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ncert solutions class 10 maths chapter 14
Exercise – 14 STATISTICS

ncert solutions class 10 maths chapter 14
ncert class 10 maths solution 14.1

There are 3 method 10th class maths 14th chapter to find mean of grouped data.
1. Direct Method
2. Assumed Mean Method
3. Step-Deviation Method

1. Direct Method

\(\displaystyle \overline{x}\,\,=\,\,\frac{{{{{\sum{{{{f}_{i}}x}}}}_{i}}}}{{\sum{{{{f}_{i}}}}}}\)

2. Assumed Mean Method

Sometimes when the numerical values of x_i and f_i are large, finding the product of x_i and f_i becomes tedious and time consuming. So, for such situations, we use assumed mean method.

\(\displaystyle \overline{x}\,\,=\,\,a\,\,+\,\,\overline{d}\)
(where a = assumed mean, d_i = deviation)

\(\displaystyle \overline{d}=\,\,\frac{{{{{\sum{{{{f}_{i}}d}}}}_{i}}}}{{\sum{{{{f}_{i}}}}}}\)
(\(\displaystyle {{d}_{i}}\,\,=\,\,{{x}_{i}}\,\,-\,\,a\))

3. Step Deviation Method

• The step-deviation method will be convenient to apply if all the d_i’s have a common factor.
• The assumed mean method and step-deviation method are just simplified forms of the direct method.

\(\displaystyle \overline{x}\,\,=\,\,a\,\,+\,\,h\left( {\frac{{{{{\sum{{{{f}_{i}}u}}}}_{i}}}}{{\sum{{{{f}_{i}}}}}}} \right)\)

To fine u_i
\(\displaystyle {{u}_{i}}\,\,=\,\,\frac{{x & {{ & }_{i}}\,\,-\,\,a}}{h}\)
(where h = class interval)

EXERCISE 14.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of houses	1	2	1	5	6	2	3

Solution :
We are use direct method because the numerical values of x_i and f_i are small.

Number of Plants (Class Interval)	Number of houses (fi)	Mid-point (xi)	fixi
0 – 2	1	1	1
2 – 4	2	3	6
4 – 6	1	5	5
6 – 8	5	7	35
8 – 10	6	9	54
10 – 12	2	11	22
12 – 14	3	13	39
	Σfi = 20		Σfixi = 162

\(\displaystyle \overline{x}\,\,=\,\,\frac{{{{{\sum{{{{f}_{i}}x}}}}_{i}}}}{{\sum{{{{f}_{i}}}}}}\)
\(\displaystyle \overline{x}\,\,=\,\,\frac{{162}}{{20}}\)
\(\displaystyle \overline{x}\) = 8.1

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs)	500 – 520	520 – 540	540 – 560	560 – 580	580 – 600
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution :
We use Assumed Mean Method to solve this –

Daily wages (in Rs.)	No. of worker (fi)	Mid-point (xi)	di = xi − a di = xi − 550	fidi
500 – 520	12	510	− 40	− 480
520 – 540	14	530	− 20	− 280
540 – 560	8	550	0	0
560 – 580	6	570	20	120
580 – 600	10	590	40	400
	Σfi = 50			Σfidi = − 780 + 520 = − 240

To find xi we use \(\displaystyle \frac{{\text{lower}\,\,\text{limit}\,\,\text{+}\,\,\text{upper}\,\,\text{limit}}}{2}\)
\(\displaystyle \overline{d}=\,\,\frac{{{{{\sum{{{{f}_{i}}d}}}}_{i}}}}{{\sum{{{{f}_{i}}}}}}\)
\(\displaystyle \overline{d}=\,\,\frac{{-\,240}}{{50}}\)
\(\displaystyle \overline{d}=\,\,-\,\,4.8\)
\(\displaystyle \overline{x}\,\,=\,\,a\,\,+\,\,\overline{d}\)
\(\displaystyle \begin{array}{l}\overline{x}\,\,=\,\,550\,\,+\,\,(-\,4.8)\\\overline{x}\,\,=\,\,550\,\,-\,\,4.8\\\overline{x}\,\,=\,\,545.20\end{array}\)