JEE Main Study Notes for Statistics: Statistics is the science that deals with collecting, analyzing, and presenting any data. Statistics takes the values of some parameters and plots them in a meaningful manner. The probability of getting at least one question from statistics in JEE Main Mathematics Question Paper is 100%. Candidates should ensure that they start their preparation early. Check JEE Main Preparation Tips.

• JEE Main Study Notes for Statistics include Mean, Median, Mode, Dispersion, Mean Deviation, and Standard Deviation.
• The topic of Statistics has a weightage of 3% in JEE Main. Despite the weightage, candidates should be well prepared as this section can fetch good marks.

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Measures of Central Tendency

A Central Tendency is a central value of a probability distribution. The measures indicate where most values fall in the distribution. In statistics, there are two types of averages under which the central measures of tendency fall:

1. Mathematical Averages
   • Arithmetic mean or Mean
   • Geometric mean
   • Harmonic mean
2. Averages of position
   • Median
   • Mode

Arithmetic Mean

1. Arithmetic Mean for Unclassified Data

If n numbers are x1, x2, x3 ........ xn then their arithmetic mean is

                     

2. Arithmetic Mean for Frequency Distribution

Let f1, f2, ........, fn is the corresponding frequencies of x1, x2, .......xn. Then

                         

3. Arithmetic Mean for Classified Data

Class Mark of the class interval a – b, 

For classified data, we take the class marks x1, x2, ...., xn of the classes as variables and

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Geometric Mean

If x1, x2, ......, xn be n values of the variable then 

         

For Frequency Distribution  , where 

or 

Harmonic Mean

        

For Frequency Distribution,  , where 

Short-cut Method to Calculate Arithmetic Mean

1. Short-cut Method for Simple Distribution

                  

Where, a = assumed mean, d = x – a, n = no. of terms

2. Short-cut Method for Unclassified Frequency Distribution

              

Where a = assumed mean, d = x – a, f = frequency of variable x

3. Short-cut Method for Classified Frequency Distribution

          

Where a = assumed mean, d = x – a, x = class-mark of the class-interval, f = frequency of the class interval

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Median

Median is the middle-most value of the data given when arranged in ascending or descending order. If the question given is in the form of ungrouped data, we first arrange the data into ascending or descending order and then identify the Median by applying the formulas in accordance with the data.

1. Determination of Median for Simple Distribution

Arrange the data provided in ascending or descending order and then find the n (number of terms).

• If the number of terms is odd then  term is the Median
• If the number of terms is even then there are two middle terms namely,

      and  term. Hence, Median = Mean of  and  terms.

2. Determination of Median for Unclassified Frequency Distribution

• First, we need to find  , where 
• Calculate the cumulative frequency of each value of the variable and then take the value of the variable which is equal to or just greater than  . The value of the chosen variable is the median.
•  Determination of Median for Classified Data (Class Limit & Boundary)

At times the questions are asked in the form of Overlapping intervals, e.g. 10 – 20, 20 – 30, 30 – 40, where the Upper limit for 10 - 20 interval = 20, and Lower limit = 10

For Non-overlapping intervals, e.g. 10 - 19, 20 - 29, Upper boundary for 10 - 19 =  = lower boundary of 20 – 29.

3. Determination of Median by Graph

To determine the median with the help of graphs, draw the “less than” ogive and “more than” ogive for the distribution. The abscissa of the point of intersection of these ogives is the median.

Mode

Mode is the variable that occurs the maximum number of times in a given series of elements. It is the value at the point in the distribution where items tend to be most heavily concentrated.

1. Mode for a Raw Data

Mode from the following numbers of a variable 70, 80, 90, 96, 70, 96, 96, 90 is 96 as 96 occurs a maximum number of times.

      

2. Mode for Unclassified Frequency Distribution

                 

In the above-given data, the frequency that has appeared a maximum number of times is 5, with 13 occurrences. Hence, the model is 5.

3. Mode for Classified Distribution

The class that has maximum frequency is called the MODAL CLASS and the middle point of the modal class is called the CRUDE MODE. The class just before the modal class is called PRE-MODAL CLASS and the class after the modal class is called the POST-MODAL CLASS

Determination of mode for classified Data (continuous distribution)

        

• l = lower limit of the modal class
• f0 = frequency of the modal class
• f–1 = frequency of the pre-modal class
• f1 = frequency of the post-modal class
• i = length of the class-interval.

Must Read: JEE Main Study Notes on Differential Calculus

Dispersion

Dispersion normally means the extent to which the numerical data tends to spread. It is also called Variation. Its measurement is called Deviation. There are four Measures (methods) of Dispersion:

• Mean Deviation
• Standard Deviation
• Range
• Quartile Deviation

Mean Deviation

Also called Mean Absolute Deviation, it is the mean of the absolute deviations of a set of data from its mean denoted by  or  or 

1. For simple (discrete) distribution

            

where n = no. of terms, z = A or M or M0

2. For Unclassified frequency distribution

    

3. For classified distribution

      

where x stands for class-mark

Standard Deviation

It measures the dispersion of the dataset relative to its mean. It is calculated as the square root of the variance. (The square of S.D., i.e., σ2 is called the Variance.)

1. For simple (discrete) distribution

      

2. For frequency distribution

   

3. For classified data

  

where x = class-mark of the interval.

Read: JEE Main Probability Study Notes

Solved Previous Year Questions on Statistics

Question: Compute the median from the following table.

Answer:

N = Σf = 159 (Odd number)

Median is [1 / 2] (n + 1) = [1 / 2] [(159 + 1)] = 80th value, which lies in the class [30 – 40] (see the row of cumulative frequency 95, which contains 80).

Hence the median class is [30 – 40].

We have l = Lower limit of median class=30

f = frequency of median class=45

C = Total of all frequencies preceding median class=50

i = width of class interval of median class=10

Required median = l + ([N / 2 − C] / f) * i

= 30 + ([159 / 2 − 50] / 45) × 10

= 3 + 295 / 45

= 36.55

Question: The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are ______________.

Answer: Corrected Σx = 40 × 200 − 50 + 40 = 7990

Corrected x bar = 7990 / 200 = 39.95

Incorrect Σx2 = n [σ2 + (\bar{x}^2xˉ2) = 200 [152 + 402] = 365000

Correct Σx2 = 365000 − 2500 + 1600 = 364100

Corrected σ = \sqrt{\frac{364100 }{ 200}}200364100 − (39.95)2

= √(1820.5 − 1596)

= √224.5

= 14.98

Question: The average of n numbers x1, x2,……. xn is M. If xn is replaced by x′, then what is the new average?

Answer: M = \frac{x_1, x_2,……. x_n}{n}nx1,x2,…….xn i.e.,

nM = x1, x2,……. xn−1 + xn

nM – xn = x1, x2,……. xn−1

[nM – xn + x′] / n = [x1, x2,……. xn−1 + x′] / n

New average = [nM − xn + x′] / n

Question: The following data gives the distribution of the height of students.

Question: What is the median of the distribution?

Answer: Arranging the data in ascending order of magnitude, we obtain

Here, the total number of items is 41 i.e., an odd number.

Hence, the median is [(41 + 1) / 2]th i.e., 21st item.

From the cumulative frequency table, we find that the median i.eThe 21st item is 155.

(All items from 16 to 22nd are equal, each 155).

Study Tips for JEE Main Mathematics

1. Make your own formula book that you can refer to as often as possible.
2. Solve a minimum of 20 questions from Mathematics every day.
3. Make use of logical thinking to solve JEE Main Mathematics questions.
4. Use NCERT, R. D Sharma, and Arihant books for learning and practice.
5. Solve as many mock tests and previous years’ questions as you can.

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