knowledge-route Maths10 Ch7


title: “Lata knowledge-route-Class10-Math1-2 Merged.Pdf(1)” type: “reveal” weight: 1

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

STATISTICS

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: $\mathbf{x} _1, \mathbf{x} _2, \mathbf{x} _3, \ldots . ., \mathbf{x} _n$ are the $\mathbf{n}$ values (or observations) the,

A.M. (Arithmetic mean) is

$\bar{x}=\frac{x_1+x_1+\ldots . .+x_n}{n}=\frac{\sum_{i=1}^{n} x_i}{n} \quad n \bar{x}$ - Sum of observations $=\frac{\sum_{i=1}^{n} x_i}{n}$

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

STATISTICS

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.

STATISTICS

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

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

$\therefore$ 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

STATISTICS

Method for Mean of Ungrouped Data

$\mathbf{x} _{\mathbf{i}}$ $\mathbf{f} _{\mathbf{i}}$ $\mathbf{F} _1 \mathbf{x} _1$
$\mathbf{x} _1$ $\mathbf{f} _1$ $\mathbf{f} _1 \mathbf{x} _1$
$\mathbf{x} _2$ $\mathbf{f} _2$ $\mathbf{f} _2 \mathbf{x} _2$
$\mathbf{x} _3$ $\mathbf{f} _3$ $\mathbf{f} _3 \mathbf{x} _3$
$\cdot$ $\cdot$ $\cdot$
$\cdot$ $\cdot$ $\cdot$
$\cdot$ $\cdot$
$\sum f_i=$ $\sum f_1 x_1=$

Grouped Frequency Distribution (Grouped)

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

STATISTICS

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

$x$ 5 8 10 12 $\mathbf{P}$ 20 25
$y$ 2 5 8 22 7 4 2

STATISTICS

Sol. Given $\overline x = 12.58$ Calculation of Mean :

$\mathbf{x} _{\mathbf{i}}$ $\mathbf{f} _{\mathbf{i}}$ $\mathbf{f} _{\mathbf{i}} \mathbf{x} _{\mathbf{i}}$
$\mathbf{5}$ $\mathbf{2}$ $\mathbf{1 0}$
$\mathbf{8}$ $\mathbf{5}$ $\mathbf{4 0}$
$\mathbf{1 0}$ $\mathbf{8}$ $\mathbf{8 0}$
$\mathbf{1 2}$ $\mathbf{2 2}$ $\mathbf{2 6 4}$
$\mathbf{P}$ $\mathbf{7}$ $\mathbf{7 P}$
$\mathbf{2 0}$ $\mathbf{4}$ $\mathbf{8 0}$
$\mathbf{2 5}$ $\mathbf{2}$ $\mathbf{5 0}$
$\sum f_i=50$ $\sum f_i x_u=524+7 P$

$$$$

$\bar{x}=\frac{\sum f_i x_i}{\sum f_i}$

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

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Ex. 3 Find the mean for the following distribution :

Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency 6 8 13 7 3 2 1

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Sol.

Marks Mid Values $\mathbf{x} _{\mathbf{i}}$ No. of students
$\mathbf{f} _{\mathbf{i}}$
$\mathbf{f} _{\mathbf{i}} \mathbf{x} _{\mathbf{i}}$
$\mathbf{1 0 - 2 0}$ $\mathbf{1 5}$ $\mathbf{6}$ $\mathbf{9 0}$
$20-30$ 25 $\mathbf{8}$ $\mathbf{2 0 0}$
$30-40$ 35 $\mathbf{1 3}$ $\mathbf{4 5 5}$
$40-50$ 45 $\mathbf{7}$ $\mathbf{3 4 5}$
$50-60$ 55 $\mathbf{3}$ $\mathbf{1 6 5}$
$60-70$ 65 $\mathbf{2}$ $\mathbf{1 3 0}$
$70-80$ 75 $\mathbf{1}$ $\mathbf{7 5}$
$\sum f_i=40$ $\sum f_i x_i=1430$

STATISTICS

$ \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{1430}{40}=\frac{143}{4}=35.75 $

(ii) Deviation Method : (Assumed Mean Method)

$ \bar{x}=A+\frac{\sum f_i d_i}{\sum f_i} $

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

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

$x_i$ 15 20 22 24 25 30 33 38
Frequency 5 8 11 20 23 18 13 2

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Sol.

$x_i$ $\mathbf{f} _i$ Let $A=25$
$d_i=x_i-25$
$\mathbf{f} _i d_i$
15 5 -10 -50
20 8 -5 -40
22 11 -3 -33
24 20 -1 -20
25 23 0 0
30 18 5 90
33 13 8 104
38 2 13 26
$\sum f_i=100$ $\sum f_i d_i=77$

STATISTICS

$ \bar{x}=A+\frac{\sum f_i d_i}{\sum f_i}=25+\frac{77}{100}=25.77 $

(iii) Step - Deviation Method : $\quad \bar{x}=A+(\frac{\sum f_i u_i}{\sum f_i}) h$ 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 :

Class $10-15$ $15-20$ $20-25$ $25-30$ $30-35$ $35-40$
Frequency 5 6 8 12 6 3

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Sol. Calculation of Mean :

Class $x_i$ $\mathbf{f} _i$ Let $A=27.5$
$u_i=\frac{x_i-27.5}{5}$
$f_i u_i$
10-15 12.5 5 -3 -15
$15-20$ 17.5 6 -2 -12
$20-25$ 22.5 8 -1 -8
$25-30$ 27.5 12 0 0
$30-35$ 32.5 6 1 6
$35-40$ 37.5 3 2 6
$\sum f_i=40$ $\sum f_i u_i=-23$

$$$$

$\Rightarrow \quad \overline{x}=A+(\frac{\sum f_i u_i}{\sum f_i}) h \quad \Rightarrow \quad \overline{x}=27.5+\times(\frac{-23}{40})=24.625$

STATISTICS

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 $f_2$

Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$
Frequency 5 $f_1$ 10 $f_2$ 7 8

STATISTICS

Sol. Let $=30, h=20$

Class $\mathbf{x} _{\mathbf{i}}$ $\mathbf{f} _{\mathbf{i}}$ $u_i=\frac{x_i-A}{h}$ $\mathbf{f} _i u_i$
$0-20$ 10 5 -1 -5
$20-40$ 30 $\mathbf{f} _1$ 0 0
$40-60$ 50 10 +1 10
$60-80$ 70 $\mathbf{f} _2$ +2 $2 f_2$
$80-100$ 90 7 +3 21
$1 0 0 - 1 2 0$ $1 1 0$ $8$ $+ 4$ $3 2$
$\sum f_i=30+f_1+f_2$ $\sum f_i u_i=58+2 f_2$

STATISTICS

Given $30+f_1+f_2=50$

$f_1+f_2=20$ …(i)

$\bar{x}=A+(\frac{\sum f_i u_i}{\sum f_i} \times h)$

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

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

$32.8 \times 5=116+4 f_2$

$164=116+4 f_2$

$4 f_2=164-116$

$4 f_2=48$

$f_2=12$

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

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

STATISTICS

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

Marks No. of Students
Below 10 5
Below 20 9
Below 30 17
Below 40 29
Below 50 45
Below 60 60
Below 70 70
Below 80 78
Below 90 83
Below 100 85

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

alt text

STATISTICS

According to step deviation formula for mean

$\bar{x}=A+(\frac{\sum f_i u_i}{\sum f_i} \times h)$

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

$\bar{x}=45+3.41$ $\quad$

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

STATISTICS

14.4 PROPERTIES OF MEAN :

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

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

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

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

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

STATISTICS

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.

STATISTICS

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.

STATISTICS

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.

STATISTICS

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.

STATISTICS

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.

STATISTICS

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

STATISTICS

Sol. Arranging the data in ascending order

  1. 696, 705, 710, 712, 715, 716, 719, 724, 725, 728, 729, 734, 745

$N=15$

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

$ =716 $

STATISTICS

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

Daily wages (in Rs.) 125 130 135 140 145 150 160 180
No. of workers 6 20 24 28 15 4 2 1

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

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Sol.

Daily wages (in Rs.) No. of workers Cumulative frequency
125 6 6
130 20 26
135 24 50
140 28 78
145 15 93
150 4 97
160 2 99
180 1 100

STATISTICS

$N=100$ (even)

$ \begin{aligned} \therefore \quad \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} \quad=137.50 \end{aligned} $

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

STATISTICS

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

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency 5 10 20 7 8 5

STATISTICS

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

Median $=26.25$

Class $f$ c.f.
$0-10$ 5 5
$10-20$ 10 15
$20-30$ 20 35
$30-40$ 7 42
$40-50$ 8 50
$50-60$ 5 55

STATISTICS

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

Variable $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Total
Frequency 12 13 $?$ 65 $?$ 25 18 229

STATISTICS

Sol.

Class Interval Frequency C.F
$10-20$ 12 12
$20-30$ 30 42
$30-40$ $f_1$ $42+f_1$
$40-50$ 65 $107+f_1$
$50-60$ $f_2$ $107+f_1+f_2$
$60-70$ 25 $132+f_1+f_2$
$70-80$ 18 $150+f_1+f_2$

STATISTICS

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

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

$\Rightarrow \quad f_1+f_2=79$

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

$ \therefore \quad \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{145-2 f_1}{13} $

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

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

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

STATISTICS

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.

STATISTICS

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.

STATISTICS

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

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

STATISTICS

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

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

STATISTICS

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

Value $x_i$ 15 16 18 19 20 21 22 24 25 34 48
Frequency $f_i$ 1 4 1 5 3 1 1 1 1 1 1

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

STATISTICS

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

Age (in Years) $5-14$ $15-24$ $25-34$ $35-44$ $45-54$ $55-64$
No. of Cases 6 11 21 23 14 5

STATISTICS

Sol.

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

Age (in Years) $4.5-14.5$ $14.5-24.5$ $24.5-34.5$ $34.5-44.5$ $44.5-54.5$ $54.5-64.5$
No. of Cases 6 11 21 23 14 5

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 $f_2=14$.

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

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

$=36.31$ Ans.

STATISTICS

Ex. 14 Find the mode of following distribution :

Daily Wages $31-36$ $37-42$ $43-48$ $49-54$ $55-60$ $61-66$
No. of workers 6 12 20 15 9 4

STATISTICS

Sol.

Daily Wages No. of workers Daily wages No of workers
$31-36$ 6 $30.5-36.5$ 6
$37-43$ 12 $36.5-42.5$ 12
$43-48$ 20 $42.5-48.5$ 20
$49-54$ 15 $48.5-54.5$ 15
$55-60$ 9 $54.5-60.6$ 9
$61-66$ 4 $60.5-66.5$ 4

Modal class frequency is 42.5 - 48.5 .

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

$\therefore$ Mode $=46.2$

STATISTICS

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.

STATISTICS

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.

STATISTICS

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.

STATISTICS

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 $\quad$ (ii) More than method

STATISTICS

In ungrouped frequency distribution :

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

Marks
Obtained
$400-450$ $450-500$ $500-550$ $550-600$ $600-650$ $650-700$ $700-750$ $750-80$
No. of
Examinees
30 45 60 52 54 67 45 47

(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.

STATISTICS

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

Marks Obtained No. of Examinees
Less than 450 30
Less than 500 75
Less than 550 135
Less than 600 187
Less than 650 241
Less than 700 308
Less than 750 353
Less than 800 400

STATISTICS

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

STATISTICS

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

Marks Obtained No. of Examinees
400 and more 400
450 and more 370
500 and more 325
550 and more 265
600 and more 213
650 and more 159
700 and more 92
750 and more 47

STATISTICS

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

STATISTICS

(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 .

STATISTICS

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

Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Number of
students
7 10 23 51 6 3

STATISTICS

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

Marks No. of
Students
Less than 10 7
Less than 20 17
Less than 30 40
Less than 40 91
Less than 50 97
Less than 60 100

STATISTICS

According to graph median $=34$ marks.

alt text

STATISTICS

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 $\quad$

(B) 520 $\quad$

(C) 190 $\quad$

(D) 35

STATISTICS

Que. 1
Ans. C

STATISTICS

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

(A) 11 $\quad$

(B) 15 $\quad$

(C) 18 $\quad$

(D) 16 $\quad$

STATISTICS

Que. 2
Ans. B

STATISTICS

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

(A) 7 $\quad$

(B) 4 $\quad$

(C) 3 $\quad$

(D) 1

STATISTICS

Que. 3
Ans. B

STATISTICS

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

Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$
Frequency $5$ $f_1$ 10 $\mathbf{f} _2$ 7 8

$$$$

(A) $0$

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

(C) $10$

(D) $25$

STATISTICS

Que. 4
Ans. A

STATISTICS

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 $\quad$

(B) 8.2 $\quad$

(C) 9.2 $\quad$

(D) 10.2

STATISTICS

Que. 5
Ans. C

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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

$\mathbf{x}$ $\mathbf{1 5}$ $\mathbf{1 7}$ $\mathbf{1 9}$ $\mathbf{2 0}+\mathbf{p}$ $\mathbf{2 3}$
$\mathbf{f}$ $\mathbf{2}$ $\mathbf{3}$ $\mathbf{4}$ $\mathbf{5 p}$ $\mathbf{6}$

STATISTICS

Sol. 1. $\quad p=1$ $\quad$

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

Class interval $50-70$ $70-90$ $90-110$ $110-130$ $130-150$ $150-170$
No. of workers 18 12 13 27 8 22

STATISTICS

Sol. 2. $\quad$ 112.20 $\quad$

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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.

STATISTICS

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

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4. Calculate the median from the following data :

Rent (in Rs.) $15-25$ $25-35$ $35-45$ $45-55$ $55-65$ $65-75$ $75-85$ $85-95$
No. of House 8 10 15 25 40 20 15 7

STATISTICS

Sol. 4. $\quad 58$ $\quad$

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

No. of accidents 0 1 2 3 4 5 Total
Frequency (No. of
days)
46 $f_1$ $f_2$ 25 10 5 200

STATISTICS

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

STATISTICS

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

Class interval : $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ Total
Frequency 5 $f_1$ 20 15 $f_2$ 5 60

STATISTICS

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

STATISTICS

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.

Class
interval :
$0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-$
100
Frequency 3 5 16 12 13 20 5 4 1 1

STATISTICS

Sol. 7. $\quad 53.17$ $\quad$

STATISTICS

8. Find the mode of following distribution :

Class
interval
$0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency 5 8 7 12 28 20 10 10

STATISTICS

Sol. 8. $\quad$ 46.67 $\quad$

STATISTICS

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

Weight (in kg) Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

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.

STATISTICS

Sol. 9. $\quad 47 .5 kg$ $\quad$

STATISTICS

10. The following table gives the height of trees :

Height Less
than 7
Les than
14
Less
than 21
Less
than 28
Less
than 35
Less
than 42
Less
than 49
Less
than 56
No. of trees 26 57 92 134 216 287 341 360

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

STATISTICS

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

$\mathbf{x}$ $\mathbf{1 0}$ $\mathbf{1 5}$ $\mathbf{P}$ $\mathbf{2 5}$ $\mathbf{3 0}$
$\mathbf{f}$ $\mathbf{5}$ $\mathbf{1 0}$ $\mathbf{7}$ $\mathbf{8}$ $\mathbf{2}$

STATISTICS

Sol. 11. $\quad 20$

STATISTICS

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

Classes $50-70$ $70-90$ $90-110$ $110-130$ $130-150$ $150-170$
Frequency 18 12 13 27 8 22

STATISTICS

Sol. 12. $\quad 20$

STATISTICS

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

Marks obtained $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency 8 10 12 22 30 18

STATISTICS

Sol. 13. $\quad 30-40$

STATISTICS

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

Classes $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Frequency 5 10 18 30 20 12 5

STATISTICS

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



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