# Ex.14.3 Q1 Statistics Solution - NCERT Maths Class 10

## Question

The following frequency distribution gives the monthly consumption of electricity of $$68$$ consumers of a locality. Find the median, mean and mode of the data and compare them.

 Monthly consumption (in units) Number of consumers $$65 – 85$$ $$4$$ $$85 – 105$$ $$5$$ $$105 – 125$$ $$13$$ $$125 – 145$$ $$20$$ $$145 – 165$$ $$14$$ $$165 – 185$$ $$8$$ $$185 – 205$$ $$4$$

Video Solution
Statistics
Ex 14.3 | Question 1

## Text Solution

What is known?

The frequency distribution of the monthly consumption of electricity of $$68$$ consumers of a locality.

What is unknown?

The median, mean and mode of the data and the comparison between them.

Reasoning:

We will find the mean by step-deviation method.

Mean,$$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$

Modal Class is the class with highest frequency

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

Where,

Class size,$$h$$

Lower limit of modal class,$$l$$

Frequency of modal class,$$f_1$$

Frequency of class preceding modal class,$$f_0$$

Frequency of class succeeding the modal class,$$f_2$$

Median Class is the class having Cumulative frequency $$(cf)$$ just greater than $$\frac{n}{2}$$

Median $$= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h$$

Class size,$$h$$

Number of observations,$$n$$

Lower limit of median class,$$l$$

Frequency of median class,$$f$$

Cumulative frequency of class preceding median class,$$cf$$

Steps:

To find the class marks, the following relation is used.

Class mark

{{x}_{i}}=\,\,\frac{\left[ \begin{align} & \text{Upper}\ \text{class}\ \text{limit}\text{+} \\ & \text{Lower}\ \text{class}\ \text{limit} \\ \end{align} \right]}{2}

Class size , $$h=20$$

Taking assumed mean,$$a=$$ $$135$$

$$d_i\;u_i$$ and $$f_i\;u_i$$ are calculated according to step deviation method as follows:

 Monthly consumption (in units) Number of consumers ${f_i}$ Class mark ${x_i}$ ${d_i} = {x_i} - a$ ${u_i} = \frac{{{d_i}}}{h}$ ${f_i}{u_i}$ $$65 – 85$$ $$4$$ $$75$$ $$-60$$ $$-3$$ $$-12$$ $$85 – 105$$ $$5$$ $$95$$ $$-40$$ $$-3$$ $$-10$$ $$105 – 125$$ $$13$$ $$115$$ $$-20$$ $$-1$$ $$-13$$ $$125 – 145$$ $$20$$ $$135$$ $$0$$ $$0$$ $$0$$ $$145 – 165$$ $$14$$ $$155$$ $$20$$ $$1$$ $$14$$ $$165 – 185$$ $$8$$ $$175$$ $$40$$ $$2$$ $$16$$ $$185 – 205$$ $$4$$ $$195$$ $$60$$ $$3$$ $$12$$ Total $$68$$ $$7$$

From the table, we obtain

\begin{align}\sum {{f_i} = 68} \\\sum {{f_i}{u_i}} = 7\end{align}

Mean, $$\overline {\rm{x}} = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$

\begin{align}&= 135 + \left( {\frac{7}{{68}}} \right) \times 20\\&= 135 + \frac{{140}}{{68}}\\&= 135 + 2.05\\&= 137.05\end{align}

To find mode

 Monthly consumption (in units) Number of consumers $$65 – 85$$ $$85 – 105$$ $$105 – 125$$ $$125 – 145$$ $$145 – 165$$ $$165 – 185$$ $$185 – 205$$ $$4$$ $$5$$ $$13$$ $$20$$ $$14$$ $$8$$ $$4$$

From the table, it can be observed that the maximum class frequency is $$20$$, belonging to class interval $$125 - 145.$$

Class size ($$h$$) $$=20$$

Modal class =$$125 - 145$$

Lower limit of modal class $$l$$=$$125$$

Frequency  of modal class $$(f_1)$$=$$20$$
Frequency  of class preceding modal class $$(f_0)$$=$$13$$
Frequency  of class succeeding the modal class $$(f_2)$$=$$14$$

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

\begin{align} &=125\!+\!\!\left(\!{\frac{{20 - 13}}{{2 \times 20 - 13 - 14}}}\!\right)\!\!\times\!20\\ &=125 + \left( {\frac{7}{{40 - 27}}} \right) \times 20\\ &=125 + \frac{7}{{13}} \times 20\\ &=125 + \frac{{140}}{{13}}\\&=125 + 10.76\\&=135.76\end{align}

To find the median of the given data, cumulative frequency is calculated as follows.

 Monthly consumption (in units) Number of consumers $$f$$ Cumulative Frequency $$cf$$ $$65 – 85$$ $$4$$ $$4$$ $$85 – 105$$ $$5$$ $$4 + 5 = 9$$ $$105 – 125$$ $$13$$ $$9 + 13 = 22$$ $$125 – 145$$ $$20$$ $$22 + 20 = 42$$ $$145 – 165$$ $$14$$ $$42 + 14 = 56$$ $$165 – 185$$ $$8$$ $$56 + 8 = 64$$ $$185 – 205$$ $$4$$ $$64 + 4 = 68$$ $$n=68$$

From the table, we obtain

$$n =68$$ $$\Rightarrow \frac{{n}}{2}=34$$

Cumulative frequency ($$cf$$ )just greater than $$\frac{{n}}{2}$$ is $$42,$$ belonding to class-interval $$125-145$$

Therefore,median class = $$125-145$$

Class size, $$h$$ $$= 20$$

Lower limit  of median class ,$$l$$ $$=125$$

Frequency  of median class ,$$f$$ $$=20$$

Cumulative frequency  of class preceding median class ,$$cf$$$$=22$$

\begin{align}\text {Median} &=l+\left(\frac{\frac{n}{2}-c f}{f}\right)\!\!\times\!h \\ &=\!\!125\!+\!\left(\frac{34-22}{20}\right)\!\!\times\!\!20 \\ &=125+12 \\ &=137\end{align}

Therefore, median, mode, mean of the given data is $$137, 135.76,$$and $$137.05$$ respectively.The three measures are approximately the same in this case.

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