Data: The word ‘data’ means information in the form of numerical figures or a set of given facts. is a
set of values of qualitative or quantitative variables.
i) The marks obtained by pupils of a class in a monthly test are: . This is quantitative data.
ii) However, a collection of colors like Red, Blue, Green would be qualitative data something that
generally cannot be measured with numerical results.
➢ Raw or Ungrouped Data: The data obtained in original form is called raw data or ungrouped data.
Clearly, the data shown above is a raw data. Normally, when we collect data, we will get raw data.
➢ Group Data: For the sake of convenience (in study and comparison), we may condense the raw data
into classes or groups. Such a presentation is known as grouped data.
➢ Array: An arrangement of raw numerical data in ascending order of magnitude is called an Array.
Example: 1,3,5,7,8,9,10,11,12,13,14,15.
➢ Tabulation or Presentation of Data: Arranging the data in the form of tables in condensed form is
called as the Tabulation or Presentation of data.
➢ Observation: Each numerical figure in a data is called observation.
➢ Frequency: The number of times a particular observation occurs is called its frequency.
➢ Frequency distribution: The tabular arrangement of data showing the frequency of each observation
is called a frequency distribution.
➢ Frequency Distribution of Ungrouped Data:
Let’s use the following data to learn Frequency distribution.
Data : 2, 1, 3, 1, 2, 3, 2, 2, 1, 3, 2, 1, 5, 4, 3, 3, 2, 1, 2, 6 (this data could be anything… may be number of
TVs in homes, number of kids in families in a community etc.)
Step 1: Arranging the above data in ascending order, we get: 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, and 6.
Step 2: Organize the data in frequency distribution. In simple terms, what it means is the number of
times the same data appears.
Thus, we may prepare a frequency table as under:
➢ Grouped Frequency Distribution:
When you have large data set, the above frequency distribution might be more cumbersome. In such
scenarios, we can simplify the data by grouping it in groups.
Inclusive Form: Let’s say for the above data, we put them in different grouping or groups or class
interval such as 0-2, 3-4 and 5-6 .
Here 0-2 means between 0 and 2 , including both 0 and 2 . So, this form is known as inclusive form.
0 is called the lower limit and 2 is upper limit of the class 0-2 .
Exclusive Form: If we do the same thing, but instead of inclusive, we have exclusive classes.
Here 0-2 means between 0 and 2 , including 0 but excluding 2 .
So, this form is known as exclusive form. 0 is called the lower limit and 2 is upper limit of the class
0-2 . Thus 0-2 would mean 0 and more but less than 2.
➢ Class-interval: Each group into which the raw data is condensed is called a class-interval.
Each class is bounded by two figures, which are called class limits.
The figure on left side called lower limit and figure on right side is called as upper limit of the class.
Thus is a class with upper limit and lower limit .
➢ Class-boundaries: In an exclusive form the lower limits and upper limits are known as class
boundaries.
Thus, if the boundaries of the class10 − 20 in exclusive forms are 10 and 20 .
The boundaries in inclusive form are obtained by subtracting from lower limit and adding to
upper limit.
Thus the boundaries in inclusive form are10 − 20 and 9.5 − 20.5.
➢ Class-size: Difference between true upper limit and true lower limit is called the class size.
➢ Class Mark or Mid-value:
Class mark= 1
2 × (𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 + 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡)
Thus the class mark of 10 − 20 is 1
2 × (20 + 10) = 15
➢ Range: The difference between the maximum value and the minimum value of the observations is called its range.
Exercise 7.1
(1) Arithmetic mean will be
10000+10250+10790+9865+5350+10110/6
= 66365/6
= 11060.08
(2) Mean of the given data
10.25+ 9 + 4.75+ 8+ 2.65+ 12+ 2.35/7
= 49/7
= 7
(3) If the mean of 8 observation is 25
Then total 25 x 8 = 200
If one observation 11 is excluded
Then total will be 200 – 11 = 189
Now the mean of remaining 189/7 = 27
(4) Total 9 x 38 = 342
If 27 is taken insted of 72.
Then total be = 342 – 72 + 27 = 207
Now the actual mean of the remaining = 297/ 9 = 33
(5) Let the member of the family be x
Sum of ages of the family = 5x
After 5 years age will be 25x + 5x = 30x
Then the present age of mean = 30x/x = 30
(6) Total age of the group = 40 x 11 = 440
2 years ago, age was = 440 + 2 x 40 = 520
Now total age of 39 is 39 x 12 = 468
The age of the person who left the group was 520 – 468 = 52 year.
(7) Mean = 5+8+12+10+15+22/6 = 12
Sum of deviations
|5 – 12| + |8 – 12| + |12 – 12| + |10 – 12| + |15 – 12| + |22 – 12|
= |-7| + |4| + |0| + |-2| + |3| + |10|
= 7 + 4 + 2 + 3 + 10
= 26
(8) We know, Mean deviations = sum deviations/ number of observation
= 100/ 20
= 5
(9) Arithmetic mean of the data = 4+21+13+5+9+10+20+19+12+20+14/ 12
= 164/12
= 13.66
10) Let, marks of Karishma = x
Mean = x+x-8+x-6+x-3+x-3+x-1+x+x+2+x+3+x+4+x+6/10= 15
=> 10x-9+6/10 = 15
=> 10x = 150+3
=> x = 153/10
=> x = 15.3
(12) Writing in orders, we get
3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8
Here median is 6 is even number
So, average of two middle number is
3.3+3.5/2 = 6.8/2
= 3.4
(13) Given 10, 12, 14, x – 3, x, x+2, 25
Median = 15 is an odd numebr
So the median = 7+1/2
So, (7+1/2) = 15
= 4th term = 15
So, x – 3 = 15
=> x = 18
The observation are 10,12, 14, 15, 18, 20, 25
(14) Arranging we get
9, 10, 10, 10, 11, 11, 12, 15, 19, 20, 21
Since 10 is used often,
so, mode = 10
(15) If mode is x
If each scare is decreased by 3.
then new mode of new series be x – 3
(16) Since there is no repeat of number in natural numbers.
So, mode will be 0.
(17) Arranging we get 5, 8, 10, 15, 15, 24, 28
Arithmetic mean = 5+8+10+15+15+24+28/7
= 105/7
= 15
Median is (7+1/2)th = 4th term = 15.
Clearly mode is 15
If increased by 1 So new mode 16.
So new data
5, 8, 10, 15, 15, 16, 16, 16, 24, 23, x
Now No. of data 11
So arithmetic mean 15 = 5 + 8 + 10 + 15 + 15 + 16 + 16 + 16 + 24 + 28 + x / 11
=> 165 = 153 + x
=> x = 12
So form new data is 12, 16, 16, 16
Exercise 7.2
(1) Max. value = 63
Minimum value = 5
Range 63 – 5 = 58
Therefore, length of interval = 58/6 = 9.6 = 10
Then
| Class | Interval frequency |
| 0 – 9 | 5 |
| 10 – 19 | 8 |
| 20 – 29 | 10 |
| 30 – 39 | 7 |
| 40 – 49 | 7 |
| 50 – 59 | 5 |
| 60 – 69 | 3 |
(2) Max number = 40
Min number = 15
Range = (40 – 15) + 1
Class number = 26/4 = 6.5 = 7
| Class | Frequency |
| 15 – 22 | 12 |
| 23 – 30 | 11 |
| 31 – 38 | 6 |
| 39 – 46 | 1 |
(3) Given distribution are
4 – 11, 12 – 19, 20 – 27, 28 – 35, 36 – 43
Next two distribution are 44 – 51, 52 – 59.
(i) Length of each class interval 11 – 4 = 7
(ii) Class boundaries 11, 19, 27, 35, 43
(iii)
| Class Interval | Class mark |
| 4 – 11 | 7.5 |
| 12 – 19 | 15.5 |
| 20 – 27 | 23.5 |
| 28 – 35 | 31.5 |
| 36 – 43 | 39.5 |
(4)
(i) Class intervals of the data 22 – 10 = 12.
(ii) Class intervals are 6- 16, 16 – 28, 28 – 40, 40 – 51, 52 – 64, 64 – 76
| Class | Intervals |
| <16 | 6 |
| <28 | 6 + 14 = 20 |
| <40 | 20 + 20 = 40 |
| <52 | 40 + 21 = 61 |
| <64 | 61 + 9 = 70 |
| <76 | 70 + 5 = 75 |
(iii)
| Class | Intervals |
| > 6 | 6 + 69 = 75 |
| > 16 | 14 + 55 = 69 |
| > 28 | 20 + 35 = 55 |
| > 40 | 21 + 14 = 35 |
| > 52 | 9 + 5 = 14 |
| > 64 | 5 |
(5)
| Class | Distribution frequency |
| 0 – 10 | 2 |
| 10 – 20 | 10 |
| 20 – 30 | 4 |
| 30 – 40 | 7 |
| 40 – 50 | 12 |
(6)
| Class | Frequency | Cumulative class than | Cumulative greater than |
| 1 – 3 | 10 | 10 | 59 |
| 4 – 6 | 12 | 10 + 12 = 22 | 49 |
| 7 – 9 | 15 | 15 + 22 = 27 | 37 |
| 10 – 12 | 13 | 37 + 13 = 50 | 22 |
| 13 – 15 | 9 | 50 + 9 = 59 | 9 |
(7)
| Runs | No. of Students | Frequency |
| 0 – 10 | 3 | 3 |
| 10 – 20 | 8 | 8 – 3 = 5 |
| 20 – 30 | 19 | 19 – 8 = 11 |
| 30 – 40 | 25 | 25 – 19 = 6 |
| 40 – 50 | 30 | 30 – 25 = 5 |
(8)
| No of Books | Greater than cumulative frequency | Frequency | Less than cumulative frequency |
| 1 – 10 | 42 | 42 – 36 = 6 | 6 |
| 11 – 20 | 36 | 36 – 23 = 13 | 6 + 13 = 19 |
| 21 – 30 | 23 | 23 – 14 = 9 | 19 + 9 = 28 |
| 31 – 40 | 14 | 14 – 6 = 8 | 28 + 8 = 36 |
| 41 – 50 | 6 | 6 | 36 |
