The marks scored by 40 candidates in an examination (out of 100) is given below:
75, 65, 57, 50, 32, 54, 75, 67, 75, 88, 80, 42, 40, 41, 34, 78, 43,
61, 42, 46, 68, 52, 43, 49, 59, 49, 67, 34, 33, 87, 97, 47, 46, 54,
48, 45, 51, 47, 41, 43.
Prepare a frequency distribution table with the class size 10.
Take the class intervals as (30-39), (40-49), ... and answer the following questions:
(i) Which class intervals have highest and lowest frequency?
(ii) Write the upper and lower limits of the class interval 30-39
(iii) What is the range of the given distribution?
Theory.
The number of times a particular observation occurs in data is called frequency.
Showing data in tabular form with showing frequency of each distribution. This representation is called Frequency distribution table.
(i) As looking in Frequency distribution table
Highest Frequency is 16
And the group of marks having highest frequency is 40-49
∴ Maximum number of students have got marks between 40 to 49.
Lowest Frequency is 1
And the group of marks having lowest frequency is 90-99
∴ Minimum number of students have got marks between 90 to 99.
(ii) The given distribution is in inclusive form . It should be converted in exclusive form
Upper limit of 1st interval is 39
Lower limit of 2nd interval is 40
= = = 0.5
Actual upper limit = Stated upper limit + = 39 + 0.5 = 39.5
Actual lower limit = Stated lower limit – = 30 – 0.5 = 29.5
∴ Upper limit of group 30-39 is 29.5
Lower limit is lower most value of group
∴ Lower limit of group 30-39 is 30.5
(iii) Range = highest – lowest
Highest marks in class is 97
Lowest marks in class is 32
Range = 97 – 32 = 65
Prepare the frequency distribution table for the given set of scores:
39, 16, 30, 37, 53, 15, 16, 60, 58, 26, 28, 19, 20, 12, 14, 24, 59,
21, 57, 38, 25, 36, 34, 15, 25, 41, 52, 45, 60, 63, 18, 26, 43, 36,
18, 27, 59, 63, 46, 48, 25, 33, 46, 27, 46, 42, 48, 35, 64, 24.
Take class intervals as (10-20), (20-30), ... and answer the following:
(i) What does the frequency corresponding to the third class interval mean?
(ii) What is the size of each class interval? Find the midpoint of the class interval 30-40.
(iii) What is the range of the given set of scores?
Theory.
The number of times a particular observation occurs in data is called frequency.
Showing data in tabular form with showing frequency of each distribution. This representation is called Frequency distribution table.
(i) Third interval is 30-40
The frequency of the Third interval is 10
(ii) Class size = upper limit – lower limit
In class interval 30-40
Upper limit of interval is 40
Lower limit of interval is 30
Class size = 40 – 30 = 10
Midpoint =
= = = 35
(iii) Range = highest – lowest
Highest Score is 64
Lowest Score is 12
Range = 64 – 12 = 52
Draw a histogram to represent the following frequency distribution.
Draw a histogram to represent the following frequency distribution.
The given distribution is in inclusive form. It should be converted in exclusive form
Upper limit of 1st interval is 19
Lower limit of 2nd interval is 20
= = = 0.5
Actual upper limit = Stated upper limit +
Actual lower limit = Stated lower limit –
∴ Frequency distribution table
Runs scored by 10 batsmen in a one day cricket match are given.
Find the average runs scored.
23, 54, 08, 94, 60, 18, 29, 44, 05, 86
Theory.
Average =
Solution.
Average =
Sum of runs scored by 10 batsmen
= 23 + 54 + 8 + 94 + 60 + 18 + 29 + 44 + 5 + 86
= 421
There are 10 batsmen
Average =
∴ Average = = 42.1 runs
Find the mean weight from the following table.
To find the mean let us prepare frequency distribution table first. We observed that some values are repeated So, to find sum of all we have to multiply weight with number of children and then find the sum.
Let Weight be x and number of children be f
Average =
= = 31.53 kg
∴ Average weight of each child is 31.53 kg
Calculate the mean for the following frequency distribution:
To find the mean let us prepare frequency distribution table first .
Calculate the mid-point of each interval
And put it as x
Mid-point =
Average =
= = = 42.75
Calculate the mean for the following frequency distribution:
To find the mean let us prepare frequency distribution table first.
Calculate the mid-point of each interval
And put it as x
Mid-point =
Average =
= = = 28.625
Find the median of the data: 15, 22, 9, 20, 6, 18, 11, 25, 14.
1st arrange data in ascending order
6, 9, 11, 14, 15, 18, 20, 22, 25
As we can count there are odd number of terms
∴ Median = term
Where N is number of terms
Median = = = 5th term = 15
Find the median of the data: 22, 28, 34, 49, 44, 57, 18, 10, 33, 41, 66, 59.
1st arrange data in ascending order
10, 18, 22, 28, 33, 34, 41, 44, 49, 57, 59, 66
As we count there are odd numbers of terms
∴ Median = Average of term and term
Where N is number of terms which is 12
∴ Median = Average of term and term
= Average of 6thterm and 7th term
= Average of 34 and 41
Average =
Sum of terms = 34 + 41 = 75
Average = = 37.5
Find the median for the following frequency distribution table:
The given distribution is in inclusive form. It should be converted in exclusive form
Upper limit of 1st interval is 119
Lower limit of 2nd interval is 120
= = = 0.5
Actual upper limit = Stated upper limit +
Actual lower limit = Stated lower limit –
As we have sum of frequency to be (N) 50
As it is an even number
It has 2 middle scores
=
For finding 25th and 26th term we need to find cumulative frequency
25 and 26 can be covered under Cumulative frequency 29
∴ 129.5 – 139.5 is Median class
⇒ Low real limit (LRL) = 129.5
⇒ Frequency of median class (fm) = 15
⇒ Cumulative Frequency of above median class (fc) = 14
⇒ Size of class interval (i) = 10
Median = LRL +
= 129.5 +
= 129.5 +
= 129.5 + = 129.5 + 7.33 = 136.83
Find the median for the following frequency distribution table:
As we have sum of frequency to be (N) 40
As it is an even number
It has 2 middle scores
=
For finding 20th and 21th term we need to find cumulative frequency
20 and 21 can be covered under Cumulative frequency 27
∴ 15-20 is Median class
⇒ Low real limit (LRL) = 15
⇒ Frequency of median class (fm) = 10
⇒ Cumulative Frequency of above median class (fc) = 17
⇒ Size of class interval (i) = 5
Median = LRL +
= 15 +
= 15 +
= 15 + = 15 + 1.66 = 16.66
Find the mode for the following data:
4, 3, 1, 5, 3, 7, 9, 6
In the given data
Only 3 is repeater is twice
∴ 3 is mode of given data
Find the mode for the following data:
22, 36, 18, 22, 20, 34, 22, 42, 46, 42
In the given data
22 is repeater thrice and 42 is repeated twice
∴ 22 is mode of given data
Find the mode for the following data:
In the given data
The maximum frequency is 16
Which is of number 20
Hence;
Number 20 is repeated maximum times
∴ 20 is the mode of the data
The size or width of the Class interval (0 - 4) is :
A. 4
B. 5
C. 3
D. 0
The width of this interval is 5, i.e., 0,1,2,3,4,5.
The midpoint of the class interval (10 -19) is:
A. 10
B. 14
C. 15
D. 14.5
We know that,
⇒ midpoint = 14.5
The difference between the highest and lowest score of a distribution gives:
A. class interval
B. class width
C. range
D. class limit
By Definition,
In a set of data, the range is the difference between the highest and the lowest observation.
The number of times a particular observation (score) occurs in a data is called its:
A. frequency
B. range
C. class interval
D. class limit
By definition,
The number of times a particular observation (score) occurs in a data is called its frequency.
In inclusive form, the actual upper limit and lower limit of class interval (0-4) are:
A. -0.5 & 3.5
B. 0.5 & 4.5
C. −1& 5
D. 1 & 5
In inclusive form,
Actual lower limit = lower limit -0.5
= 0-0.5
= -0.5
And, Actual upper limit = upper limit-0.5
= 4-0.5
= 3.5
The height of a rectangle in a histogram represents:
A. class interval
B. midpoint
C. frequency density
D. frequency
During Representation,
The height of a rectangle in a histogram represents the frequency.
In a histogram, the width of the rectangle indicates:
A. class interval
B. midpoint
C. frequency density
D. frequency
During the representation,
In a histogram, the width of the rectangle indicates the class interval.
The mean of scores 10, 15, 12, 15, 15 is:
A. 15
B. 13
C. 13.4
D. 14.3
⇒ Mean = 13.4
Class interval grouping of data is done when:
A. the range of data is small
B. the range of data is large
C. the class intervals are small
D. class intervals are large
Class interval grouping of data is done when the range of data is large.
The mean of 6, 4, 7, x and 10 is 8. The value of x is:
A. 10
B. 12
C. 14
D. 13
⇒ 40 = x+27
⇒ x = 13
If n = 10 and Mean = 12, then Σ fx is:
A. 120
B. 1200
C. 12
D. 13
We know that,
⇒ = 120
The mean of first three multiples of 5 is :
A. 5
B. 10
C. 15
D. 30
The first three multiples of 5 are-5, 10, 15.
And, their mean will be-
⇒ Mean = 5
The median of 37, 83, 70, 29, 32, 42, 40 is:
A. 29
B. 30
C. 40
D. 42
Arranging the data in ascending order, we get-
29, 32, 37, 40, 42, 70, 83
And, since the no. of observations(n) is 7 and which is odd
⇒ Median = 4th term
⇒ Median = 40
In an inclusive class interval (10 -14), the lower real limit is:
A. 9.5
B. 10.5
C. 13.5
D. 14.5
In inclusive form,
Lower real limit = lower Limit-0.5
= 10-0.5
= 9.5
In an exclusive class interval (10 -20), the lower real limit is:
A. 20
B. 10
C. 10.5
D. 20.5
In inclusive form,
Lower real limit = lower Limit
= 10
The mode of 2, 3, 3, 5, 3, 5, 7, 3,5is:
A. 3
B. 5
C. 3 and 5
D. 3,5,7
Mode of observations is the data with highest frequency.
Here, 3 appears 3 times -
⇒ mode = 3
For given two values of x, 16, 18 the frequencies are respectively 12 and 20. Then the mode is:
A. 16
B. 18
C. 12
D. 20
Mode of observations is the data with highest frequency and here 18 has highest frequency i.e., 20.
⇒ Mode = 18
A collection of data having more than 3 modes is said to be:
A. uni-mode
B. bi-mode
C. tri-mode
D. multi-mode
A collection of data having more than 3 modes is said to be multimode.
Prepare a frequency distribution table for the scores given:
42,22,55,18,50,10,33,29,17,29,29,27,34,15,40,42,40,41,35,27,
44,31,38,19,54,55,38,19,20,30,42,59,15,19,27,23,40,32,28,51.
Take the class intervals as 10-20, 20-30, 30-40, 40-50, 50-60. From
the frequency distribution table answer the following questions:
(i) What does the frequency corresponding to the class interval 20-30 indicate?
(ii) In which class intervals are the scores 10, 20 and 30 included?
(iii) Find the range of the scores.
(i) The frequency corresponding to the class interval 20-30 indicate that there are 10 values lying between 20 and 30 and which is highest of all.
(ii) 10 will be included in 10-20, 20 will be included in 20-30 and 30 will be included in 30-40.
(iii) Since the intervals includes data values from 10 to 59.
⇒ Range = 59-10
⇒ Range = 49
The following are the marks scored in a unit test (out of 25). Prepare a frequency distribution table, taking the class intervals as 0-4,
5-9, 10-14, 15-19, 20-24:
21,14,3,7,23,18,24,16,18,17,20,10,17,18,21,23,19,12,14,9,16,1 8,12,14,11.
From the table (i) find the mid-points of each class interval (ii) find the class interval having a maximum frequency (iii) find the range of the scores.
(i) Mid-points of 0-4, 5-9, 10-14, 15-19, 20-24 are 2, 7, 12, 17, 22 respectively.
(ii) Here maximum frequency is 12 which is corresponding to the 15-19 class interval.
(iii) Range = Highest data value-lowest data value
⇒ Range = 24-3 = 21
Draw a histogram for the following frequency distribution.
Draw a histogram for the following frequency distribution.
The marks obtained by 12 students in a mathematics examination are given below.
48,78,93,90,66,54,83,58,60,75,89,84.
Find (i) the mean of the marks (ii) the mean mark of the students if each student is given 4 grace marks.
(i)
⇒ Mean = 73.17
(ii)
Since, each student got 4 grace marks implies we need to add 4 marks 12 times.
⇒Mean = 77.17
If the mean of 8,12,21,42, x is 20, find the value of x.
⇒ 100 = x+83
⇒ x = 17
Find the mean for the following distribution:
12,14,10,12,15,12,18,10,15,11,19,20,12,15,19,10,18,16,20,17.
⇒ Mean = 14.75