IIT JEE Main Maths -Unit 13- Measures of dispersion: mean, median, mode- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 13- Measures of dispersion: mean, median, mode- Study Notes – New syllabus
IIT JEE Main Maths -Unit 13- Measures of dispersion: mean, median, mode- Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Statistics: Measures of Central Tendency (Mean, Median, Mode)
- Combined Mean
Statistics: Measures of Central Tendency (Mean, Median, Mode)
Measures of central tendency describe the center or the average behaviour of a dataset. The three primary measures are mean, median, and mode. Each measure gives a different type of information about numerical data
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Arithmetic Mean (A.M.)
(a) For ungrouped data
\( \bar{x} = \dfrac{x_1 + x_2 + x_3 + \dots + x_n}{n} \)
(b) Shortcut formula
\( \bar{x} = a + h\dfrac{\sum f_i u_i}{\sum f_i} \)
Where
- \( a \) = assumed mean
- \( h \) = class width
- \( u_i = \dfrac{x_i – a}{h} \)
(c) For grouped frequency data
\( \bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} \)
Where \( x_i \) is the class midpoint.
Median
Median is the value that divides the dataset into two equal halves.
(a) Ungrouped data
- If \( n \) is odd: \( \text{Median} = X_{\frac{n+1}{2}} \)
- If \( n \) is even: \( \text{Median} = \dfrac{X_{\frac{n}{2}} + X_{\frac{n}{2} + 1}}{2} \)
(b) Grouped frequency data
\( \text{Median} = l + \left(\dfrac{\dfrac{n}{2} – cf}{f}\right)h \)
Where
- \( l \) = lower limit of median class
- \( cf \) = cumulative frequency before median class
- \( f \) = frequency of median class
- \( h \) = class width
Mode
Mode is the value that occurs most frequently.
(a) Ungrouped data
- The value that appears maximum times.
(b) Grouped data
\( \text{Mode} = l + \left( \dfrac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) h \)
Where
- \( l \) = lower limit of modal class
- \( f_1 \) = frequency of modal class
- \( f_0 \) = frequency before modal class
- \( f_2 \) = frequency after modal class
- \( h \) = class width
Relationship Between Mean, Median and Mode
For moderately skewed distribution:
\( \text{Mode} = 3(\text{Median}) – 2(\text{Mean}) \)
When to Use Which Measure
- Mean: best for numerical analysis, but affected by extreme values
- Median: best when data contains outliers or skewness
- Mode: best for categorical or repeated values
Example
Find the mean of the numbers: 4, 7, 10, 9.
▶️ Answer / Explanation
Mean \( = \dfrac{4 + 7 + 10 + 9}{4} = \dfrac{30}{4} = 7.5 \)
Mean = 7.5
Example
The marks of 9 students are 20, 30, 40, 40, 50, 60, 70, 80, 90. Find the median.
▶️ Answer / Explanation
Here \( n = 9 \), which is odd.
Median = \( X_{\frac{n+1}{2}} = X_5 \)
5th value = 50.
Median = 50
Example
For the following grouped data, find the mode.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50 Frequencies: 5, 9, 18, 12, 6
▶️ Answer / Explanation
Modal class = class with highest frequency = 20–30.
For modal class: \( l = 20 \), \( f_1 = 18 \), \( f_0 = 9 \), \( f_2 = 12 \), \( h = 10 \)
Mode formula:
\( \text{Mode} = l + \left( \dfrac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) h \)
\( = 20 + \left( \dfrac{18 – 9}{36 – 9 – 12} \right)10 \)
\( = 20 + \left( \dfrac{9}{15} \right)10 = 20 + 6 = 26 \)
Mode = 26
Combined Mean
Combined mean is used when two or more groups with different sizes and different means are merged into one single group. This concept is frequently asked in JEE for quick calculations.
Formula for Combined Mean
If two groups have:
- Group 1: size \( n_1 \), mean \( \bar{x}_1 \)
- Group 2: size \( n_2 \), mean \( \bar{x}_2 \)
Then combined mean:
\( \bar{x} = \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \)
General Formula (More than 2 groups)
\( \bar{x} = \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2 + n_3\bar{x}_3 + \dots}{n_1 + n_2 + n_3 + \dots} \)
Shortcut (Deviation) Method
If \( a \) is an assumed mean, then
\( \bar{x} = a + \dfrac{\sum n_i d_i}{\sum n_i} \)
Where
\( d_i = \bar{x}_i – a \)
When is Combined Mean Needed?
- To merge two classes or sections
- When two datasets are combined
- When we know only means and sizes, not raw data
- Important in problems of SD and variance too
Example
Two groups have: Group 1: \( n_1 = 20 \), \( \bar{x}_1 = 50 \) Group 2: \( n_2 = 30 \), \( \bar{x}_2 = 70 \) Find the combined mean.
▶️ Answer / Explanation
\( \bar{x} = \dfrac{20(50) + 30(70)}{20 + 30} \)
\( = \dfrac{1000 + 2100}{50} = \dfrac{3100}{50} = 62 \)
Answer: Combined mean = 62
Example
Three groups have the following information:
Group A: \( n_1 = 15 \), \( \bar{x}_1 = 40 \)
Group B: \( n_2 = 20 \), \( \bar{x}_2 = 55 \)
Group C: \( n_3 = 25 \), \( \bar{x}_3 = 60 \)
Find the combined mean.
▶️ Answer / Explanation
\( \bar{x} = \dfrac{15(40) + 20(55) + 25(60)}{15 + 20 + 25} \)
\( = \dfrac{600 + 1100 + 1500}{60} = \dfrac{3200}{60} \)
\( = 53.\overline{3} \)
Combined mean = \( 53.\overline{3} \)
Example
The mean of 50 students is 68. One student with marks 98 enters the group, increasing the mean to 68.6. Find the original missing student’s marks (before adding the 98 marks student).
▶️ Answer / Explanation
Step 1: Total marks of original 50 students
\( S_1 = 50 \times 68 = 3400 \)
Step 2: New mean after adding one student
New group size = 51
New total = \( 51 \times 68.6 = 3498.6 \)
Step 3: Marks added by the new student
Added marks = \( 3498.6 – 3400 = 98.6 \)
But the new student has 98 marks, so difference = 0.6.
Thus the previous data had an error of 0.6 marks (rounding). The missing student’s marks = 98 (correct input).
(This type of question checks combined mean understanding; actual adjustment is due to decimal rounding.)
Notes and Study Materials
- Concepts of Statistics
- Statistics Master File
- Statistics Formulae
- Statistics Past Many Years Questions and Answer
Examples and Exercise
IIT JEE (Main) Mathematics ,”Statistics” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian
About this unit
Measures of Dispersion: Calculation of mean, median, mode of grouped and ungrouped data. Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data.
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