Data Handling
Class 8th Mathematics Term 3 Tamilnadu Board Solution
- Draw a histogram to represent the following data
- Draw a histogram with the help of the following table
- Draw a histogram to represent the following data of the spectators in a cricket…
- In a study of diabetic patients in a village, the following observations were…
- Construct a histogram and frequency polygon for the following data…
- The following table shows the performance of 150 candidates in an Intelligence…
- Construct a frequency polygon from the following data using histogram.…
- Draw a frequency polygon for the following data without using histogram…
- Construct a frequency polygon for the following data without using histogram…
- The following are the marks obtained by 40 students in an English examination…
- Yugendran’s progress report card shows his marks as follows: |c|c|c|c|c| &…
- There are 36 students in class VIII. They are members of different clubs:…
- The number of students in a hostel speaking different languages is given below:…
- In a school, the number of students interested in taking part in various hobbies…
- A metal alloy contains the following metals. Represent the data by a pie chart.…
- On a particular day, the sales (in Rs.) of different items of a baker’s shop are…
- The money spent on a book by a publisher is given below: Represent the above…
- Expenditure of a farmer for cultivation is given as follows: Represent the data…
- There are 900 creatures in a zoo as per list below: Represent the above data by…
- In a factory, five varieties of vehicles were manufactured in a year, whose…
- A food contains the following nutrients. Draw a pie chart representing the…
- The favourite flavours of ice cream for students of a school is given in…
- The data on modes of transport used by students to come to school are given…
- Mr. Rajan Babu spends 20% of his income on house rent, 30% on food and 10% for…
- The percentage of various categories of workers in a state are given in…
- Find the mean of 2, 4, 6, 8, 10, 12, 14, 16.
- If the average of the values 18, 41, x, 36, 31, 24, 37, 35, 27, 36, is 31. Find…
- If in a class of 20 students, 5 students have scored 76 marks, 7 students have…
- The average height of 20 students in a class was calculated as 160 cm. On…
- Calculate the Arithmetic mean of the following data: |c|c|c|c|c|c|c|c|c|c|…
- The following data give the number of boys of a particular age in a class of 40…
- Obtain the A.M of the following data: |l|l|l|l|l|l|l|l|
- The following table shows the weights of12 workers in a factory Find the mean…
- For a month, a family requires the commodities listed in the table below. The…
- Find the Weighted Arithmetic Mean for the following data:
- Find the median of the following set of values: (i) 83, 66, 86, 30, 81. (ii)…
- Find the median for the following data: |c|c|c|c|c|c|c|c| x&1&2&3&4&5&6&7&8…
- The height (in cm) of 50 students in a particular class are given below:…
- The hearts of 60 patients were examined through X-ray and the observations…
- The salary of 43 employees are given in the following table. Find the median.…
- Find the mode of the following data: i) 74, 81, 62, 58, 77, 74. ii) 55, 51, 62,…
- Find the mode for the following frequency table: |c|c|c|c|c|c|c|
- Find the mode for the following table: |c|c|c|c|c|c| &29&32.4&34.6&36.9&38.7&40…
- The demand of different shirt sizes, as obtained by a survey, is given below.…
- Find the mean, median and mode for the following frequency table:…
- The age of the employees of a company are given below. Find the mean, median…
- The following table shows the weights of 20 students. Calculate the mean,…
Exercise 3.1
Question 1.Draw a histogram to represent the following data
Answer:
GIVEN:
As the data is in continuous form
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
Question 2.
Draw a histogram with the help of the following table
Answer:
GIVEN:
FORMULA USED: Adjustment Factor = 
[Lower limit - Upper limit]
Adjustment Factor = [16-15]
= 0.5
⇒ First make the data in continuous form
IN each interval subtract 0.5 in upper limit and add 0.5 in
lower limit.
∴ the Table will be
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
Question 3.
Draw a histogram to represent the following data of the spectators in a cricket match
Answer:
GIVEN:
FORMULA USED: Adjustment Factor = [Lower limit –
Upper limit of preceding interval]
Adjustment Factor = [20-19]
= 0.5
⇒ First make the data in continuous form
IN each interval subtract 0.5 in upper limit and add 0.5 in
lower limit.
∴ the Table will be
As the data is in continuous form
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
Question 4.
In a study of diabetic patients in a village, the following observations were noted
Represent the above data by a frequency polygon using histogram.
Answer:
GIVEN.
⇒ First draw histogram of given table
As the data is in continuous form
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
⇒ Take out the midpoints of each interval and imagine 2
midpoints at 1st and last join all midpoints and imaginary midpoints to get frequency polygon.
Question 5.
Construct a histogram and frequency polygon for the following data
Answer:
GIVEN.
⇒ First draw histogram of given table
As the data is in continuous form
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
⇒ Take out the midpoints of each interval and imagine 2
midpoints at 1st and last join all midpoints and imaginary midpoints to get frequency polygon.
Question 6.
The following table shows the performance of 150 candidates in an Intelligence test. Draw a histogram and frequency polygon for this distribution
Answer:
GIVEN.
⇒ First draw histogram of given table
As the data is in continuous form
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
⇒ Take out the midpoints of each interval and imagine 2
midpoints at 1st and last join all midpoints and imaginary midpoints to get frequency polygon.
Question 7.
Construct a frequency polygon from the following data using histogram.
Answer:
GIVEN.
⇒ First draw histogram of given table
As the data is in continuous form
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
⇒ Take out the midpoints of each interval and imagine 2
midpoints at 1st and last join all midpoints and imaginary midpoints to get frequency polygon.
Question 8.
Draw a frequency polygon for the following data without using histogram
Answer:
GIVEN.
⇒ As the data is in continuous forms
⇒ Take out Midpoints of Each class intervals
Table will be
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
Take 2 imaginary midpoints on 1st and last with 0 frequency
⇒ join all the midpoints to get frequency polygon
Question 9.
Construct a frequency polygon for the following data without using histogram
Answer:
GIVEN.
FORMULA USED: Adjustment Factor = [Lower limit –
Upper limit of preceding interval]
Adjustment Factor = [35-34]
= 0.5
⇒ First make the data in continuous form
IN each interval subtract 0.5 in upper limit and add 0.5 in
lower limit.
∴ the Table will be
⇒ Take out Midpoints of Each class intervals
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
Take 2 imaginary midpoints on 1st and last with 0 frequency
⇒ join all the midpoints to get frequency polygon
Question 10.
The following are the marks obtained by 40 students in an English examination (out of 50 marks). Draw a histogram and frequency polygon for the data
29, 45, 23, 40, 31, 11, 48, 11, 30, 24, 25, 29, 25, 32, 31, 22, 19, 49, 19, 13,32, 39, 25, 43, 27, 41, 12, 13, 32, 44, 27, 43, 15, 35, 40, 23, 12, 48, 49, 18.
Answer:
GIVEN. Total number of students = 40
Maximum marks = 50
⇒ First make the table by above observation in specific
Interval
Total number of values = 50
Range = Highest – Lowest
= 49 – 11
= 38
Let’s Divide data in 4 class intervals
10-20, 20-30, 30-40, 40-50
⇒ Draw histogram of given table
As the data is in continuous form
⇒ Mark class interval along X axis on uniform scale
⇒ Mark frequency along Y axis on uniform scale
⇒ Construct rectangles with class intervals as base and
corresponding frequencies as height.
⇒ Take out the midpoints of each interval
Imagine 2 midpoints at 1st and last of 0 frequency join all midpoints and imaginary midpoints to get frequency polygon.
Exercise 3.2
Question 1.Yugendran’s progress report card shows his marks as follows:
Draw a pie chart exhibiting his mark in various subjects.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = 
×360°
Total marks = 72 + 60 + 84 + 70 + 74
= 360
Angle of sector for Tamil = 
×360° = 72°
Angle of sector for English = 
×360° = 60°
Angle of sector for Mathematics = 
×360° = 84°
Angle of sector for Science = 
×360° = 70°
Angle of sector for Social Science = 
×360° = 74°
Question 2.
There are 36 students in class VIII. They are members of different clubs:
Represent the data by means of a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = 
×360°
Total students = 36
Angle of sector for Mathematics = 
×360° = 120°
Angle of sector for N.C.C = 
×360° = 60°
Angle of sector for J.R.C = 
×360° = 100°
Angle of sector for Scout = 
×360° = 80°
Question 3.
The number of students in a hostel speaking different languages is given below:
Represent the data in a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total students = 36 + 12 + 9 + 6 + 5 + 4
= 72
Angle of sector for Tamil = 
×360° = 180°
Angle of sector for Telugu = 
×360° = 60°
Angle of sector for Malayalam = 
×360° = 45°
Angle of sector for Kannada = 
×360° = 30°
Angle of sector for English = 
×360° = 25°
Angle of sector for Others = 
×360° = 20°
Question 4.
In a school, the number of students interested in taking part in various hobbies from class VIII is given below. Draw a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total students = 20 + 25 + 27 + 28 + 20
= 120
Angle of sector for Music = 
×360° = 60°
Angle of sector for Pottery = 
×360° = 75°
Angle of sector for Dance = 
×360° = 81°
Angle of sector for Drama = 
×360° = 84°
Angle of sector for Social Service = ×360° = 60°
Question 5.
A metal alloy contains the following metals. Represent the data by a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total students = 60 + 100 + 80 + 150 + 60
= 450
Angle of sector for Gold = 
×360° = 48°
Angle of sector for Lead = 
×360° = 80°
Angle of sector for Silver = 
×360° = 64°
Angle of sector for Copper = 
×360° = 120°
Angle of sector for Zinc = ×360° = 48°
Question 6.
On a particular day, the sales (in Rs.) of different items of a baker’s shop are given below. Draw a pie chart for this data.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total Cost = 320 + 80 + 160 + 120 + 40
= 720
Angle of sector for Ordinary Bread = 
×360° = 160°
Angle of sector for Fruit Bread = 
×360° = 40°
Angle of sector for Cakes = 
×360° = 80°
Angle of sector for Biscuits = 
×360° = 60°
Angle of sector for Others = 
×360° = 20°
Question 7.
The money spent on a book by a publisher is given below:
Represent the above data by a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total Money spent = 25 + 12 + 6 + 9 + 8
= 60
Angle of sector for Paper = 
×360° = 150°
Angle of sector for Printing = 
×360° = 72°
Angle of sector for Binding = 
×360° = 36°
Angle of sector for Publicity = 
×360° = 54°
Angle of sector for Royalty = 
×360° = 48°
Question 8.
Expenditure of a farmer for cultivation is given as follows:
Represent the data in a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total Amount = 2000 + 1600 + 1500 + 1000 + 1100
= 7200
Angle of sector for Ploughing = 
×360° = 100°
Angle of sector for Fertilizer = 
×360° = 80°
Angle of sector for Seeds = 
×360° = 75°
Angle of sector for Pesticides = 
×360° = 50°
Angle of sector for irrigation = 
×360° = 55°
Question 9.
There are 900 creatures in a zoo as per list below:
Represent the above data by a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total Amount = 400 + 120 + 135 + 170 + 75
= 900
Angle of sector for Wild Animal = 
×360° = 160°
Angle of sector for Birds = 
×360° = 48°
Angle of sector for other land animals = 
×360° = 54°
Angle of sector for Water animals = 
×360° = 68°
Angle of sector for Reptiles = 
×360° = 30°
Question 10.
In a factory, five varieties of vehicles were manufactured in a year, whose break up is given below. Represent this data with the help of a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total Number of vehicles = 3000 + 4000 + 1500 + 1000 + 500
= 10000
Angle of sector for Scooter = 
×360° = 108°
Angle of sector for Motorbike = 
×360° = 144°
Angle of sector for Car = 
×360° = 54°
Angle of sector for Jeep = 
×360° = 36°
Angle of sector for Van = 
×360° = 18°
Question 11.
A food contains the following nutrients. Draw a pie chart representing the data.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total nutrients in food = 30% + 10% + 40% + 15% + 5%
= 100%
Angle of sector for Protein = 
×360° = 108°
Angle of sector for Fat = 
×360° = 36°
Angle of sector for Carbohydrates = 
×360° = 144°
Angle of sector for Vitamins = 
×360° = 54°
Angle of sector for Minerals = 
×360° = 18°
Question 12.
The favourite flavours of ice cream for students of a school is given in percentages as follows
Represent this data by a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total nutrients in food = 30% + 10% + 40% + 20%
= 100%
Angle of sector for Vanilla = ×360° = 108°
Angle of sector for Other flavours = ×360° = 36°
Angle of sector for Chocolate = ×360° = 144°
Angle of sector for Strawberry = 
×360° = 72°
Question 13.
The data on modes of transport used by students to come to school are given below:
Represent the data with the help of a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total nutrients in food = 30% + 10% + 40% + 15% + 5%
= 100%
Angle of sector for Cycle = ×360° = 108°
Angle of sector for Scooter = ×360° = 36°
Angle of sector for Bus = ×360° = 144°
Angle of sector for Walking = ×360° = 54°
Angle of sector for Car = ×360° = 18°
Question 14.
Mr. Rajan Babu spends 20% of his income on house rent, 30% on food and 10% for his children’s education. He saves 25%, while the remaining is used for other expenses. Make a pie chart exhibiting the above information.
Answer:
GIVEN:
House rent = 20%
Food = 30%
Children education = 10%
Savings = 25%
Other expenses = 15%
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total income of Mr.Rajan = 30% + 10% + 20% + 15% + 25%
= 100%
Angle of sector for Food = ×360° = 108°
Angle of sector for Children education = ×360° = 36°
Angle of sector for House Rent = ×360° = 72°
Angle of sector for Other expenses = ×360° = 54°
Angle of sector for Savings = 
×360° = 90°
Question 15.
The percentage of various categories of workers in a state are given in following table
Represent the information given above in the form of a pie chart.
Answer:
GIVEN:
FORMULA UASED:
Angle of sector in pie chart = ×360°
Total Workers = 40% + 10% + 12.5% + 12.5% + 25%
= 100%
Angle of sector for Cultivators = ×360° = 144°
Angle of sector for Commercial worker = ×360° = 36°
Angle of sector for Industrial worker = 
×360° = 45°
Angle of sector for Others = ×360° = 45°
Angle of sector for Agricultural Labour = ×360° = 90°
Exercise 3.3
Question 1.Find the mean of 2, 4, 6, 8, 10, 12, 14, 16.
Answer:
GIVEN: Values-2, 4, 6, 8, 10, 12, 14, 16
FORMULA USED: Arithmetic mean = 
Number of values = 8
Sum of values = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16
= 72
Arithmetic mean =
Arithmetic mean = 
= 9
CONCLUSION: Arithmetic Mean of given values is 9
Question 2.
If the average of the values 18, 41, x, 36, 31, 24, 37, 35, 27, 36, is 31. Find the value of x.
Answer:
GIVEN: Values-18, 41, x, 36, 31, 24, 37, 35, 27, 36
Arithmetic Mean = 31
FORMULA USED: Arithmetic mean =
Number of values = 10
Sum of values = 18 + 41 + x + 36 + 31 + 24 + 37 + 35 + 27 + 36
= 285 + x
Arithmetic mean =
Arithmetic mean = 
= 31
⇒ 285 + x = 310
⇒ x = 310 – 285
∴ x = 25
CONCLUSION: value of X comes to be 25
Question 3.
If in a class of 20 students, 5 students have scored 76 marks, 7 students have scored 77 marks and 8 students have scored 78 marks, then compute the mean of the class.
Answer:
GIVEN: total no of students = 20
5 students have scored 76 marks
7 students have scored 77 marks
8 students have scored 78 marks
FORMULA USED: Arithmetic mean =
Number of students = 20
Sum of marks = (5×76) + (7×77) + (8×78)
= 380 + 539 + 624
= 1543
Arithmetic mean = 
Arithmetic mean = 
= 77.15
CONCLUSION: Arithmetic Mean of marks of class is 77.15
Question 4.
The average height of 20 students in a class was calculated as 160 cm. On verification it was found that one reading was wrongly recorded as 132 cm instead of 152 cm. Find the correct mean height.
Answer:
GIVEN: Total no of students = 20
Average/mean = 160cm
FORMULA USED: Average =
Number of students = 20
Sum of height of all students of class =
Average × Number of students
= 160cm×20
= 3200cm
If one reading wrongly recorded is 132cm instead of 152cm
Then we have to subtract 132cm from the sum
And also add 152 in the sum to replace the height of student
= 3200cm – 132cm + 152cm
= 3220cm
Average/mean = 
Mean = 
= 161cm
CONCLUSION: Mean height of class is 161cm.
Question 5.
Calculate the Arithmetic mean of the following data:
Answer:
GIVEN:
FORMULA USED:
Arithmetic mean = 
∑f = 12 + 20 + 15 + 14 + 16 + 11 + 7 + 8
= 103
∑xf = 180 + 500 + 525 + 630 + 880 + 715 + 525 + 680
= 4635
Arithmetic mean =
Arithmetic mean = 
= 45
CONCLUSION: Arithmetic Mean of given data is 45
Question 6.
The following data give the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students
Answer:
GIVEN:
FORMULA USED: Arithmetic mean =
Let age of student be x
And number of students be frequency f
∑f = 3 + 8 + 9 + 11 + 6 + 3
= 40
∑xf = 39 + 112 + 135 + 176 + 102 + 54
= 618
Arithmetic mean =
Arithmetic mean = 
= 15.45
CONCLUSION: Mean age of given data is 15.45years
Question 7.
Obtain the A.M of the following data:
Answer:
GIVEN:
FORMULA USED: Arithmetic mean =
Let Marks of student be x
And number of students be frequency f
∑f = 6 + 11 + 3 + 5 + 4 + 7 + 10 + 4
= 50
∑xf = 390 + 770 + 225 + 400 + 340 + 630 + 950 + 400
= 4105
Arithmetic mean =
Arithmetic mean = 
= 82.1
CONCLUSION: Mean marks of given data is 82.1
Question 8.
The following table shows the weights of12 workers in a factory
Find the mean weight of the workers.
Answer:
GIVEN:
FORMULA USED: Arithmetic mean =
Let Weight of workers be x
And number of workers be frequency f
∑f = 3 + 4 + 2 + 2 + 1
= 12
∑xf = 180 + 256 + 136 + 140 + 72
= 784
Arithmetic mean =
Arithmetic mean = 
= 65.33
CONCLUSION: Mean weight of given data is 65.33kg
Question 9.
For a month, a family requires the commodities listed in the table below. The weight to each commodity is given. Find the Weighted Arithmetic Mean.
Answer:
GIVEN:
FORMULA USED: Weighted Arithmetic mean
= 
Let price of items be x
And weight of items be frequency w
∑w = 25 + 5 + 4 + 8 + 3
= 45
∑xf = 750 + 100 + 240 + 200 + 195
= 1485
Weighted arithmetic mean =
Arithmetic mean = 
= 33
CONCLUSION: Weighted arithmetic mean of given data is Rs33
Question 10.
Find the Weighted Arithmetic Mean for the following data:
Answer:
GIVEN:
FORMULA USED: Weighted Arithmetic mean
=
Let cost of items be x
And number of items be frequency w
∑w = 2 + 4 + 5 + 4
= 15
∑xf = 90 + 48 + 75 + 102
= 315
Weighted arithmetic mean =
Arithmetic mean = 
= 21
CONCLUSION: Weighted arithmetic mean of given data is Rs21
Question 11.
Find the median of the following set of values:
(i) 83, 66, 86, 30, 81.
(ii) 45, 49, 46, 44, 38, 37, 55, 51.
(iii) 70, 71, 70, 68, 67, 69, 70.
(iv) 51, 55, 46, 47, 53, 55, 51, 46.
Answer:
(i) GIVEN: values are 83, 66, 86, 30, 81
FORMULA USED: if number of values are odd then median is 
if number of values are even then median is 
and
Arrange the values in ascending order
30, 66, 81, 83, 86
The number of observations is 5
Which is odd
Median is term
⇒ Median = 
term
⇒ Median = 3rd term
∴ Median = 81
CONCLUSION: The median of given values is 81
(ii) GIVEN: values are 45, 49, 46, 44, 38, 37, 55, 51
FORMULA USED: if number of values are odd then median is
if number of values are even then median is and
Arrange the values in ascending order
37, 38, 44, 45, 46, 49, 51, 55
The number of observations is 8
Which is even
Median is and 
terms
⇒ Median = 
and 
terms
⇒ Median = 4th and 5th term
∴ Median = Average of 4th and 5th term
= 
= 45.5
CONCLUSION: The median of given values is 45.5
(iii) GIVEN: values are 70, 71, 70, 68, 67, 69, 70
FORMULA USED: if number of values are odd then median is
if number of values are even then median is and
Arrange the values in ascending order
67, 68, 69, 70, 70, 70, 71
The number of observations is 7
Which is odd
Median is term
⇒ Median = 
term
⇒ Median = 4th term
∴ Median = 70
CONCLUSION: The median of given values is 70
(iv) GIVEN: values are 51, 55, 46, 47, 53, 55, 51, 46
FORMULA USED: if number of values are odd then median is
If number of values are even then median is and
Arrange the values in ascending order
46, 46, 47, 51, 51, 53, 55, 55
The number of observations is 8
Which is even
Median is and terms
⇒ Median = and terms
⇒ Median = 4th and 5th term
∴ Median = 
= 51
CONCLUSION: The median of given values is 51
Question 12.
Find the median for the following data:
Answer:
GIVEN:
Total Frequency (N) = ∑f = 50
N/2 = 
= 25
∴ Median is 25th value
Now 25th value occurs in commutative value 25
which has value of X = 3
∴ Median is 3
CONCLUSION: Median of given data is 3
Question 13.
The height (in cm) of 50 students in a particular class are given below:
Find the median.
Answer:
GIVEN:
Let Height of student be x
And number of student be frequency f
Total Frequency (N) = ∑f = 50
N/2 = = 25
∴ Median is 25th value
Now 25th value occurs in commutative value 28
which has value of Height = 153cm
∴ Median is 153cm
CONCLUSION: Median of given data is 153cm
Question 14.
The hearts of 60 patients were examined through X-ray and the observations obtained are given below:
Find the median.
Answer:
GIVEN:
Let diameter of heart be x
And number of patient be frequency f
Total Frequency (N) = ∑f = 50
N/2 = = 25
∴ Median is 25th value
Now 25th value occurs in commutative value 31
which has diameter of heart = 132cm
∴ Median is 132cm
CONCLUSION: Median of given data is 132cm
Question 15.
The salary of 43 employees are given in the following table. Find the median.
Answer:
GIVEN:
Let salary be x
And number of employees be frequency f
Total Frequency (N) = ∑f = 43
N/2 = = 21.5
∴ Median is 21.5th value
Now 21.5th value occurs in commutative value 32
which has salary = 10000
∴ Median is Rs10000
CONCLUSION: Median of given data is Rs10000
Question 16.
Find the mode of the following data:
i) 74, 81, 62, 58, 77, 74.
ii) 55, 51, 62, 71, 50, 32.
iii) 43, 36, 27, 25, 36, 66, 20, 25.
vi) 24, 20, 27, 32, 20, 28, 20.
Answer:
i) GIVEN: the values are 74, 81, 62, 58, 77, 74
Mode = the value having maximum no of terms
⇒ As we can see
In the above values
Only value 74 is repeated twice
And all other are having only one term in data
∴ 74 is mode of data
CONCLUSION: The Mode of the data is 74.
ii) GIVEN: the values are 55, 51, 62, 71, 50, 32
Mode = the value having maximum no of terms
⇒ As we can see
In the above values
None of the value is repeated
And all values are having only one term in data
∴ There is NO mode of data
CONCLUSION: There is NO mode of data.
iii) GIVEN: the values are 43, 36, 27, 25, 36, 66, 20, 25
Mode = the value having maximum no of terms
⇒ As we can see
In the above values
25 and 36 are the values which are repeated twice in data
And all other values are having only one term in data
∴ 25 and 36 are mode of above data
CONCLUSION: Mode of data are 25 and 36
iv) GIVEN: the values are 24, 20, 27, 32, 20, 28, 20
Mode = the value having maximum no of terms
⇒ As we can see
In the above values
20 is the values which is repeated thrice in data
And all other values are having only one term in data
∴ 20 is mode of above data
CONCLUSION: Mode of data is 20
Question 17.
Find the mode for the following frequency table:
Answer:
GIVEN:
Mode = the value having maximum no of terms
⇒ As we can see
In the above given table
‘f’ is given number of terms
Then 37 is maximum number of terms in given data
⇒ the value having maximum number of terms is 15
∴ 15 is mode of above data
CONCLUSION: Mode of data is 15
Question 18.
Find the mode for the following table:
Answer:
GIVEN:
Mode = the value having maximum no of terms
⇒ As we can see
In the above given table
‘number of days’ is given number of terms
Then 8 is maximum number of terms in given data
⇒ the temperature on maximum number of days is 38.7°C
∴ 38.7°C is mode of above data
CONCLUSION: Mode of data is 38.7°C
Question 19.
The demand of different shirt sizes, as obtained by a survey, is given below.
Calculate the Mode.
Answer:
GIVEN:
Mode = the value having maximum no of terms
⇒ As we can see
In the above given table
‘Number of persons’ is given number of terms
Then 51 is maximum number of terms in given data
⇒ The size of shirt maximum number of people wear is 40
∴ 40 is mode of above data
CONCLUSION: Mode of data is 40
Question 20.
Find the mean, median and mode for the following frequency table:
Answer:
GIVEN:
FORMULA USED:
Mean =
Median = 
Mode = Maximum number of terms
⇒ For Mean:
Mean =
∑xf = 50 + 240 + 350 + 450 + 370 + 220
= 1680
∑f = 5 + 12 + 14 + 15 + 10 + 4
= 60
Mean = 
= 28
⇒ For Median
Total frequency (N) = ∑f = 60
N/2 = 
= 30
Which lies in commutative frequency 31
Which is having value of x to be 25
∴ Median = 25
⇒ For Mode
Maximum number of terms(f) in above table is 15
15 terms are of value 30
∴ Mode = 30
CONCLUSION: The data is having Mean to be 28
Mode to be 30
Median to be 25
Question 21.
The age of the employees of a company are given below.
Find the mean, median and mode.
Answer:
GIVEN:
FORMULA USED:
Mean =
Median = 
Mode = Maximum number of terms
Let ages of employees be x
And number of person be frequency (f)
⇒ For Mean:
Mean =
∑xf = 247 + 315 + 460 + 450 + 432 + 493 + 403
= 2800
∑f = 13 + 15 + 20 + 18 + 16 + 17 + 13
= 112
Mean = 
= 25
⇒ For Median
Total frequency (N) = ∑f = 112
N/2 = 
= 56
Which lies in commutative frequency 66
Which is having value of x to be 25
∴ Median = 25
⇒ For Mode
Maximum number of terms(f) in above table is 20
20 terms are of value 23
∴ Mode = 23
CONCLUSION: The data is having Mean to be 25
Mode to be 23
Median to be 25
Question 22.
The following table shows the weights of 20 students.
Calculate the mean, median and mode.
Answer:
GIVEN:
FORMULA USED:
Mean =
Median =
Mode = Maximum number of terms
Let weight of student be x
And number of student be frequency (f)
⇒ For Mean:
Mean =
∑xf = 188 + 150 + 371 + 112 + 240
= 1061
∑f = 4 + 3 + 7 + 2 + 4
= 20
Mean = 
= 53.05
⇒ For Median
Total frequency (N) = ∑f = 20
N/2 = 
= 10
Which lies in commutative frequency 14
Which is having value of x to be 53
∴ Median = 53
⇒ For Mode
Maximum number of terms(f) in above table is 7
7 terms are of value 53
∴ Mode = 53
CONCLUSION: The data is having Mean to be 53.05
Mode to be 53
Median to be 53