the 100 students in Solved Example 16.

Solution:

The highest frequency is 40, so the modal class is 20-30.
$l = 20$ , $f_{1} = 40$ , $f_{0} = 30$ , $f_{2} = 12$ , $h = 10$ .
Mode $= 20 + \frac{40 - 30}{2 ( 40 ) - 30 - 12} \times 10 = 20 + \frac{100}{38} \approx 22.63$ .

Answer: about 22.6 marks (see Figure 3)

Solved Example 20

Find the median of the following discrete distribution.

$x$	2	3	4	5	6
$f$	2	4	8	4	1

Solution:

Cumulative frequencies: 2, 6, 14, 18, 19, so $N = 19$ (odd).
Median $= (\frac{19 + 1}{2})^{th} = 1 0^{th}$ observation.
Observations 7 to 14 have the value 4 ( $c f$ goes from 6 to 14), so the 10th observation is 4.

Answer: Median = 4

Practice Questions

The algebraic sum of the deviations of a set of values from their arithmetic mean is:Answer: 0
The numbers of children in 25 families are 3, 1, 4, 0, 2, 1, 1, 2, 3, 3, 2, 2, 2, 5, 0, 1, 4, 1, 2, 1, 2, 3, 0, 1, 4. Find the mean number of children per family.Answer: 2
The mean of 12 numbers is 24. If 5 is added to every number, the new mean is:Answer: 29
The mean of $n$ numbers is $\overset{x}{ˉ}$ . If the sum of the first $n - 1$ numbers is $k$ , the $n^{th}$ number is:Answer: $n \overset{x}{ˉ} - k$
If a variate $X = a U + bV$ , where $U$ and $V$ are variates, the mean of $X$ is:Answer: $a \overset{ˉ}{U} + b \overset{ˉ}{V}$
The mean of a set of observations is $\overset{x}{ˉ}$ . If each observation is divided by $β$ and then increased by 12, the new mean is:Answer: $\frac{x ˉ}{β} + 12$
A variable takes the values $0, 1, 2, \dots, n$ with frequencies $q^{n},^{n} C_{1} q^{n - 1} p,^{n} C_{2} q^{n - 2} p^{2}, \dots, p^{n}$ , where $p + q = 1$ . The mean is:Answer: $n p$
The weighted mean of the first $n$ natural numbers when all weights are equal is:Answer: $\frac{n + 1}{2}$
The sum of the squares of the deviations of a set of values, taken from their mean, is: (A) least (B) zero (C) maximum (D) none of theseAnswer: (A) least. It is smaller than the sum of squares about any other point, and zero only when all values are equal.
The points scored by a basketball team in a series of matches are 15, 3, 8, 10, 22, 5, 27, 11, 12, 19, 18, 21, 13, 14. The median is:Answer: 13.5
The best average for qualitative data that can be ranked (such as intelligence) is:Answer: Median
The median of 16, 10, 14, 11, 9, 8, 12, 6, 5 is:Answer: 10
The arithmetic mean of $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ is:Answer: $\frac{2 ^{n}}{n + 1}$
The mean wage of 1000 workers in two shifts of 700 and 300 workers is ₹500. The mean wage of the 700 day-shift workers is ₹450. The mean wage of the night-shift workers is:Answer: ₹616.67
The mean of $n$ items is $\overset{x}{ˉ}$ . If the items are increased by $2, 2^{2}, 2^{3}, \dots, 2^{n}$ respectively, the new mean is:Answer: $\overset{x}{ˉ} + \frac{2 ^{n + 1} - 2}{n}$
$\overset{x}{ˉ}_{1}$ and $\overset{x}{ˉ}_{2}$ are the means of two distributions with $\overset{x}{ˉ}_{1} < \overset{x}{ˉ}_{2}$ , and $\overset{x}{ˉ}$ is the mean of the combined distribution. Then:Answer: $\overset{x}{ˉ}_{1} < \overset{x}{ˉ} < \overset{x}{ˉ}_{2}$
A boy goes to school at $x$ km/h and returns by the same route at $y$ km/h. His average speed is:Answer: $\frac{2 x y}{x + y}$ km/h
Ram spends equal amounts of money on three kinds of pens priced at ₹5, ₹10 and ₹15 per pen. The average cost of each pen is:Answer: ₹ $\frac{90}{11}$
The mean of $x_{1}, x_{2}, \dots, x_{10}$ is 20. The mean of $x_{1} + 4, x_{2} + 8, x_{3} + 12, \dots, x_{10} + 40$ is:Answer: 42
The mean of a set of numbers is $\overset{x}{ˉ}$ . If each number is increased by $λ$ , the new mean is:Answer: $\overset{x}{ˉ} + λ$
If the mean of the first $n$ natural numbers is $\frac{n + 7}{3}$ , then $n$ is:Answer: 11
A salesman's monthly sales for the first 11 months were ₹12000 each. The average for the whole year fell to ₹11375. The sale in the last month was:Answer: ₹4500
The mean of the first three terms is 14 and the mean of the next two terms is 18. The mean of all five terms is:Answer: 15.6
The arithmetic mean of $1, 2, 4, 8, \dots, 2^{n}$ is:Answer: $\frac{2 ^{n + 1} - 1}{n + 1}$
The mean of 50 observations is 36. If the observations 30 and 42 are deleted, the mean of the remaining observations is:Answer: 36
The mean of the first $n$ terms of the A.P. $a, a + d, a + 2 d, \dots$ is:Answer: $a + \frac{( n - 1 ) d}{2}$
If the mean of the $n$ observations $1^{2}, 2^{2}, 3^{2}, \dots, n^{2}$ is $\frac{46 n}{11}$ , then $n$ is:Answer: 11
The mean of a set of numbers is $\overset{x}{ˉ}$ . If each number is multiplied by $λ$ , the new mean is:Answer: $λ \overset{x}{ˉ}$
The mean of discrete observations $y_{1}, y_{2}, \dots, y_{n}$ is given by:Answer: $\frac{\sum y _{i}}{n}$
The mean of $n$ values is $\overset{x}{ˉ}$ . If the first value is increased by 1, the second by 2, and so on, the new mean is:Answer: $\overset{x}{ˉ} + \frac{n + 1}{2}$
Find the mode of the distribution with $x$ : 4, 5, 6, 7, 8 and $f$ : 6, 7, 10, 8, 3.Answer: 6
The mean of 27, 31, 89, 107, 156 is 82. The mean of 130, 126, 68, 50, 1 is:Answer: 75
A group has 40 observations. The mean of the first 10 is 4.5 and the mean of the remaining 30 is 3.5. The mean of the group is:Answer: $\frac{15}{4}$
The average age of a group of men and women is 30 years. The average age of the men is 32 and of the women is 27. The percentage of women in the group is:Answer: 40%
If the mean of $x_{1}, x_{2}, \dots, x_{n}$ is $\overset{x}{ˉ}$ , the mean of $a x_{1} + b, a x_{2} + b, \dots, a x_{n} + b$ is:Answer: $a \overset{x}{ˉ} + b$
The mean of 7 observations of one group is 10 and that of 3 observations of another group is 5. The mean of the combined group is:Answer: 8.5
The mean Mathematics marks of 100 students is 72. There are 70 boys with mean 75. The mean marks of the girls is:Answer: 65
The arithmetic mean of $1 \cdot 3 \cdot 5, 3 \cdot 5 \cdot 7, 5 \cdot 7 \cdot 9, \dots, (2 n - 1) (2 n + 1) (2 n + 3)$ is:Answer: $2 n^{3} + 8 n^{2} + 7 n - 2$
The median of 21 observations is 40. If the observations greater than the median are increased by 6, the new median is:Answer: 40
The mean of $x, x + 2, x + 4, x + 6, x + 8$ is 11. The mean of the last three observations is:Answer: 13
The median of 19 observations is 30. If two observations 8 and 32 are added, the median of the 21 observations is:Answer: 30
The average age of a teacher and three students is 20 years. All students are of equal age and the teacher is 20 years older than each student. The age of the teacher is:Answer: 35 years
The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations is increased by 2, the median of the new set: [AIEEE 2003]Answer: remains the same
The average marks of boys in a class is 52 and of girls is 42. The average of the whole class is 50. The percentage of boys is: [AIEEE 2007]Answer: 80%

Common Mistakes to Avoid

Watch out

Using class limits instead of class marks for $x_{i}$ when finding the mean of grouped data.
Finding the median without first arranging the data in order.
Using the cumulative frequency of the median class in the median formula. The formula needs the $c f$ of the class before it.
Taking $f_{0}$ and $f_{2}$ as the smallest frequencies in the table. They are the frequencies of the classes just before and just after the modal class.
Forgetting to multiply by $h$ in the step-deviation method.
Taking the simple mean of speeds or prices. Average speed and average cost always come from total ÷ total.
Assuming the mean of $x_{i} + 2 i$ is $\overset{x}{ˉ} + 2$ . The amount added changes with $i$ , so add the whole series $2 (1 + 2 + \dots + n)$ .
For a corrected mean, dividing the correction by the wrong count. The number of observations does not change when a misread value is replaced.

Frequently Asked Questions

What is the difference between mean, median and mode?

The mean is the total of all observations divided by their number. The median is the middle value when the data is arranged in order. The mode is the value that occurs most often. The mean uses every value, so extreme values affect it; the median and mode are positional and resist extreme values.

How do you find the median of grouped data?

Make a cumulative frequency column and find $\frac{N}{2}$ . The median class is the first class whose cumulative frequency reaches $\frac{N}{2}$ . Then median $= l + (\frac{N /2 - c f}{f}) \times h$ , where $c f$ is the cumulative frequency of the class before the median class, $f$ its frequency and $h$ its width.

What is the formula for the mode of grouped data?

Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h$ . Here $l$ is the lower limit of the modal class, $f_{1}$ its frequency, $f_{0}$ and $f_{2}$ the frequencies of the classes just before and after it, and $h$ the class width. All classes must have equal width.

Why is the sum of deviations from the mean always zero?

Because $\sum (x_{i} - \overset{x}{ˉ}) = \sum x_{i} - n \overset{x}{ˉ}$ , and $n \overset{x}{ˉ}$ is exactly the total $\sum x_{i}$ . The positive deviations of values above the mean cancel the negative deviations of values below it, which is why the mean is called the balance point of the data.

When should the median be used instead of the mean?

Use the median when the data has extreme values or is skewed, such as incomes, house prices or waiting times. One very large value can pull the mean far from the typical value, but the median depends only on the middle position and hardly changes. The median also works for data that can only be ranked.

What happens to mean, median and mode when every observation is changed to ax + b?

All three averages follow the same rule: the new mean is $a \overset{x}{ˉ} + b$ , the new median is $a M + b$ and the new mode is $a \times mode + b$ . Adding $b$ shifts every average by $b$ , and multiplying by $a$ scales every average by $a$ .

How are mean, median and mode asked in JEE Main?

JEE Main questions usually test combined and corrected means, missing frequencies for a given mean, means of series such as $x_{i} + 2 i$ , the median after some observations change, and the effect of adding or multiplying every value by a constant. Most can be solved in under a minute using totals and the shift and scale rules.

Is statistics part of the JEE Advanced syllabus?

Yes. JEE Advanced includes measures of central tendency and dispersion: mean, median and mode of grouped and ungrouped data, along with mean deviation, variance and standard deviation. Advanced questions often combine these with series sums, binomial coefficients or algebra, as in the weighted mean and binomial frequency examples on this page.

Previous year questions on Basic Concepts of Statistics

4 questions from past papers, each with a step-by-step solution.

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