Frequency Distribution Tables And Graphs Solution of TS & AP Board Class 8 Mathematics
Exercise 7.1
Question 1.
Find the arithmetic mean of the sales per day in a fair price shop in a week.
Rs. 10000, Rs. .10250, Rs. .10790, Rs. .9865, Rs. .15350, Rs. .10110
Answer:
Given, observations 10000, 10250, 10790, 9865, 15350, 10110
Clearly, No of observations, n = 6
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
Therefore, mean of above data
Hence, mean is Rs. 110060.83
Question 2.
Find the mean of the data;10.25, 9, 4.75, 8, 2.65, 12, 2.35
Answer:
Given, observations 10.25, 9, 4.75, 8, 2.65, 12, 2.35
Clearly, No of observations, n = 7
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
Therefore, mean of above data
Hence, mean is 7
Question 3.
Mean of eight observations is 25. If one observation 11 is excluded, find the mean of the remaining.
Answer:
Given, Mean of 8 observations is 25.
Clearly, No of observations, n = 8 and mean, 
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
⇒ sum of terms is 200.
If one observation 11 is excluded,
No of terms = 7
Sum of terms = 200 - 11 = 189
Hence, new mean is 27.
Question 4.
Arithmetic mean of nine observations is calculated as 38. But in doing so, an observation 27 is mistaken for 72. Find the actual mean of the data.
Answer:
Given, mean, 
No of observations, n = 9
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
⇒ sum of observations = x1 + x2 + …+ x9 = 342
As, 27 is mistaken for 72, therefore, actual sum of observations will increase by (72 - 27) = 45
⇒ Actual sum of observations = 342 + 45 = 387
Hence, new mean is 43.
Question 5.
Five years ago mean age of a family was 25 years. What is the present mean age of the family?
Answer:
Let the no of members in family be 'n'.
Mean age = 25
Let x1, x2, …, xn be ages of family members.
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
Therefore, 5 years ago
⇒ mean age, 
⇒ x1 + x2 + …+ xn = 25n [1]
In present, age of each family member will be increase by 5, therefore ages will be
x1 + 5, x2 + 5, …, xn + 5
⇒ Present mean age, 
[From 1]
Hence, Present mean age is 30.
Question 6.
Two years ago the mean age of 40 people was 11 years. Now a person left the group and the mean age is changed to 12 years. Find the age of the person who left the group.
Answer:
no of members = 40
Mean age two years ago = 11
Let x1, x2, …, x40 be ages of family members.
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
Therefore, two years ago
⇒ mean age, 
⇒ x1 + x2 + …+ x40 = 440 [1]
In present, age of each family member will be increase by 2, therefore ages will be
x1 + 2, x2 + 2, …, x40 + 2
But as a person left the group, let his present age be 'x' years
In this case, Sum of ages will be decrease by 'x' and No of persons will be 39.
⇒ Present mean age, 
⇒ 468 = 440 + 80 - x [From 1]
⇒ x = 520 - 468
⇒ x = 52
Present age of that person is 52 years.
Question 7.
Find the sum of deviations of all observations of the data 5, 8, 10, 15, 22 from their mean.
Answer:
Given, observations 5, 8, 10, 15, 22
Clearly, No of observations, n = 5
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
Therefore, mean of above data
Hence, mean is 12
Now, we know deviation
di = xi - a
in this case, a = 12
d1 = x1 - a = 5 - 12 = -7
d2 = x2 - a = 8 - 12 = -4
d3 = x3 - a = 10 - 12 = -2
d4 = x4 - a = 15 - 12 = 3
d5 = x5 - a = 22 - 12 = 10
Sum of deviations = (-7) + (-4) + (-2) + 3 + 10 = 0
Hence, sum of deviations from their mean is 0.
Question 8.
If sum of the 20 deviations from the mean is 100, then find the mean deviation.
Answer:
Given, No of deviations = 20
Sum of deviations = 100
We know,
Mean deviation 
⇒ Mean deviation 
⇒ Mean deviation = 5
Hence, mean deviation is 5.
Question 9.
Marks of 12 students in a unit test are given as 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14.
Assume a mean and calculate the arithmetic mean of the data. Assume another number as mean and calculate the arithmetic mean again. Do you get the same result? Comment.
Answer:
Given, observations are 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14
Let assumed mean, a = 17
We know, mean by assumed mean method is
Where, a is assumed mean and n is no of observations and xi's are observations.
Therefore, mean for given data
Now, let assumed mean a = 14
Yes, we get same result in both cases !
This shows that, actual mean doesn't depend on assumed mean !
Question 10.
Arithmetic mean of marks (out of 25) scored by 10 students was 15. One of the student, named Karishma enquired the other 9 students and find the deviations from her marks are noted as − 8, − 6, − 3, − 1, 0, 2, 3, 4, 6. Find Karishma’s marks.
Answer:
Let Karishma got 'a' marks.
And assumed mean be 'a'.
We know,
Arithmetic mean, 
Where, a is assumed mean.
As, deviations given are -8, -6, -3, -1, 0, 2, 3, 4, 6
Sum of deviations = -8 + (-6) + (-3) + (-1) + 0 + 2 + 3 + 4 + 6
= -3
As, arithmetic mean of marks is 15
Putting values in formula, We have,
⇒ 15 = a - 0.3
⇒ a = 15.3
⇒ Karishma got 15.3 marks
Question 11.
The sum of deviations of ‘n’ observations from 25 is 25 and sum of deviations of the same ‘n’ observations from 35 is − 25. Find the mean of the observations.
Answer:
Given, no of observations = n
And sum of deviation from 25 is 25 and sum of deviation from 35 is -25.
Let the actual mean be x,
We know that
Arithmetic mean,
Where, a is assumed mean.
In first case, where assumed mean, a = 25 and sum of deviation is 25, Putting values in formula
We have,
[1]
In first case, where assumed mean, a = 35 and sum of deviation is -25, Putting values in formula
We have,
[2]
Adding [1] and [2], we get
⇒ 2x = 60
⇒ x = 30
⇒ arithmetic mean of required data is 30.
Question 12.
Find the median of the data: 3.3, 3.5, 3.1, 3.7, 3.2, 3.8
Answer:
For finding median, first we arrange data either in ascending or descending order,
3.1, 3.2, 3.3, 3.5, 3.7, 3.8
No of terms (n) = 6
As, n is even
Median is arithmetic mean of 
term and 
term.
⇒ Median is arithmetic mean of 3rd and 4th term.
⇒ Median 
⇒ Median = 3.4
Question 13.
The median of the following observations, arranged in ascending order is 15.
10, 12, 14, x − 3, x, x + 2, 25. Then find x.
Answer:
Given, median = 15
As the terms are in ascending order, and
No of terms, n = 7
As, n is odd
Median is 
term.
⇒ Median is 4th term
⇒ median = x - 3
⇒ 15 = x - 3
⇒ x = 18
Hence, value of x is 18.
Question 14.
Find the mode of 10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10.
Answer:
As, mode is the most frequently occurring value.
Clearly, 10 is most frequent
Therefore, mode is 10.
Question 15.
Mode of certain scores is x. If each score is decreased by 3, then find the mode of the new series.
Answer:
As, mode is the most frequently occurring value.
'x' is most frequent score, now if each score is decreased by 3.
Each 'x' in scores will also be decreased by 3.
And mode will be (x - 3).
Question 16.
Find the mode of all digits used in writing the natural numbers from 1 to 100.
Answer:
From 1 to 100,
As, mode is the most frequent occurring value.
Mode of above data is 1, as 1 occurs 21 times.
Question 17.
Observations of a raw data are 5, 28, 15, 10, 15, 8, 24. Add four more numbers so that mean and median of the data remain the same, but mode increases by 1.
Answer:
Given, data is 5, 28, 15, 10, 15, 8, 24
No of terms,
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
Therefore, mean of above data
For finding median, let's arrange data in increasing order
5, 8, 10, 15, 15, 24, 28
As, n is even
Median is term.
⇒ Median is 4th term
⇒ median = 15
Also, mode is the most frequently occurring value.
Clearly, 15 occurs two times and is most frequent
Therefore, mode is 15.
Now, we have to add four terms to this data such that mean and median remains same, but mode increase by 1.
So, mode of new data = 16
But as 16 is mode, its frequency must be greater than frequency of 2 i.e. minimum 3.
So, we have three terms out of four as 16, 16 and 16.
Now, let 4th term be 'x'.
New data,
5, 8, 10, 15, 15, 16, 16, 16, 24, 28, x
No of terms = 11
As, mean remains same, mean of above data = 15
By using formula, mean of above data
⇒ 165 = 153 + x
⇒ x = 12
So, four terms are 12, 16, 16, 16.
Question 18.
If the mean of a set of observations x1, x2, ...., ...., x10 is 20. Find the mean of x1 + 4, x2 + 8, x3 + 12, .... ...., x10 + 40.
Answer:
Given,
No of observations, n = 10
Mean, 
We know,
Arithmetic mean of x1, x2, x3, …, xn (n observations) is
For given data,
⇒ x1 + x2 + x3 + …+x10 = 200 [1]
Now, to find the mean of
x1+4, x2 + 8, …, xn + 40
No of terms, n = 10
Mean using above formula,
Hence, mean of new data is 42.6
Question 19.
Six numbers from a list of nine integers are 7, 8, 3, 5, 9 and 5. Find the largest possible value of the median of all nine numbers in this list.
Answer:
No of terms, n = 9
As, n is odd
Median is term.
⇒ Median is 5th term
Let's first arrange given no in ascending order.
3, 5, 5, 7, 8, 9
Clearly, 9 is not possible as it is already at 6th place and adding any term or terms can only shift its position ahead.
Also, 8 is possible if other three terms are greater than 8.
Therefore, 8 is the largest value possible.
Question 20.
The median of a set of 9 distinct observations is 20. If each of the largest 4 observations of the set is increased by 2, find the median of the resulting set.
Answer:
No of terms, n = 9
As, n is odd
Median is term.
⇒ Median is 5th term
If each of the largest 4 observations of set is increased by 2, it does not affect 5th term or the order of terms.
Therefore, median remains same i.e. 20.
Exercise 7.2
Question 1.
Given below are the ages of 45 people in a colony.
33 8 7 25 31 26 5 50 25 48 56
33 28 22 15 62 59 16 14 19 24 35
26 9 12 46 15 42 63 32 5 22 11
42 23 52 48 62 10 24 43 51 37 48
36
Construct grouped frequency distribution for the given data with 6 class intervals.
Answer:
Question 2.
Number of students in 30 class rooms in a school are given below. Construct a frequency distribution table for the data with a exclusive class interval of 4 (students).
25 30 24 18 21 24 32 34 22 20 22
32 40 28 30 22 26 31 34 15 38 28
20 16 15 20 24 30 25 18
Answer:
Question 3.
Class intervals in a grouped frequency distribution are given as 4 – 11, 12 – 19, 20 – 27, 28 – 35, 36 – 43. Write the next two class intervals. (i) What is the length of each class interval? (ii) Write the class boundaries of all classes, (iii) What are the class marks of each class?
Answer:
Next two class intervals 44 - 51 and 52 - 59.
(i) length of class interval = difference b/w upper and lower limit = 7
(ii)
(iii)
We know,
Class-mark 
Question 4.
In the following grouped frequency distribution table class marks are given.
(i) Construct class intervals of the data. (Exclusive class intervals)
(ii) Construct less than cumulative frequencies and
(iii) Construct greater than cumulative frequencies.
Answer:
(i) As the difference between two consecutive class intervals is 12, we can evaluate the exclusive class intervals as,
Lower limit = class mark - 6
Upper limit = class mark + 6
(ii)
(iii)
Question 5.
The marks obtained by 35 students in a test in statistics (out of 50) are as below.
35 1 15 35 45 23 31 40 21 13 15
20 47 48 42 34 43 45 33 37 11 13
27 18 12 37 39 38 16 13 18 5 41
47 43
Construct a frequency distribution table with equal class intervals, one of them being 10-20 (20 is not included).
Answer:
Question 6.
Construct the class boundaries of the following frequency distribution table. Also construct less than cumulative and greater than cumulative frequency tables.
Answer:
Question 7.
Cumulative frequency table is given below. Which type of cumulative frequency is given. Try to build the frequencies of respective class intervals.
Answer:
Question 8.
Number of readers in a library are given below. Write the frequency of respective classes. Also write the less than cumulative frequency table.
Answer:
Exercise 7.3
Question 1.
The following table gives the distribution of 45 students across the different levels of Intelligent Quotient. Draw the histogram for the data.
Answer:
Question 2.
Construct a histogram for the marks obtained by 600 students in the VII class annual examinations.
Answer:
Question 3.
Weekly wages of 250 workers in a factory are given in the following table. Construct the histogram and frequency polygon on the same graph for the data given.
Answer:
Question 4.
Ages of 60 teachers in primary schools of a Mandal are given in the following frequency distribution table. Construct the Frequency polygon and frequency curve for the data without using the histogram. (Use separate graph sheets)
Answer:
For above problem, let us first calculate class marks for each interval.
Class mark
Now using points, we can make frequency curve and frequency polygon.
Frequency Polygon
Frequency Curve
Question 5.
Construct class intervals and frequencies for the following distribution table. Also draw the ogive curves for the same.
Answer:
Less than ogive
More than ogive
