Exercise-21.6 ML Aggarwal Class-10 Histogram Measures of Central Tendency Solutions ICSE Chapter-21 . We Provide Step by Step Answer of Exe-21.1,  Exe-21.2,  Exe-21.3,  Exe-21.4,  Exe-21.5, and Exe-21.6  with MCQs and Chapter-Test of Measures of Central Tendency Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  . Visit official Website CISCE  for detail information about ICSE Board Class-10.

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 10th Chapter-21 Measures of Central Tendency Writer ML Aggarwal Book Name Understanding Topics Solution of Exe – 21.5 (Ogive), and Exe-21.6 with MCQ Questions   and Chapter Test Academic Session 2021-2022

## How to Solve Measures of Central Tendency Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Answer of Chapter-21 Measures of Central Tendency of ML Aggarwal Solutions. Read the Chapter Carefully with formula and then solve all example of Exe-21.1,  Exe-21.2,  Exe-21.3,  Exe-21.4,  Exe-21.5,  given in  your text book .

For more practice on Measures of Central Tendency related problems /Questions / Exercise try to solve Measures of Central Tendency exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE) Measure of Central Tendency (Mean) and also  Concise Selina Publications ICSE Mathematics  Get the to understand the topic more clearly in effective way.

Exe – 21.6 , ML Aggarwal Class-10

## Histogram Measures of Central Tendency Solutions ICSE

Page 517-519

Question 1 : Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median. Answer 1 : Representing the given data in cumulative frequency distributions :  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 = $\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$ $\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } +\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$ $\frac { 1 }{ 2 } \left[ 30th\quad term+31th\quad term \right]$
hence = 30.5 observation
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 2  Exercise-21.6 ML Aggarwal Class-10 Histogram Measures

The weight of 50 workers is given below: 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)

The cumulative frequency table of the given distribution table is as follows: 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 $\frac { 3n }{ 4 } =\frac { 3\times 50 }{ 4 } =37.5$
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 3

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) 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%. 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 $\frac { n }{ 2 } =\frac { 160 }{ 2 } =80$

##### (i)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.

Page 518

#### Question 4 Exercise-21.6 ML Aggarwal Class-10 Histogram Measures

The daily wages of 80 workers in a project are given below Use a graph paper to draw an ogive for the above distribution. ( a scale of 2 cm = Rs 50 on x- axis and 2 cm = 10 workers on y-axis).  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) Number of workers = 80
(i) Median = ${ \left( \frac { n }{ 2 } \right) }^{ th }$ term = 40 th 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) = ${ \left( \frac { 80 }{ 4 } \right) }^{ th }$ term = 20 th 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 5

Marks obtained by 200 students in an examination are given below : 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.  (i) Median is 57.
(ii) 44 students failed.
(iii) No. of students who secured grade one = 200 – 188 = 12.

#### Question 6  Exercise-21.6 ML Aggarwal Class-10 Histogram Measures

The monthly income of a group of 320 employees in a company is given below : Read Next 👇 Click on Page Number Given Below 👇