STATISTICS

STATISTICS

14.1 INTRODUCTION :

The branch of science known as statistics has been used in India from ancient times. Statistics deals with

collection of numerical facts. i.e., data, their classification & tabulation and their interpretation.

14.2 MEASURES OF CENTRAL TENDANCY :

The commonly used measure of central tendency (or averages) are :

(i) Arithmetic Mean (AM) or Simply Mean (ii) Median (iii) Mode

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14.3 ARITHMETIC MEAN : Arithmetic mean of a set of observations is equal to their sum divided by the total number of observations.

Mean of raw data: ${\mathbf{x}}_1,{\mathbf{x}}_2,{\mathbf{x}}_3,\dots ..,{\mathbf{x}}_n$ are the $\mathbf{n}$ values (or observations) the,

A.M. (Arithmetic mean) is

$\overline{x}=\frac{x_1+x_1+\dots ..+x_n}{n}=\frac{\sum _{i=1}^n{x}_i}{n}n\overline{x}$ - Sum of observations $=\frac{\sum _{i=1}^n{x}_i}{n}$

i.e. product of mean & no. of items gives sum of observation.

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Ex. 1 The mean of marks scored by 100 students was found to be 40. Later on its was discovered that a score of 56 was misread as 83 . Find the correct mean.

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Sol. $n=100,\overline{x}=40$

$\bar{x}=\frac{1}{n}(\sum x_i)\Rightarrow 40=\frac{1}{100}(\sum x_i)$

$\therefore$ Incorrect value of $\sum x_i=4000$. Now, Correct value of $\sum x_i=4000-83+83=3970$

$\therefore$ Correct mean $=\frac{\text{ correct value of }\sum x_i}{n}=\frac{3970}{100}=39.7$

So, the correct mean is 39.7

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Method for Mean of Ungrouped Data

Grouped Frequency Distribution (Grouped)

(i) Direct method : for finding mean mean $\overline{x}=\frac{\sum f_i{x}_i}{\sum f_u}$

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Ex. 2 Find the missing value of $P$ for the following distribution whose mean is 12.58

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Sol. Given $\bar{x}=12.58$ Calculation of Mean :

$\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}$

$12.58=\frac{524+7P}{50}629=524+7P;7P=105;P=15$.

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Ex. 3 Find the mean for the following distribution :

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Sol.

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$\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{1430}{40}=\frac{143}{4}=35.75$

(ii) Deviation Method : (Assumed Mean Method)

$\overline{x}=A+\frac{\sum f_i{d}_i}{\sum f_i}$

where, $A=$ Assumed mean $d_i=$ Deviation from mean $(x_i-A)$

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Ex 4. Find the mean for the following distribution by using deviation method :

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Sol.

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$\overline{x}=A+\frac{\sum f_i{d}_i}{\sum f_i}=25+\frac{77}{100}=25.77$

(iii) Step - Deviation Method : $\overline{x}=A+(\frac{\sum f_i{u}_i}{\sum f_i})h$ where,

$A=\text{ Assumed mean }{u}_i=\frac{x_i-A}{h},h=\text{ Width of class interval }$

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Ex. 5 Find the mean of following distribution with step - deviation method :

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Sol. Calculation of Mean :

$\Rightarrow \bar{x}=A+(\frac{\sum f_i{u}_i}{\sum f_i})h\Rightarrow \bar{x}=27.5+\times (\frac{-23}{40})=24.625$

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Ex. 6 The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequency $f_1$ and $f_2$

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Sol. Let $=30,h=20$

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Given $30+f_1+f_2=50$

$f_1+f_2=20$ …(i)

$\overline{x}=A+(\frac{\sum f_i{u}_i}{\sum f_i}\times h)$

$62.8=30=(\frac{58+2f_2}{50}\times 20)$

$62.8=(58+2f_2)\times \frac{2}{5}$

$32.8\times 5=116+4f_2$

$164=116+4f_2$

$4f_2=164-116$

$4f_2=48$

$f_2=12$

Now, $f_1=f_2=20$ $$ $f_1+12=20$ $$ $f_1=8$

So, the missing frequencies are $f_1=8$ and $f_2=12$.

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Ex. 7 Find the mean marks from the following data :

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Sol. Charging less than type frequency distribution in general frequency distribution.

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According to step deviation formula for mean

$\overline{x}=A+(\frac{\sum f_i{u}_i}{\sum f_i}\times h)$

$\overline{x}=45+(\frac{29}{85}\times 10)$

$\overline{x}=45+3.41$ $$

$\overline{x}=48.41$ So, the mean marks is 48.41

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14.4 PROPERTIES OF MEAN :

(i) Sum of deviations from mean is zero. i.e. $\sum _{i=1}^n(x_i-\overline{x})=0$

(ii) If a constant real number ’ $a$ ’ is added to each of the observation than new mean will be $\overline{x}+a$.

(iii) If a constant real number ‘a’ is subtracted from each of the observation then new mean will be $\overline{x}-a$.

(iv) If constant real number ’ $a$ ’ is multiplied with each of the observation then new mean will be $a\overline{x}$.

(v) If each of the observation is divided by a constant no ’ $a$ ‘, then new mean will be $\frac{\overline{x}}{a}$.

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14.5 MERITS OF ARITHETIC MEAN :

(i) It is rigidly defined, simple, easy to understand and easy to calculate.

(ii) It is based upon all the observations.

(iii) Its value being unique, we can use it to compare different sets of data.

(iv) It is least affected by sampling fluctuations.

(v) Mathematical analysis of mean is possible. So, It is relatively reliable.

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14.6 DEMERITS OF ARITHMETCI MEAN :

(i) It can not be determined by inspection nor it can be located graphically.

(ii) Arithmetic mean cannot be used for qualities characteristics such as intelligence, honesty, beauty etc.

(iii) It cannot be obtained if a single observation is missing.

(iv) It is affected very much by extreme values. In case of extreme items, A.M. gives a distorted picture of the distribution and no longer remains representative of the distribution.

(v) It may lead to wrong conclusions if the details of the data from which it is computed are not given.

(vi) It can not be calculated if the extreme class is open, e.g. below 10 or above 90.

(vii) It cannot be used in the study of rations, rates etc.

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14.7 USES OF ARITHMETIC MEAN :

(i) It is used for calculating average marks obtained by a student.

(ii) It is extensively used in practical statistics and to obtain estimates.

(iii) It is used by businessman to find out profit per unit article, output per machine, average monthly income and expenditure etc.

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14.8 MEDIAN :

Median is the middle value of the distribution. It is the value of variable such that the number of observations above it is equal to the number of observations below it.

Median of raw data

(i) Arrange the data in ascending order.

(ii) Count the no. of observation (Let there be ’ $n$ ’ observation)

(A) if $n$ be odd then median $=$ value of $(\frac{n+1}{2}{)}^{\text{th }}$ observation.

(B) if $n$ is even then median is the arithmetic mean of $(\frac{n}{2}{)}^{\text{th }}$ observation and $(\frac{n}{2}+1{)}^{\text{th }}$ observation.

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Median of class - interval data (Grouped)

Median $=\ell +\frac{\frac{N}{2}-C}{f}\times h$

$\ell =$ lower limit of median class, $N=$ total no of observation

$C=$ cumulative frequency of the class preceding the median class

$h=$ size of the median class

$f=$ frequency of the median class.

What is median class :

The class in which $(\frac{N}{2}{)}^{\text{th }}$ item lie is median class.

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Ex.8. Following are the lives in hours of 15 pieces of the components of air craft engine. Fin the median : $715,724,725,710,729,745,649,699,696,712,734,728,716,705,719$

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Sol. Arranging the data in ascending order

644. 696, 705, 710, 712, 715, 716, 719, 724, 725, 728, 729, 734, 745

$N=15$

So, Median $=(\frac{N+1}{2}{)}^{\text{th }}$ observation $=(\frac{15+1}{2}{)}^{\text{th }}$ observation

$=716$

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Ex. 9 The daily wages (in rupees) of 100 workers in a factory are given below :

Find the median wage of a worker for the above date.

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Sol.

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$N=100$ (even)

$\begin{array}{rl}\therefore \text{ Median }& =\frac{(\frac{N^{th}}{2})\text{ observation }+(\frac{N}{2}+1{)}^{\text{th }}\text{ observation }}{2}\\ \text{ Median }& =\frac{{50}^{\text{th }}\text{ observation }+{51}^{\text{th }}\text{ observation }}{2}\\ & =\frac{135+140}{2}=137.50\end{array}$

$\therefore$ Median wage of a workers in the factory is Rs 137.50 .

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Ex. 10 Calculate the median for the following distribution class :

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Sol. (i) First we find $(\frac{N}{2}{)}^{\text{th }}$ value i.e. $(\frac{55}{2}{)}^{\text{th }}={27.5}^{\text{th }}$ which lies in 20-30.

$\therefore 20-30$ class in median class here $\ell =20$

$\begin{array}{rl}& \frac{N}{2}=27.5,C=15,f=20,h=10\\ \therefore & \text{ Median }=20+\frac{275-15}{20}\times 10\end{array}$

Median $=26.25$

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Ex. 11 in the median of the following frequency distribution is 46, find the missing frequencies :

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Sol.

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Let the frequency of the class $30-40$ be $f_1$ and that of the class $50-60$ be $f_2$. The total frequency is 229

$12+30+f_1+65+f_2+25+18=229$

$\Rightarrow f_1+f_2=79$

It is given that median is 46 ., clearly, 46 lies in the class $40-50$. So, $40-50$ is the median class

$\therefore \ell =40,h=10,f=65\text{ and }C=42+f_1,N=229$

Median $=\ell +\frac{\frac{N}{2}-C}{f}\times h$

$46=40+\frac{\frac{229}{2}-(42+f_1)}{65}\times 10$

$46=40+\frac{145-2f_1}{13}$

$\Rightarrow 6=\frac{145-2f_1}{13}$ $\Rightarrow 2f_1=67$ $f_1=33.5\text{ or }34$

$\text{ Since, }{f}_1+f_2=79\therefore f_1=45$

$Hence,f_1=34$ and $f_2=45.$

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Merits of Median :

(i) It is rigidly defined, easily, understood and calculate.

(ii) It is not all affected by extreme values.

(iii) It can be located graphically, even if the class - intervals are unequal.

(iv) It can be determined even by inspection is some cases.

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Demerits of Median :

(i) In case of even numbers of observations median cannot be determined exactly.

(ii) It is not based on all the observations.

(iii) It is not subject to algebraic treatment.

(iv) It is much affected by fluctuations of sampling.

Uses of Median :

(i) Median is the only average to be used while dealing with qualitative data which cannot be measured quantitatively but can be arranged in ascending or descending order of magnitude.

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14.9 MODE:

Mode or modal value of the distribution is that value of variable for which the frequency is maximum.

Mode of ungrouped data : - (By inspection only)

Arrange the data in an array and then count the frequencies of each variate.

The variant having maximum frequency is the mode.

Mode of continuous frequency distribution

Mode $=\ell +\frac{f_1+f_0}{2f_1-f_0-f_2}\times h$

Where $\ell =$ lower limit of the modal class

$f_1=$ frequency of the class i.e. the largest frequency.

$f_0=$ frequency of the class preceding the modal class.

$f_2=$ frequency of the class succeeding the modal class.

$h=$ width of the modal class

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Ex.12. Fin the mode of the following data :

$25,16,19,48,19,20,34,15,19,20,21,24,19,16,22,16,18,20,16,19$.

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Sol. Frequency table for the given data as given below :

19 has the maximum frequency of 5 . So, Mode $=19$.

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Ex.13. The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital.

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Sol.

Here class intervals are not is inclusive form. So, Converting the above frequency table in inclusive form.

Class 34.5 - 44.5 has maximum frequency. So it is the modal class.

$\ell 34.5,h=10,f_0=21,f_1=23$ and $f_2=14$.

$\begin{array}{rl}\therefore \text{ Mode }& =\ell +\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\\ \text{ Mode }& =34.5+\frac{23-21}{46-21-14}\times 10\end{array}$

$=34.5+\frac{2}{11}\times 10$

$=36.31$ Ans.

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Ex. 14 Find the mode of following distribution :

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Sol.

Modal class frequency is 42.5 - 48.5 .

$\begin{array}{rl}l& =42.5\\ f_1& =20f_0=12,f_2=15,h=6\\ \therefore \text{ Mode }& =42.5+\frac{20-12}{2(20)-12-15}\times 6\end{array}$

$\therefore$ Mode $=46.2$

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Merits of Mode

(i) It can be easily understood and is easy to calculate.

(ii) It is not affected by extreme values and can be found by inspection is some cases.

(iii) It can be measured even if open - end classes and can be represented graphically.

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Demerits of Mode:

(i) It is ill - fined. It is not always possible to find a clearly defined mode.

(ii) It is not based upon all the observation.

(iii) It is not capable of further mathematical treatment. it is after indeterminate.

(iv) It is affected to a greater extent by fluctuations of sampling.

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Uses of Mode :

Mode is the average to be used to find the ideal size, e.g., in business forecasting, in manufacture of ready-made garments, shoes etc.

Relation between Mode, Median & Mean : Mode = 3 median - 2 mean.

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14.10 CUMULATIVE FREQUENCY CURVE OR OGIVE :

In a cumulative frequency polygon or curves, the cumulative frequencies are plotted against the lower and upper limits of class intervals depending upon the manner in which the series has been cumulated. There are two methods of constructing a frequency polygon or an Ogive.

(i) Less than method $$ (ii) More than method

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In ungrouped frequency distribution :

Ex. 15 The marks obtained by 400 students in medical entrance exam are given in the following table.

(i) Draw Ogive by less than method.

(ii) Draw Ogive by more than method.

(iii) Find the number of examinees, who have obtained the marks less than 625.

(iv) Find the number of examinees, who have obtained 625 and more than marks.

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Sol. (i) Cumulative frequency table for less than Ogive method is as following.

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Following are the O-give for the above cumulative frequency table by applying the given method and the assumed scale.

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(ii) Cumulative frequency table for more than Ogive method is as following : -

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Following are the O-give for the above cumulative frequency table.

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(iii) So, the number of examinees, scoring marks less than 625 are approximately 220.

(iv) So, the number of examinees, scoring marks 625 and more will be approximately 190 .

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Ex. 16 Draw on O-give for the following frequency distribution by less than method and also find its median from the graph.

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Sol. Converting the frequency distribution into less than cumulative frequency distribution.

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According to graph median $=34$ marks.

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DAILY PRACTIVE PROBLEMS 14

OBJECTIVE DPP - 14.1

1. The median of following series if $520,20,340,190,35,800,1210,50,80$

(A) 1210 $$

(B) 520 $$

(C) 190 $$

(D) 35

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2. If the arithmetic mean of 5,7,9, $x$ is 9 then the value of $x$ is

(A) 11 $$

(B) 15 $$

(C) 18 $$

(D) 16 $$

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3. The mode of the distribution 3, 5, 7, 4, 2, 1, 4, 3, 4 is

(A) 7 $$

(B) 4 $$

(C) 3 $$

(D) 1

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4. If the first five elements of the set $x_1,x_2,\dots x_{10}$ are replaced by $x_i+5,i=1,2,3,4,5$ and next five elements are replaced by

(A) $0$

(B) $\frac{n+1}{2}$

(C) $10$

(D) $25$

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5. If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be

(A) 7.2 $$

(B) 8.2 $$

(C) 9.2 $$

(D) 10.2

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SUBJECTIVE DPP - 14.2

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1. Find the value of $p$, if the mean of the following distribution whose mean is 20

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Sol. 1. $p=1$ $$

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2. Find the mean of following distribution by step deviation method :

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Sol. 2. $$ 112.20 $$

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3. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50 . Compute the missing frequency.

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Sol. 3. $f_1=8,f_2=12$ $$

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4. Calculate the median from the following data :

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Sol. 4. $58$ $$

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5. Find the missing frequencies and the median for the following distribution if the mean is 1.46.

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Sol. 5. $f_1=76,f_2=38$, and median $=1$ $$

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6. If the median of the following frequency distribution is 28.5 find the missing frequencies :

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Sol. 6. $f_1=8,f_2=7$ $$

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7. The marks is science of 80 students of class $X$ are given below : Find the mode of the marks obtained by the students in science.

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Sol. 7. $53.17$ $$

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8. Find the mode of following distribution :

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Sol. 8. $$ 46.67 $$

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9. During the medical check - up of 35 students of a class, their weights were recorded as follows :

Draw a less than type ogive for the given data. Hence, obtain median weight from the graph and verify the result by using the formula.

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Sol. 9. $47.5kg$ $$

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10. The following table gives the height of trees :

Draw “less than” ogive and “more than” ogive.

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11. If the mean of the following data is 18.75 , find the value of $p$ : [CBSE - 2005]

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Sol. 11. $20$

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12. Find the mean of following frequency distribution [CBSE - 2006]

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Sol. 12. $20$

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13. Find the median class of the following data : [CBSE - 2008]

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Sol. 13. $30-40$

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14. Find the mean, mode and median of the following data : [CBSE - 2008]

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Sol. 14. $$ Mean $=35.6$, Median $=35.67$ and mode $=35.45$