ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency Exercise 21.6

 Exercise 21.6

Question 1

The following table shows the distribution of the heights of a group of a factory workers.

Height (in cm)150-155155-160160-165165-170170-175175-180180-185
No. of workers61218201386
(i) Determine the cumulative frequencies.
(ii) Draw the cumulative frequency curve on a graph paper.
Use 2 cm = 5 cm height on one axis and 2 cm = 10 workers on the other.
(iii) From your graph, write down the median height in cm.
Sol :
Representing the distribution in cumulative frequency distribution :
Height(in cm)No. of workers
(f)		c.f
150-15566
155-1601218
160-1651836
165-1702056
170-1751369
175-180877
180-185683

Here, n = 83 which is even.

Now taking points (155, 6), (160, 18), (165, 36), (170, 56),

(175, 69), (180, 77) and (185, 83) on the graph.




















Now join them with free hand to form the ogive
or cumulative frequency curve as shown.
Here n = 83 which is odd

Median$=\frac{n+1}{2}$ th observation
$=\frac{83+1}{2}=42$ th observation

Take a point A (42) on y-axis and from A,

draw a horizontal line parallel to x-axis meeting the curve at P.

From P draw a line perpendicular on the x-axis which meets it at Q.

∴Q is the median which is 166.5 cm. Ans.


Question 2

Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median.

Marks0-1010-2020-3030-4040-5050-6060-7070-80
No. of students38121410652
Sol :

Representing the given data in cumulative frequency distributions :


















MarksNo. of workers
(f)		c.f
0-1033
10-20811
20-301223
30-401437
40-501047
50-60653
60-70558
70-80260
Taking points (10, 3), (20, 11), (30, 23), (40, 37),
(50,47), (60,53), (70, 58) and (80, 60) on the graph.
Now join them in a free hand to form an ogive as shown.
Here n = 60 which is even

Median $=\frac{1}{2}\left[\frac{60}{2} t h \right.\text{ term }+\left(\frac{60}{2}+1\right) t h\text{ term ]}$

$=\frac{1}{2}\left[\frac{60}{2}+\left(\frac{60}{2}+1\right)\right.\text{ th term ]}$

$=\frac{1}{2}$[30th term+31th term]

=30.5 observations

Now take a point A (30.5) on y-axis and from A,

draw a line parallel to x-axis meeting the curve at P

and from P, draw a perpendicular to x-axis meeting is at Q.

∴ Q is the median which is 35.


Question 3

Use graph paper for this question.

The following table shows the weights in gm of a sample of 100 potatoes taken from a large consignment:

Weight(gm)50-6060-7070-8080-9090-100100-110110-120120-130
Frequency310121618141210
(i) Calculate the cumulative frequencies.
(ii) Draw the cumulative frequency curve and from it determine the median weight of the potatoes. (1996)
Sol :
Representing the given data in cumulative frequency table :

Weight (gm)Frequency
(f)		c.f
50-6088
60-701018
70-801230
80-901646
90-1001864
100-1101478
110-1201290
120-13010100

Now plot the points (60, 8), (70, 18), (80, 30), (90, 46), (100, 64),

(110, 78), (120, 90), (130, 100) on the graph and join them

in a free hand to form an ogive as shown






















Here n =100 which is even.
Median $=\frac{1}{2}\left[\frac{n}{2} t h \text{ term }+\left(\frac{n}{2}+1\right) t h\text{ term ]}$

$=\frac{1}{2}\left[\frac{100}{2}+\left(\frac{100}{2}+1\right) t h\right.\text{ term ]}$

$=\frac{1}{2}$[50th term+51th term]

=50.5

Now take a point A (50.5) on they-axis and from A

draw a line parallel to x-axis meeting the curve at R

From P, draw a perpendicular on x-axis meeting it at Q.

Q is the median which is = 93 gm.


Question 4

Attempt this question on graph paper.

Age (years)5-1515-2525-3535-4545-5555-6565-75
No. of casualties due to accidents61015132487

(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other.

(ii) From your graph determine (1) the median and (2) the upper quartile

Sol :

Representing the given data in less than cumulative frequency.

AgeNo. of Casualties
Cumulative Frequency
Less than 1566
Less than 251216
Less than 351531
Less than 451344
Less than 552468
Less than 55876
Less than 75783

Now plot the points (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65,76)

and (75, 83) on the graph and join these points in free hand

to form a cumulative frequency curve (ogive) as shown.

Here n = 83, which is odd.






















(i) Median = $=\frac{n+1}{2}=\frac{83+1}{2}+\frac{84}{2}=42$
Now we take point A (42) on y-axis and from A,
draw a line parallel to x-axis meeting the curve at P
and from P, draw a perpendicular to x-axis meeting it at Q.
Q is the median which is = 43

(ii) Upper quartile $=\frac{3(n+1)}{4}=\frac{3 \times(83+1)}{4}=\frac{252}{4}=63$

Take a point B 63 on y-axis and from B,

draw a parallel line to x-axis meeting the curve at L.

From L, draw a perpendicular to x-axis meeting it at M which is 52.

∴ Upper quartile = 52 years


Question 5

The weight of 50 workers is given below:

Weight in kg50-6060-7070-8080-9090-100100-110110-120
No. of workers471114653
Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5 workers along the other axis. Use a graph to estimate the following:

(i) the upper and lower quartiles.

(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight. (2015)

Sol :

The cumulative frequency table of the given distribution table is as follows:

Height(in cm)No. of workers
(f)		c.f
50-6044
60-70711
70-801122
80-901436
90-100642
100-110547
110-120350

The ogive is as follows:

Plot the points (50, 0), (60, 4), (70, 11), (80, 22), (90, 36), (100, 42),
(110, 47), (120, 50) Join these points by using freehand drawing.
The required ogive is drawn on the graph paper.
Here n = number of workers = 50
(i) To find upper quartile:
Let A be the point on y-axis representing a frequency
$\frac{3 n}{4}=\frac{3 \times 50}{4}=37.5$
Through A, draw a horizontal line to meet the ogive at B.
Through B draw a vertical line to meet the x-axis at C.
The abscissa of the point C represents 92.5 kg.
The upper quartile = 92.5 kg To find the lower quartile:
Let D be the point on y-axis representing frequency$=\frac{n}{4}=\frac{50}{4}=12.5$
Through D, draw a horizontal line to meet the ogive at E.
Through E draw a vertical line to meet the x-axis at F.
The abscissa of the point F represents 72 kg.
∴ The lower quartile = 72 kg
(ii) On the graph point, G represents 95 kg.
Through G draw a vertical line to meet the ogive at H.
Through H, draw a horizontal line to meet y-axis at 1.
The ordinate of point 1 represents 40 workers on the y-axis .
∴The number of workers who are 95 kg and above
= Total number of workers – number of workers of weight less than 95 kg
= 50 – 40 = 10

Question 6

The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution.
(Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis)

Scores0-1010-2020-3030-4040-5050-6060-7070-8080-9090-100
No. of shooters91320263022151087

Use your graph to estimate the following:

(i) The median.

(ii) The interquartile range.

(iii) The number of shooters who obtained a score of more than 85%.

Sol :

ScoresNo. of Shooter
c.f
0-1099
10-201322
20-302042
30-402668
40-503098
50-6022120
60-7015135
70-8010145
80-908153
90-1007160

Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98),

(60, 120), (70, 135), (80, 145), (90, 153), (100, 160)

on the graph and join them with free hand to get an ogive as shown:

































Here n = 160

$\frac{n}{2}=\frac{160}{2}=80$

Median : Take a point 80 on 7-axis and through it,

draw a line parallel to x-axis-which meets the curve at A.

Through A, draw a perpendicular on x-axis which meet it at B.

B Is median which is 44.


(ii) Interquartile range (Q1)

$\frac{n}{4}=\frac{160}{4}=40$

From a point 40 ony-axis, draw a line parallel to x-axis

which meet the curve at C and from C draw a line perpendicular to it

which meet it at D. which is 31.

The interquartile range is 31.


(iii) Number of shooter who get move than 85%.

Scores : From 85 on x-axis, draw a perpendicular to it meeting the curve at P.

From P, draw a line parallel to x-axis meeting y-axis at Q.

Q is the required point which is 89.

Number of shooter getting more than 85% scores = 160 – 149 = 11.


Question 7

The daily wages of 80 workers in a project are given below

Wages (in Rs)400-450450-500500-550550-600600-650650-700700-750
No. of workers26121824135
Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs 50 on x- axis and 2 cm = 10 workers on y-axis). Use your ogive to estimate:
(i) the median wage of the workers.
(ii) the lower quartile wage of the workers.
(iii) the number of workers who earn more than Rs 625 daily. (2017)
Sol :

Wages (in Rs)No. of workers
c.f
400-45022
450-50068
500-5501220
550-6001838
600-6502462
650-7001375
700-750580

Number of workers = 80

(i) Median$=\left(\frac{n}{2}\right)$th term
=40th term

Through mark 40 on the y-axis, draw a horizontal line
which meets the curve at point A.

























Through point A, on the curve draw a vertical line which meets the x-axis at point B.
The value of point B on the x-axis is the median, which is 604.

(ii) Lower Quartile (Q1) $=\left(\frac{80}{4}\right)^{t h}$ term
=20th term=550

(iii) Through mark 625 on x-axis, draw a vertical line which meets the graph at point C.

Then through point C, draw a horizontal line which meets the y-axis at the mark of 50.

Thus, number of workers that earn more Rs 625 daily = 80 – 50 = 30


Question 8

Marks obtained by 200 students in an examination are given below :

Marks0-1010-2020-3030-4040-5050-6060-7070-8080-9090-100
No. of shooters511102028374029146

Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph, determine

(i) The median marks.

(ii) The number of students who failed if minimum marks required to pass is 40.

(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.

Sol :

Marksfc.f
0-1055
10-201116
20-301026
30-402046
40-502874
50-6037111
60-70402151
70-8029180
80-9014194
90-1006200
N=200














(i) Median is 57.

(ii) 44 students failed.

(iii) No. of students who secured grade one = 200 – 188 = 12.


Question 9

The monthly income of a group of 320 employees in a company is given below

Monthly IncomeNo. of Employees
6000-700020
7000-800045
8000-900065
9000-100095
1000-1100060
11000-1200030
12000-130005

Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine

(i) the median wage.

(ii) the number of employees whose income is below Rs. 8500.

(iii) If the salary of a senior employee is above Rs. 11500, find the number of senior employees in the company.

(iv) the upper quartile.

Sol :
Monthly IncomeNo. of Employees
(f)		c.f.
6000-70002020
7000-80004565
8000-900065130
9000-1000095225
10000-1100060285
11000-1200030315
12000-130005320


















Now plot the points (7000,20), (8000,65), (9000,130),
(10000,225), (11000,285), (12000,315) and(13000, 320)
on the graph and join them in order with a free hand
to get an ogive as shown in the figure

(i) Total number of employees = 320
$\frac{N}{2}=\frac{320}{2}=160$

From 160 on the y-axis, draw a line parallel to x-axis meeting the curve at P.
From P. draw a perpendicular on x-axis meeting it at M, M is the median which is 9300

(ii) From 8500 on the x-axis, draw a perpendicular which meets the curve at Q.
From Q, draw a line parallel to x-axis meeting y-axis at N. Which is 98

(iii) From 11500 on the x-axis, draw a line perpendicular to x-axis meeting the curve at R.
From R, draw a line parallel to x-axis meeting y-axis at L. Which is 300
No. of employees getting more than Rs. 11500 = 320

(iv) Upper quartile (Q1)
$\frac{3 N}{4}=\frac{320 \times 3}{4}=240$

From 240 on y-axis, draw a line perpendicular on the x-axis which meets the curve at S.
From S, draw a perpendicular on x-axis meeting it at T. Which is 10250.
Hence Q3 = 10250

Question 10

Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students
Weight40-4545-5050-5555-6060-6565-7070-7575-80
Frequency51722455131209
Use your ogive to estimate the following:
(i) The percentage of students weighing 55 kg or more.
(ii) The weight above which the heaviest 30% of the students fall,
(iii) The number of students who are :
1. under-weight and
2. over-weight, if 55.70 kg is considered as standard weight.
Sol :
Weight40-4545-5050-5555-6060-6565-7070-7575-80
Frequency51722455131209
c.f5224439140171191200





















Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140),
(70, 171), (75, 191) and (80, 200) on the graph
and join them in free hand to get an ogive as shown From the graph,
number of students weighing 55 kg or more = 200 – 44 = 156

Percentage $=\frac{156}{200} \times 100=78 \%$

(ii) 30% of 200 $=\frac{200 \times 30}{100}=60$

∴ Heaviest 60 students in weight = 9 + 20 + 31 = 60

(From the graph, the required weight is 65 kg or more but less than 80 kg)


(iii) Total number of students who are

1. under weight = 47 and

2. over weight = 152

(∴ Standard weight is 55.70 kg)


Question 11

The marks obtained by 100 students in a Mathematics test are given below :

Marks0-1010-2020-3030-4040-5050-6060-7070-8080-9090-100
No. of shooters37121723149654

Draw an ogive on a graph sheet and from it determine the :

(i) median

(ii) lower quartile

(iii) number of students who obtained more than 85% marks in the test.

(iv) number of students who did not pass in the test if the pass percentage was 35. We represent the given data in cumulative frequency table as given below :

Sol :

We represent the given data in the

cumulative frequency table as given below:

N=100

Median $=\frac{100}{2}=50^{\text {th }}$ term=45

∴Median=45


(ii) Lower quartile : $(Q_1)$

$\mathrm{N}=100 \Rightarrow \frac{100}{4}=25^{\mathrm{th}}$ term=32

$\therefore Q_{1}=32$


(iii) Number of students with 85%

or less=70

∴More than 85% marks

=100-70=30


(iv) Number of students who did not pass=38

Marksfc.f
0-1033
10-20710
20-301222
30-401739
40-502362
50-601476
60-70985
70-80691
80-90596
90-1004100













Question 12

The marks obtained by 120 students in a Mathematics test are-given below

Marks0-1010-2020-3030-4040-5050-6060-7070-8080-9090-100
No. of shooters591622261811643
Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive to estimate the following:
(i) the median
(ii) the lower quartile
{iii) the number of students who obtained more than 75% marks in the test.
(iv) the number of students who did not pass in the test if the pass percentage was 40. (2002)
Sol :
We represent the given data in cumulative frequency table as given below :

Marksfc.f
0-1055
10-20914
20-301630
30-402252
40-502678
50-601896
60-7011107
70-806113
80-904117
90-1003120









Now we plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107),

(80, 113), (90, 117) and (100, 120) on the graph

and join the points in a free hand to form an ogive as shown.

Here n = 120 which is an even number


(i) Median $=\frac{1}{2}\left[\frac{120}{2}+\left(\frac{120}{2}+1\right)\right]$

$=\frac{1}{2}(60+61)=60.5$

Now take a point A (60.5) on y-axis and from A

draw a line parallel to x- axis meeting the curve in P

and from P, draw a perpendicular to x-axis meeting it at Q.

∴ Q is the median which is 43.00 (approx.)


(ii) Lower quartile $=\frac{n}{4}=\frac{120}{4}=30$

Now take a point B (30) on y-axis and from B,

draw a line parallel to x-axis meeting the curve in L

and from L draw a perpendicular to x-axis meeting it at M.

M is the lower quartile which is 30.

(iii) Take a point C (75) on the x-axis

and from C draw a line perpendicular to it meeting the curve at R.

From R, draw a line parallel to x-axis meeting y-axis at S.

∴S shows 110 students getting below 75%

and 120 – 110 = 10 students getting more than 75% marks.

(iv) Pass percentage is 40%

Now take a point D (40) on x-axis and from D

draw a line perpendicular to x-axis meeting the curve at E

and from E, draw a line parallel to x-axis meeting the y-axis at F.

∴ F shows 52

∴ No of students who could not get 40% and failed in the examination are 52.


Question 13

The following distribution represents the height of 160 students of a school.

Height (in cm)140-145145-150150-155155-160160-165165-170170-175175-180
No. of Students122030382416128
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :
(i)The median height.
(ii)The inter quartile range.
(iii) The number of students whose height is above 172 cm.
Sol :
The cumulative frequency table may be prepared as follows:
Height
(in cm)		No. of Students
(f)		c.f.
140-1451212
145-1502032
150-1553062
155-16038100
160-16524124
165-17016140
170-17512152
175-1808160



























Now, we take height along x-axis and number of students along the y-axis.
Now, plot the point (140, 0), (145, 12), (150, 32), (155, 62), (160, 100), (165, 124),
(170, 140), (175, 152) and (180, 160). Join these points by a free hand curve to get the ogive.

(i) Here N = 160
$\frac{N}{2}=80$
On the graph paper take a point A on the y- axis representing 80.
A draw horizontal line meeting the ogive at B.
From B, draw BC ⊥ x-axis, meeting the x-axis at C. The abscissa of C is 157.5
So, median = 157.5 cm

(ii) Proceeding in the same way as we have done in above,
we have, Q1 = 152 and Q3 = 164 So, interquartile range = Q3 – Q1 = 164 – 152 = 12 cm

(iii) From the ogive, we see that the number of students whose height is less than 172 is 145.
No. of students whose height is above 172 cm = 160 – 145 = 15

Question 14

100 pupils in a school have heights as tabulated below :
Height in cm121-130131-140141-150151-160161-170171-180
No. of pupils12163020148
Draw the ogive for the above data and from it determine the median (use graph paper).
Sol :
Representing the given data in cumulative frequency table (in continuous distribution):
Height
(in cm)		No. of pupilsc.f.
120.5-130.51212
130.5-140.51628
140.5-150.53058
150.5-160.52078
160.5-170.51492
170.5-180.58100






















∴ Here n = 100 which is an even number
∴ Median = 
$\frac{1}{2}\left[\frac{n}{2}+\left(\frac{n}{2}+1\right)\right]$
$=\frac{1}{2}\left[\frac{100}{2}+\left(\frac{100}{2}+1\right)\right]$
$=\frac{1}{2}(50+51)=\frac{101}{2}$
=50.5
Now plot points (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92)
and (180.5, 100) on the graph and join them in free hand to form an ogive as shown.
Now take a point A (50-5) on y-axis and from A
draw a line parallel to x-axis meeting the curve at P
and from P, draw a line perpendicular to x-axis meeting it at Q
∴ Q (147.5) is the median.

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