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JEE Main 2025 Syllabus PDF - Subject-Wise Detailed Syllabus

Central Tendency - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Median is considered one the most difficult concept.

  • Central Values, Mean, Mode is considered one of the most asked concept.

  • 78 Questions around this concept.

Solve by difficulty

In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?

In a set of 2n distinct observations, each of the observations below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3. Then the mean of the new set of observations :

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks are given?

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively.  Later on, three observations, 3, 4 and 5, were found to be incorrect.  If the incorrect observations are omitted, then the variance of the remaining observations is :

Let  x_{1},x_{2},......................,x_{n} be n observations, and let \bar{x}  be their arithmetic mean and \sigma ^{2} be their variance.

Statement 1: Variance of 2x_{1},2x_{2},..............,2x_{n} \: is \: 4\sigma ^{2}.

Statement 2: Arithmetic mean of 2x_{1},2x_{2},..............,2x_{n} \: is \: 4 \bar{x}.

The mean of the numbers a,b,8,5,10\; is\; 6 and the variance is 6.80. Then which one of the following gives possible values of a\; \; and\; \; b ?

The mean age of 25 teachers in a school is 40 years.  A teacher retires at the age of 60 years and a new teacher is appointed in his place.  If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :

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Let M denote the median of the following frequency distribution

Class 0-4 4-8 8-12 12-16 16-20
Frequency 3 9 10 8 6

Then 20M is equal to :

Concepts Covered - 4

Central Values

Central value of Data

A measure of central tendency (or central value) is a single value that attempts to describe a set of data by identifying the central position within that set of data. Apart from mean (often called the average), there are other central values such as the median and the mode.

The mean, median and mode are all valid measures of central tendency, but under different conditions, some measures of central tendency become more appropriate to use than others. 

Mean

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set. If we have n values in a data set, i.e. x_1,x_2,x_3,\ldots,x_n , then its mean, usually denoted by \bar x (pronounced "x bar"), is:

\bar{x}=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}

Median

The median is the middle value for a set of data that has been arranged in asscending or descending order. 

For example, to find the median of the following data

\begin{array}{llllllllllll}{65} & {55} & {89} & {56} & {35} & {14} & {56} & {55} & {87} & {45} & {92}\end{array}

We first rearrange that data into order (ascending)

\begin{array}{llllllllllll}{14} & {35} & {45} & {55} & {55} &\mathbf {56} & {56} & {65} & {87} & {89} & {92}\end{array}

Median mark is the value exactly at the middle - in this case, 56

When the n is even in the data set, then simply you have to take the middle two scores and average them.

Mode

The mode is the most frequent value in our data set.

Normally, the mode is used for categorical data where we wish to know which is the most common category, 

\begin{array}{llllllllllll}{65} & {55} & {89} & {56} & {35} & {14} & {56} & {55} & {87} & {45} & {92}&{55}\end{array}

in the above case, mode of the data set is 55

Mean

MEAN (Arithmetic Mean)

The mean is the sum of the value of each observation in a dataset divided by the number of observations.

For example, to calculate the mean weight of 50 people, add the 50 weights together and divide by 50. Technically this is the arithmetic mean. 

 

Mean of the Ungrouped Data 

If n observations in data are x1, x2, x3, ……, xn, then arithmetic mean \mathit{\bar x} is given by    

\mathit{\bar x}=\frac{x_1+x_2+x_3+\ldots\dots +x_n }{n}=\frac{1}{n}\sum^{n}_{i=1}x_i

 

Mean of Ungrouped Frequency Distribution

If observations in data are x1, x2, x3, ……, xn with respective frequencies f1, f2, f3, ……, fn ; then 

Sum of the value of the observations = f1x1 + f2​​​​​​x2 + f3​​​​​x3 + …….. + fnxn

and Number of observations = f1 + f2 + f3 + ……+ fn

The mean in this case is given by 

\bar x=\frac{f_1x_1+f_2x_2+f_3x_3+\ldots\ldots +f_nx_n}{f_1+f_2+f_3+\ldots\ldots +f_n}=\frac{\sum_{i=1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}

 

Grouped Frequency Distribution

x_i is taken as mid-point of respective classes (or interval).

i.e.,

\\m=\frac{\text {lower boundary}+\text {upper boundary}}{2}\\\text{then, }\;\bar x=\frac{\sum_{i=1}^{n}f_im_i}{\sum_{i=1}^{n}f_i}

For example,

A frequency table displaying professor’s last statistic test is shown, the best estimate of the class mean is 

\begin{array}{|c|c|}\hline \text { Grade Interval } & {\text { Number of Students }} \\ \hline 10-12 & {1} \\ \hline 12-14 & {2} \\ \hline 14-16 & {0} \\ \hline 16-18 & {4} \\ \hline 18-20 & {1} \\ \hline\end{array}

First find the midpoints for all intervals

\begin{array}{|c|c|}\hline \text { Grade Interval } & {\text { Midpoint }} \\ \hline 10-12 & {11} \\ \hline 12-14 & {13} \\ \hline 14-16 & {15} \\ \hline 16-18 & {17} \\ \hline 18-20 & {19} \\\hline\end{array}

Now calculate the sum of the product of each interval frequency and midpoint,\sum_{i=i}^{n}f_im_i

11(1) + 13(2) + 15(0) + 17(4) + 19(1) = 124

\bar x=\frac{\sum_{i=1}^n f_i m_i}{\sum_{i=1}^{n} f_i}=\frac{124}{8}=15.5

Median

MEDIAN

The median is the "middle value" of the data when arranged in ascending or descending order.

It is a number that separates ordered data into 2 equal halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger.

 

Median of Ungrouped Data

If the number of observations is n,

First arrange the observations in ascending or descending order.

If n is odd :       

                     Median=\left ( \frac{n+1}{2} \right )^{th}\;\text{observation}

If n is even :             

                     Median=\frac{\text{Value of}\left ( \frac{n}{2} \right )^{th}\;\text{observation}+\text{Value of}\left ( \frac{n}{2}+1 \right )^{th}\;\text{observation}}{2}

For example,

Consider the following data: 1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1 

Ordered from smallest to largest: : 1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5

Since there are 14 observations, the median is average of (n/2)th = 7th and (n/2 + 1)th = 8th term. So median is the average of 6.8 and 7.2, which equals 7. 

The median is seven. Half of the values are smaller than seven and half of the values are larger than seven.

Median of Ungrouped Frequency Distribution

To find the median, first arrange the observations in ascending order. After this the cumulative frequencies are obtained.

Let the sum of frequencies is denoted by N.

Now if N is odd, then identify the observation whose cumulative frequency equal to or just greater than \frac{N+1}{2}. This value of the observation lies in the middle of the data and therefore, it is the required median.

If N is even, then find two observations, first whose cumulative frequency equal to or just greater than (N/2) and second whose cumulative frequency equal to or just greater than (N/2 + 1). The median is the average of these two observations

 

Median of Continuous Frequency Distribution

In this case, the following formula can be used

  1. When observations arranged in ascending order   

           \text { Median }=l+\frac{\left(\frac{N}{2}-cf\right)}{f}\times h

           where,

           l = lower limit of median class, 

           N = number of observations, 

           cf = cumulative frequency of class preceding the median class, 

           f = frequency of median class, 

           h = class size (width) (assuming class size to be equal).

Mode

MODE

Mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency. 

In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula:

\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h

where 

l = lower limit of the modal class, 

h = size of the class interval (assuming all class sizes to be equal), 

f1 = frequency of the modal class, 

f0 = frequency of the class preceding the modal class, 

f2 = frequency of the class succeeding the modal class.

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Central Values
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