Statistics — Exercise 19.2

State which of the following variables are continuous and which are discrete:

(i) marks scored (out of 50) in a test.

(ii) daily temperature of your city.

(iii) sizes of shoes.

(iv) distance travelled by a man.

(v) time.

Answer

(i) Discrete

(ii) Continuous

(iii) Discrete

(iv) Continuous

(v) Continuous

Using class intervals 0-4, 5-9, 10-14, ....... construct the frequency distribution for the following data :

13, 6, 10, 5, 11, 14, 2, 8, 15, 16, 9, 13, 17, 11, 19, 5, 7, 12, 20, 21, 18, 1, 8, 12, 18.

Answer

The frequency distribution table for given grouped data is :

Given below are the marks obtained by 27 students in a test:

21, 3, 28, 38, 6, 40, 20, 26, 9, 8, 14, 18, 20, 16, 17, 10, 8, 5, 22, 27, 34, 2, 35, 31, 16, 28, 37.

(i) Using the class intervals 1-10, 11-20 etc. construct a frequency table.

(ii) State the range of these marks.

(iii) State the class mark of the third class of your frequency table.

Answer

(i) The frequency distribution table for given grouped data is :

(ii) Range = Highest mark - Lowest mark = 40 - 2 = 38.

Hence, range of marks = 38.

(iii) Class mark = $\dfrac{\text{Lower limit + Upper limit}}{2}$

Third class of frequency table = 21-30.

Substituting values we get :

Class mark = $\dfrac{21 + 30}{2} = \dfrac{51}{2}$ = 25.5

Hence, class mark = 25.5

Explain the meaning of the following terms:

(i) variate

(ii) class size

(iii) class mark

(iv) class limits

(v) true class limits

(vi) frequency of a class

(vii) cumulative frequency of a class.

Answer

(i) Variant — A particular value of a variable is called variate.

(ii) Class size — The difference between the actual upper limit and the actual lower limit of a class is called its class size.

(iii) Class mark — The class mark of a class is the value midway between its actual lower limit and actual upper limit.

(iv) Class limits — In the frequency table the class interval is called class limits.

(v) True class limits — In a continuous distribution, the class limits are called true or actual class limits.

(vi) Frequency of a class — The number of tally marks opposite to a variate is its frequency and it is written in the next column opposite to tally marks of the variate.

(vii) Cumulative frequency of a class — The sum of frequency of all previous classes and that particular class is called the cumulative frequency of the class.

Fill in the blanks :

(i) The number of observations in a particular class is called .......... of the class.

(ii) The difference between the class marks of two consecutive classes is the .......... of the class.

(iii) The range of the data 16, 19, 23, 13, 11, 25, 18 is .......... .

(iv) The mid-point of the class interval is called its .......... .

(v) The class mark of the class 4 – 9 is .......... .

Answer

(i) The number of observations in a particular class is called frequency of the class.

(ii) The difference between the class marks of two consecutive classes is the size of the class.

(iii) Range = Highest value - Lowest value

= 25 - 11 = 14.

The range of the data 16, 19, 23, 13, 11, 25, 18 is 14.

(iv) The mid-point of the class interval is called its class marks.

(v) Class mark = $\dfrac{4 + 9}{2} = \dfrac{13}{2}$ = 6.5

The class mark of the class 4 – 9 is 6.5.

The marks obtained (out of 50) by 40 students in a test are given below:

28, 31, 45, 03, 05, 18, 35, 46, 49, 17, 10, 28, 31, 36, 40, 44, 47, 13, 19, 25, 24, 31, 38, 32, 27, 19, 25, 28, 48, 15, 18, 31, 37, 46, 06, 01, 20, 10, 45, 02.

(i) Taking class intervals 1 – 10, 11 – 20, .., construct a tally chart and a frequency distribution table.

(ii) Convert the above distribution to continuous distribution.

(iii) State the true class limits of the third class.

(iv) State the class mark of the fourth class.

Answer

(i) The frequency distribution table for given grouped data is :

(ii) Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

$= \dfrac{11 - 10}{2} = \dfrac{1}{2}$ = 0.5

Subtracting the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for the given data is :

(iii) Third class : 20.5 - 30.5

Hence, lower limit : 20.5 and upper limit : 30.5

(iv) Class mark = $\dfrac{\text{Lower limit + Upper limit}}{2}$

= $\dfrac{30.5 + 40.5}{2} = \dfrac{71}{2}$ = 35.5

Hence, class mark of fourth class = 35.5

Use the adjoining table to find:

(i) upper and lower limits of fifth class.

(ii) true class limits of the fifth class.

(iii) class boundaries of the third class.

(iv) class mark of the fourth class.

(v) width of sixth class.

Answer

(i) Fifth class : 48 - 52

Hence, upper limit = 52 and lower limit = 48.

(ii) Adjustment factor = (Lower limit of class - Upper limit of previous class) / 2

= $\dfrac{48 - 47}{2} = \dfrac{1}{2}$ = 0.5

True lower limit = 48 - 0.5 = 47.5

True upper limit = 52 + 0.5 = 52.5

Hence, true class limits of fifth class = 47.5 and 52.5

(iii) Third class : 38 - 42

True lower limit = 38 - 0.5 = 37.5

True upper limit = 42 + 0.5 = 42.5

Hence, class boundaries of third class = 37.5 and 42.5

(iv) Fourth class : 43 - 47

Class mark = $\dfrac{\text{Lower class limit + Upper class limit}}{2}$

= $\dfrac{43 + 47}{2} = \dfrac{90}{2}$

= 45.

Hence, class mark of fourth class = 45.

(v) Fifth class : 53 - 57

True lower limit = 53 - 0.5 = 52.5

True upper limit = 57 + 0.5 = 57.5

Hence, width = 5.

The marks of 200 students in a test were recorded as follows :

Draw the cumulative frequency table.

Answer

The cumulative frequency table is shown below :

Given below are the marks secured by 35 students in a test:

41, 32, 35, 21, 11, 47, 42, 00, 05, 18, 25, 24, 29, 38, 30, 04, 14, 24, 34, 44, 48, 33, 36, 38, 41, 46, 08, 34, 39, 11, 13, 27, 26, 43, 03.

Taking class intervals 0 - 10, 10 - 20, 20 - 30 ...., construct frequency as well as cumulative frequency distribution table. Find the number of students obtaining below 20 marks.

Answer

The cumulative frequency distribution table is given below:

From table,

The no. of students obtaining below 20 marks = 10.

The marks out of 100 of 50 students in a test are given below:

5 35 6 35 18 36 12 36 85 32

20 36 22 38 24 50 22 39 74 31

25 54 25 64 25 70 28 66 58 25

29 72 31 82 31 84 31 82 37 21

32 84 32 92 35 95 34 92 35 5

(i) Taking a class interval of size 10, construct a frequency as well as cumulative frequency table for the given data.

(ii) Which class has the largest frequency?

(iii) How many students score less than 40 marks?

(iv) How many students score first division (60% or more) marks?

Answer

(i) The cumulative frequency table for given data is :

(ii) From table,

Class 30 - 40 has the largest frequency distribution table.

(iii) From table,

34 students score less than 40 marks.

(iv) Total marks = 100

60% marks = $\dfrac{60}{100} \times 100$ = 60.

From tables, students scoring less than 60 = 37

Total students = 50.

Students scoring more than 60% = 50 - 37 = 13.

Hence, 13 students scored more than 60% marks.

Construct the frequency distribution table from the following data :

State the number of children in the age group 10 - 13.

Answer

The frequency distribution table for given data is :

From table,

Frequency of class 10 - 13 = 73.

Hence, the no. of children in the age group 10 - 13 are 73.

Rewrite the following cumulative frequency distribution into frequency distribution:

Less than or equal to 10 = 2

Less than or equal to 20 = 7

Less than or equal to 30 = 18

Less than or equal to 40 = 32

Less than or equal to 50 = 43

Less than or equal to 60 = 50

Answer

The frequency distribution table for given data is :

The maximum temperatures (in degree celsius) for Delhi for the month of April, 2014, as reported by the Meteorological Department, are given below:

27.4, 28.3, 23.9, 23.6, 25.4, 27.5, 28.1, 28.4, 30.5, 29.7, 30.6, 31.7, 32.2, 32.6, 33.4, 35.7, 36.1, 37.2, 38.4, 40.1, 40.2, 40.5, 41.1, 42.0, 42.1, 42.3, 42.4, 42.9, 43.1, 43.2.

Construct a frequency distribution table.

Answer

Here, maximum = 43.2 and minimum = 23.6.

Range = 43.2 - 23.6 = 19.6

Let us form 5 classes of each size 4.

Since, we want to score 43.2 in last class and 43.5 is the upper limit of the last class, so the lower limit of first class is 23.5

The frequency distribution table of the given data is as follows :

The class marks of a distribution are 94, 104, 114, 124, 134, 144 and 154. Determine the class size and the class limits of the fourth class.

Answer

Class size = Difference between two successive class marks = 104 - 94 = 10.

Class mark of fourth class = 124.

Lower limit of class = Class mark - (Class size/2)

= 124 - $\dfrac{10}{2}$

= 124 - 5

= 119.

Upper class limit = Class mark + (Class size/2)

= 124 + $\dfrac{10}{2}$

= 124 + 5

= 129.

Hence, class size = 10, lower limit = 119 and upper limit = 129.

The class marks of a distribution are 9.5, 16.5, 23.5, 30.5, 37.5 and 44.5. Determine the class size and the class limits of the third class.

Answer

Class size = Difference between two successive class marks = 16.5 - 9.5 = 7.

Class mark of third class = 23.5

Lower limit of class = Class mark - (Class size/2)

= 23.5 - $\dfrac{7}{2}$

= 23.5 - 3.5

= 20.

Upper class limit = Class mark + (Class size/2)

= 23.5 + $\dfrac{7}{2}$

= 23.5 + 3.5

= 27.

Hence, class size = 7, lower limit = 20 and upper limit = 27.