CBSE - Statistics

In our day to day life, we can collect the following data.

1. Number of females per 1000 males in various states of our country

2. Weights of students of our class

3. Production of wheat in the last 10 years in our country

4. Number of plants in our locality

5. Rainfall in our city in the last 10 years

The information which is collected by the investigator himself with a definite objective in his mind is called as primary data
when the information is gathered from a source which already had the information stored, it is called as secondary data.
It can be observed that the data in 1, 3, and 5 is secondary data and the data in 2 and 4 is primary data.

The frequency means the number of students having same blood group.
It can be observed that 9 students have their blood group as A,
6 as B,
3 as AB, and
12 as O.

Therefore, the blood group of 30 students of the class can be represented as follows.

It can be observed clearly that 
Most common Blood Group (Highest frequency): O -(maximum number of students)
Rarest Blood Group (Lowest frequency): AB - (minimum number of students)

It is given that a grouped frequency distribution table of class size 5 has to be constructed. Therefore, the class intervals will be 0 - 5, 5 - 10, 10 - 15, 15 - 20…

By observing the data given as above, a grouped frequency distribution table can be constructed as follows.

It can be observed that there are very few engineers whose homes are at more than or equal to 20 km distance from their work place. Most of the engineers have their workplace up to 15 km distance from their homes.

(i) A grouped frequency distribution table of class size 2 has to be constructed. The class intervals will be 84 - 86, 86 - 88, and 88 - 90…

By observing the data given above, the required table can be constructed as follows.

(ii) It can be observed that the relative humidity is high. Therefore, the data is about a month of rainy season.

(iii) Range of data = Maximum value - Minimum value

= 99.2 - 84.9 = 14.3

(i) A grouped frequency distribution table has to be constructed taking class intervals 160 - 165, 165 - 170, etc. By observing the data given above, the required table can be constructed as follows.

(ii) It can be concluded that more than 50% of the students are shorter than 165 cm.

Taking class intervals as 0.00, - 0.04, 0.04, - 0.08, and so on, a grouped frequency table can be constructed as follows.

The number of days for which the concentration of SO₂ is more than 0.11 is the number of days for which the concentration is in between 0.12 - 0.16, 0.16 - 0.20, 0.20 - 0.24.

Required number of days = 2 + 4 + 2 = 8

Therefore, for 8 days, the concentration of SO₂ is more than 0.11 ppm.

By observing the data given above, the required frequency distribution table can be constructed as follows.

(i) By observation of the digits after decimal point, the required table can be constructed as follows.

(ii) It can be observed from the above table that the least frequency is 2 of digit 0, and the maximum frequency is 8 of digit 3 and 9. Therefore, the most frequently occurring digits are 3 and 9 and the least frequently occurring digit is 0.

(i) Our class intervals will be 0 - 5, 5 - 10, 10 - 15…..

The grouped frequency distribution table can be constructed as follows.

(ii) The number of children who watched TV for 15 or more hours a week is 2 (i.e., the number of children in class interval 15 - 20).

A grouped frequency table of class size 0.5 has to be constructed, starting from class interval 2 - 2.5.

Therefore, the class intervals will be 2 - 2.5, 2.5 - 3, 3 - 3.5…

By observing the data given above, the required grouped frequency distribution table can be constructed as follows.

(i) By representing causes on x-axis and family fatality rate on y-axis and choosing an appropriate scale (1 unit = 5% for y axis), the graph of the information given above can be constructed as follows.

All the rectangle bars are of the same width and have equal spacing between them.

(ii) Reproductive health condition is the major cause of women's ill health and death worldwide as 31.8% of women are affected by it.

(iii) The factors are as follows.

1. Lack of medical facilities

2. Lack of correct knowledge of treatment

(i) By representing section (variable) on x-axis and number of girls per thousand boys on y-axis, the graph of the information given above can be constructed by choosing an appropriate scale (1 unit = 100 girls for y-axis)

Here, all the rectangle bars are of the same length and have equal spacing in between them.

(ii) It can be observed that maximum number of girls per thousand boys (i.e., 970) is for ST and minimum number of girls per thousand boys (i.e., 910) is for urban.

Also, the number of girls per thousand boys is greater in rural areas than that in urban areas, backward districts than that in non-backward districts, SC and ST than that in non-SC/ST.

(i) By taking polling results on x-axis and seats won as y-axis and choosing an appropriate scale (1 unit = 10 seats for y-axis), the required graph of the above information can be constructed as follows.

Here, the rectangle bars are of the same length and have equal spacing in between them.

(ii) Political party 'A' won maximum number of seats.

(i) It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of 1 in between them. Therefore,
1/2 =0.5  has to be added to each upper class limit and also have to subtract 0.5 from the lower class limits so as to make the class intervals continuous.

Taking the length of leaves on x-axis and the number of leaves on y-axis, the histogram of this information can be drawn as above.

Here, 1 unit on y-axis represents 2 leaves.

(ii) Other suitable graphical representation of this data is frequency polygon.

(iii) No, as maximum number of leaves (i.e., 12) has their length in between 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.

(i) By taking life time (in hours) of neon lamps on x-axis and the number of lamps on y-axis, the histogram of the given information can be drawn as follows.

Here, 1 unit on y-axis represents 10 lamps.

(ii) It can be concluded that the number of neon lamps having their lifetime more than 700 is the sum of the number of neon lamps having their lifetime as 700 - 800, 800 - 900, and 900 - 1000.

Therefore, the number of neon lamps having their lifetime more than 700 hours is 184. (74 + 62 + 48 = 184)

It can be observed that the class intervals of the given data are not continuous. There is a gap of 1 in between them. Therefore, 1/2 =0.5 has to be added to the upper class limits and 0.5 has to be subtracted from the lower class limits.

Also, class mark of each interval can be found by using the following formula.

Class mark 

Continuous data with class mark of each class interval can be represented as follows.

By taking class marks on x-axis and runs scored on y-axis, a frequency polygon can be constructed as follows.

The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 1 is selected and the length of the rectangle is proportionate to it.

Taking the age of children on x-axis and proportion of children per 1 year interval on y-axis, the histogram can be drawn

(i) The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 2 is selected and the length of the rectangle is proportionate to it.

The proportion of the surnames per 2 letters interval can be calculated as:

 By taking the number of letters on x-axis and the proportion of the number of surnames per 2 letters interval on y-axis and choosing an appropriate scale (1 unit = 4 students for y axis), the histogram can be constructed as follows.

(ii) The class interval in which the maximum number of surnames lies is 6 - 8 as it has 44 surnames in it i.e., the maximum for this data.

We can find the class marks of the given class intervals by using the following formula.

Class mark 

Taking class marks on x-axis and frequency on y-axis and choosing an appropriate scale (1 unit = 3 for y-axis), the frequency polygon can be drawn as follows.

It can be observed that the performance of students of section 'A' is better than the students of section 'B' in terms of good marks.

The number of goals scored by the team is

2, 3, 4, 5, 0, 1, 3, 3, 4, 3

Arranging the number of goals in ascending order,

0, 1, 2, 3, 3, 3, 3, 4, 4, 5

The number of observations is 10, which is an even number. Therefore, median score will be the mean of 10/2 i.e., 5^th and 10/2 +1 
i.e., 6^th observation while arranged in ascending or descending order.

Mode of data is the observation with the maximum frequency in data.

Therefore, the mode score of data is 3 as it has the maximum frequency as 4 in the data.

The marks of 15 students in mathematics test are

41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60

Arranging the scores obtained by 15 students in an ascending order,

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

As the number of observations is 15 which is odd, therefore, the median of data will be 15+1 / 2 = 8^th observation whether the data is arranged in an ascending or descending order.

Therefore, median score of data = 52

Mode of data is the observation with the maximum frequency in data. Therefore, mode of this data is 52 having the highest frequency in data as 3.

It can be observed that the total number of observations in the given data is 10 (even number). Therefore, the median of this data will be the mean of 10 / 2 i.e., 5^th and 10/2 +1 i.e., 6^th observation.

Arranging the data in an ascending order,

14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28

It can be observed that 14 has the highest frequency, i.e. 4, in the given data. Therefore, mode of the given data is 14.

Salary (xi)	Number of workers (fi)	fixi
3000	16	48000
4000	12	48000
5000	10	50000
6000	8	48000
7000	6	42000
8000	4	32000
9000	3	27000
10000	1	10000
Total	Σfi = 60	Σfixi = 305000

= 5083.33

Therefore, mean salary of 60 workers is Rs 5083.33.

When any data has a few observations such that these are very far from the other observations in it, it is better to calculate the median than the mean of the data as median gives a better estimate of average in this case.

(i) Consider the following example - the following data represents the heights of the members of a family.

154.9 cm, 162.8 cm, 170.6 cm, 158.8 cm, 163.3 cm, 166.8 cm, 160.2 cm

In this case, it can be observed that the observations in the given data are close to each other. Therefore, mean will be calculated as an appropriate measure of central tendency.

(ii) The following data represents the marks obtained by 12 students in a test.

48, 59, 46, 52, 54, 46, 97, 42, 49, 58, 60, 99

In this case, it can be observed that there are some observations which are very far from other observations. Therefore, here, median will be calculated as an appropriate measure of central tendency.