# NCERT Solution class 10 maths chapter 14 | ncert class 10 maths solution STATISTICS

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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 xi and fi are large, finding the product of xi and fi becomes tedious and time consuming. So, for such situations, we use assumed mean method.

$$\displaystyle \overline{x}\,\,=\,\,a\,\,+\,\,\overline{d}$$
(where a = assumed mean, di = 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 di’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 ui
$$\displaystyle {{u}_{i}}\,\,=\,\,\frac{{x & {{ & }_{i}}\,\,-\,\,a}}{h}$$
(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 xi and fi 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 $$\displaystyle \frac{{\text{lower}\,\,\text{limit}\,\,\text{+}\,\,\text{upper}\,\,\text{limit}}}{2}$$
$$\displaystyle \overline{d}=\,\,\frac{{{{{\sum{{{{f}_{i}}d}}}}_{i}}}}{{\sum{{{{f}_{i}}}}}}$$
$$\displaystyle \overline{d}=\,\,\frac{{-\,240}}{{50}}$$
$$\displaystyle \overline{d}=\,\,-\,\,4.8$$
$$\displaystyle \overline{x}\,\,=\,\,a\,\,+\,\,\overline{d}$$
$$\displaystyle \begin{array}{l}\overline{x}\,\,=\,\,550\,\,+\,\,(-\,4.8)\\\overline{x}\,\,=\,\,550\,\,-\,\,4.8\\\overline{x}\,\,=\,\,545.20\end{array}$$

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