CBSE Class 10 Mathematics Statistics Notes Set 04

Mathematics Concept Notes for Class 10: Chapter 13 Statistics

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Mode

Mode is the value of the observation having the maximum frequency.

Example 1. Find the mode from the following data: 110, 120, 130, 120, 110, 140, 130, 120, 140, 120.
Answer:
Arranging the data in the form of a frequency table:
ValueFrequency
1102
1204
1302
1402

The value 120 occurs the maximum number of times, i.e. 4. Hence the mode of this data is 120.

Mode of Grouped Data

In a grouped frequency distribution it is not possible to find the mode by looking at the frequencies. We can only locate the class having the maximum frequency, called the modal class. The mode is a value inside the modal class and is calculated by

\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]

where \( l \) = lower limit of the modal class, \( h \) = width of the modal class, \( f_1 \) = frequency of the modal class, \( f_0 \) = frequency of the class preceding the modal class, \( f_2 \) = frequency of the class succeeding the modal class.

Example 2. Find the mode of the following distribution.
Height (in cm)above 30above 40above 50above 60above 70above 80
No. of plants3430271982

Answer:
To convert to a frequency distribution, write above 30 as 30 - 40, above 40 as 40 - 50 and so on. The frequencies are found by subtracting the next number of plants from the current one.
Height (in cm)30 - 4040 - 5050 - 6060 - 7070 - 8080 - 90
No. of plants4381162

Maximum frequency = 11, so the modal class is 60 - 70.
\( l = 60,\ h = 10,\ f_1 = 11,\ f_0 = 8,\ f_2 = 6 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 60 + \frac{11 - 8}{2(11) - 8 - 6} \times 10 = 63.75 \).
Example 3. The following distribution shows the income of 500 employees (in terms of hundreds of rupees).
Income (in hundreds of rupees)No. of employees
0 - 5090
50 - 100150
100 - 150100
150 - 20080
200 - 25070
250 - 30010
Find the mode of the data.

Answer:
Maximum frequency = 150, so the modal class is 50 - 100.
\( l = 50,\ h = 50,\ f_1 = 150,\ f_0 = 90,\ f_2 = 100 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 50 + \frac{150 - 90}{2(150) - 90 - 100} \times 50 = 77.27 \).

Median

The middle-most value of the given data when it is arranged in ascending or descending order of magnitude is called the median. The median is also called a positional average.
1. If the total number of observations n is odd, Median = value of the \( \left( \frac{n+1}{2} \right)^{th} \) observation.
2. If n is even, Median \( = \frac{\text{value of the } \left( \frac{n}{2} \right)^{th} \text{ term} + \text{value of the } \left( \frac{n}{2} + 1 \right)^{th} \text{ term}}{2} \).

Example 4. Find the median of the following data: 25, 34, 31, 23, 22, 26, 35, 28, 20, 24, 32, 33.
Answer:
Arranging the data in ascending order: 20, 22, 23, 24, 25, 26, 28, 31, 32, 33, 34, 35.
Here the number of observations \( n = 12 \) (even).
Median \( = \frac{\text{value of the 6th observation} + \text{value of the 7th observation}}{2} = \frac{26 + 28}{2} = 27 \).

Example 5. Find the median of the following distribution: 37, 31, 42, 43, 46, 25, 39, 45, 32, 35, 49.
Answer:
Arranging the data in ascending order: 25, 31, 32, 35, 37, 39, 42, 43, 45, 46, 49.
Here the number of observations \( n = 11 \) (odd).
Median = value of the \( \left( \frac{11 + 1}{2} \right)^{th} \) observation = value of the 6th observation = 39.

Median of Grouped Data

In a grouped frequency distribution, the class corresponding to the cumulative frequency (c.f.) which is just greater than \( \frac{N}{2} \) is called the median class, and the median is obtained by the formula

\[ \text{Median} = l + \left( \frac{\frac{N}{2} - c.f.}{f} \right) \times h \]

where \( l \) = lower limit of the median class, \( c.f. \) = cumulative frequency of the class preceding the median class, \( f \) = frequency of the median class, \( h \) = class size and \( N = \sum f_i \).
Cumulative frequency is the running total of the frequencies. It is of two types: less than type, in which the frequencies of all the preceding classes are added to the frequency of the class, and more than type, in which the frequencies of all the succeeding classes are added to the frequency of the class.

Class intervalFrequencyCumulative frequency (less than type)Cumulative frequency (more than type)
\( b_0 - b_1 \)\( f_1 \)\( f_1 \)\( f_1 + f_2 + \cdots + f_n \)
\( b_1 - b_2 \)\( f_2 \)\( f_1 + f_2 \)\( f_2 + f_3 + \cdots + f_n \)
\( b_2 - b_3 \)\( f_3 \)\( f_1 + f_2 + f_3 \)\( f_3 + f_4 + \cdots + f_n \)
\( \vdots \)\( \vdots \)\( \vdots \)\( \vdots \)
\( b_{n-1} - b_n \)\( f_n \)\( f_1 + f_2 + \cdots + f_n \)\( f_n \)
Example 6. The following frequency distribution gives the daily wages of labourers (in Rs.). Find the median of the distribution.
Wages (in Rs.)No. of labourers
200 - 3003
300 - 4005
400 - 50020
500 - 60010
600 - 7006

Answer:
ClassFrequencyCumulative frequency
200 - 30033
300 - 40058
400 - 5002028
500 - 6001038
600 - 700644

\( N = 44 \), so \( \frac{N}{2} = 22 \). The cumulative frequency just greater than this is 28, so the median class is 400 - 500.
\( l = 400,\ c.f. = 8,\ f = 20,\ h = 100 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 400 + \frac{22 - 8}{20} \times 100 = 470 \).
Example 7. The following table gives the amount of money (in thousands of rupees) deposited in bank accounts by 80 customers. Find the median of the distribution.
Amount of money (in thousand)0 - 88 - 1616 - 2424 - 3232 - 4040 - 48
No. of customers8101624157

Answer:
ClassFrequencyCumulative frequency
0 - 888
8 - 161018
16 - 241634
24 - 322458
32 - 401573
40 - 48780

\( N = 80 \), so \( \frac{N}{2} = 40 \). The cumulative frequency just greater than this is 58, so the median class is 24 - 32.
\( l = 24,\ c.f. = 34,\ f = 24,\ h = 8 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 24 + \frac{40 - 34}{24} \times 8 = 26 \).

Knowledge Hub

1. The sum of the absolute deviations taken from the median is less than the sum of the absolute deviations taken from any other observation in the data.
2. The median is not affected by extreme values (outliers).
3. Empirical relation: Mode = 3 Median − 2 Mean.
4. If the median class happens to be the first class interval, the c.f. of the preceding class is taken as zero.

Summary

1. Mean is the sum of all the observations divided by the total number of observations. For grouped data: direct method \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \); assumed mean method \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} \); step-deviation method \( \bar{x} = a + \frac{\sum f_iu_i}{\sum f_i} \times h \). Here \( \sum f_i \) is the total frequency, a is the assumed mean, \( d_i = x_i - a \), \( u_i = \frac{x_i - a}{h} \) and h is the class size.
2. Mode is the value of the observation having the maximum frequency. For grouped data, \( \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \), where the class with the maximum frequency is the modal class.
3. Median is the middle-most value of the data when arranged in order. For n odd it is the \( \left( \frac{n+1}{2} \right)^{th} \) observation; for n even it is the mean of the \( \left( \frac{n}{2} \right)^{th} \) and \( \left( \frac{n}{2} + 1 \right)^{th} \) observations.
4. For grouped data, \( \text{Median} = l + \frac{\frac{N}{2} - c.f.}{f} \times h \), where the median class is the class whose cumulative frequency is just greater than \( \frac{N}{2} \).

Chapter 13 Statistics Concepts and Summary for Class 10 Mathematics

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