Frequency Distribution — Exercise 15

Define statistics as a subject.

Answer

Statistics is the science which deals with the collection, presentation, analysis and interpretation of numerical data.

What are primary data and secondary data ? Which of the two are more reliable and why?

Answer

Primary Data : The data collected by the investigator himself with a definite plan in mind are known as primary data.

Secondary Data : The data collected by someone, other than the investigator are known as secondary data.

Primary data is more reliable than secondary data because primary data collected by the user himself.

Fill in the blanks

(i) The difference between the maximum and minimum observations in a data is called the ............ of the data.

(ii) The number of observations in a class-interval is called the ............ of the interval.

(iii) The mid-point of a class-interval is called the ............ of the interval.

(iv) Lower-limit of the class-interval 24-30 is ............

(v) Upper limit of the class-interval 16-20 is ............

(vi) The class-mark of the class-interval is 20-30 is ............

(vii) The class-mark of the class-interval 9.5-19.5 is ............

Answer

(i) Range

(ii) Frequency

(iii) Class mark

(iv) 24

(v) 20

(vi) 25

(vii) 14.5

Find the range of the data

(a) 5, 7, 16, 21, 8, 10

(b) 11, 13, 17, 14, 19, 14, 15, 18

Answer

(a) By formula,

Range = Highest value - lowest value

Range = 21 - 5 = 16

Hence, the range of the data = 16.

(b) By formula,

Range = Highest value - Lowest value

Range = 19 - 11 = 8

Hence, the range of the data = 8.

The class marks of a frequency distribution are 28, 34, 40, 46, 52. Find the class-size and all the class intervals.

Answer

By formula,

Class Size = Difference between two consecutive class marks

= 34 - 28 = 6.

Lower limit = Class mark - $\dfrac{\text{Class size}}{2}$

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

Class intervals for class mark 28 :

Lower Limit = 28 - 3 = 25

Upper limit = 28 + 3 = 31

Class = 25 - 31

Class intervals for class mark 34 :

Lower Limit = 34 - 3 = 31

Upper limit = 34 + 3 = 37

Class = 31 - 37

Class intervals for class mark 40 :

Lower Limit = 40 - 3 = 37

Upper limit = 40 + 3 = 43

Class = 37 - 43

Class intervals for class mark 46 :

Lower Limit = 46 - 3 = 43

Upper limit = 46 + 3 = 49

Class = 43 - 49

Class intervals for class mark 52 :

Lower Limit = 52 - 3 = 49

Upper limit = 52 + 3 = 55

Class = 49 - 55.

Hence, class size = 6 and class intervals : 25 - 31, 31 - 37, 37 - 43, 43 - 49, 49 - 55.

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

(i) Marks obtained by the students of a class in a test.

(ii) Daily maximum temperature of a city.

(iii) I.Q. of a students of a class.

(iv) Weights of players of a Volley-ball team.

(v) Number of car-accidents in a city.

(vi) Distance travelled by a train.

(vii) Time taken by runners in a race.

(viii) Sizes of shoes sold in a shoe-store.

(ix) Number of patients in a hospital per day.

Answer

(i) Marks obtained by the students in a test will always be in whole number or natural number (e.g : 17, 20, ... ) hence, it is discrete variable.

(ii) Temperature can be in any value (e.g : 27⁰ C, 21.5⁰ C). So, it is continuous variable.

(iii) I.Q. of a person can be any number. Hence it is a continuous variable.

(iv) Weight can be measured in any value (e.g : 47.5 kg, 98 kg). Hence it is a continuous variable.

(v) Accidents will always be in whole number (e.g : 2, 5, 15). We cannot say total accidents are 2.5, 7.8. Hence, it is a discrete variable.

(vi) Distance can be of any value (like 78 km, 117.95 km). Hence, it is a continuous variable.

(vii) Time can be measured in any value (like 18.7 sec, 34.7 min). Hence, it is a continuous variable.

(viii) Shoes size always be in whole number or natural number(like 7, 9, 10). We cannot have the shoes size of 8.6, 9.3. Hence, it is a discrete variable.

(ix) Hospitals will count the number of patietns as whole number or natural number(like 12 patients, 19 patients). We cannot say there are 15.6 patients in the hospital. Hence, it is a discrete variable.

Define the following terms:

(i) Variable

(ii) Class-interval

(iii) Class-size

(iv) Class-mark

(v) Class-limits

(vi) True class-limits

(Vii) Frequency of a class

(viii) Cumulative frequency of a class

Answer

(i) Variable : A quantity which can take different values is called a variable.

(ii) Class-interval : A range of values into which data is grouped in a frequency distribution.

(iii) Class-size : It is the difference between the true upper limit and the true lower limit of class.

(iv) Class-mark : It is the midpoint( middle value) of a class-interval.

(v) Class-limits : These are the smallest and largest values of the class interval.

(vi) True class-limits : These are the actual limits of a class interval after removing the gap between two consecutive classes.

(vii) Frequency of a class : It is the number of observations that fall within that class-interval.

(viii) Cumulative frequency of a class : It is the sum of the frequencies of all the previous classes and that particular class.

Following data gives the number of children in 40 families :

1, 2, 6, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 1, 6, 2, 2

Represent it in the form of a frequency distribution.

Answer

Frequency distribution table :

The marks obtained by 40 students of a class in an examination are given below. Present the data in the form of a frequency distribution using equal class-size, one such class being 10 - 15(15 not included).

3, 20, 13, 1, 21, 13, 3, 23, 16, 13, 18, 12, 5, 12, 5, 24, 9, 2, 7, 18, 20, 3, 10, 12, 7, 18, 2, 5, 7, 10, 16, 8, 16, 17, 8, 23, 21, 6, 23, 15

Answer

Since the given class is 10 - 15 (15 not included)

so, the class size = 5

Since, the maximum marks of the student equals to 24, thus the class intervals will be :

0 - 5, 5 - 10, 10 - 15, 15 - 20, 20 - 25

Frequency distribution table :

Construct a frequency table for the following ages (in years) of 30 students using equal class-intervals, one of them being 9-12, where 12 is not included.

18, 12, 7, 6, 11, 15, 21, 9, 8, 13, 15, 17, 22, 19, 14, 21, 23, 8, 12, 17, 15, 6, 18, 23, 22, 16, 9, 21, 11, 16

Answer

Given class is 9 - 12 (12 is not included)

so class size = 12 - 9 = 3

Minimum age = 6

Maximum age = 23

So the class intervals will be

6 - 9, 9 - 12, 12 - 15, 15 - 18, 18 - 21, 21 - 24

Frequency distribution table :

The weekly wages (in rupees) of 30 workers in a factory given below :

630, 635, 690, 610, 635, 636, 639, 645, 698, 690, 620, 660, 632, 633, 655, 645, 604, 608, 612, 640, 685, 635, 636, 678, 640, 668, 690, 606, 640, 690

Represent the data in the form of a frequency distribution with class size 10.

Answer

Minimum age = 604

Maximum age = 698

Given class size = 10

so we can take the class intervals as

600 - 610, 610 - 620, 620 - 630, 630 - 640, 640 - 650, 650 - 660, 660 - 670, 670 - 680, 680 - 690, 690 - 700

Frequency distribution table :

The weights in grams of 50 apples picked at random from a consignment are as follows :

131, 113, 82, 75, 204, 81, 84, 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90, 115, 110, 98, 106, 99, 107, 84, 76, 186, 82, 100, 109, 128, 115, 107, 115, 119, 93, 187, 139, 129, 130, 68, 195, 123, 125, 111, 92, 86, 70, 126

Form the grouped frequency table by dividing the variable range into intervals of equal width of 20 g.

Answer

Minimum weight = 68 g

Maximum weight = 204 g

Given class width = 20 g

So we can take the class intervals as

60 - 80, 80 - 100, 100 - 120, 120 - 140, 140 - 160, 160 - 180, 180 - 200, 200 - 220

Frequency distribution table :

The marks obtained by 35 students in an examination are given below :

370, 290, 318, 175, 170, 410, 378, 405, 380, 375, 315, 305, 325, 275, 241, 288, 261, 355, 402, 380, 178, 253, 428, 240, 210, 175, 154, 405, 380, 370, 306, 460, 328, 440, 425

Form cumulative frequency table with class intervals of length 50.

Answer

Minimum mark = 154

Maximum mark = 460

Given class length = 50

So we can take class intervals as

150 - 200, 200 - 250, 250 - 300, 300 - 350, 350 - 400, 400 - 450, 450 - 500,

Frequency distribution table :

Construct the cumulative frequency table from the frequency table given below:

Answer

Cumulative frequency table :

Construct a frequency distribution table from the following cumulative frequency distribution:

Answer

Frequency = Current C.F - Previous C.F

Frequency distribution table :

Construct a frequency table from the following data :

Answer

Frequency distribution table :

Convert the following frequency distribution to exclusive form :

Use this table to find :

(i) The true class-limits of the fourth class-interval.

(ii) The class-boundaries of the fifth class-interval.

(iii) The class-mark of the third class-interval.

(iv) The class-size of the sixth class-interval.

Answer

Adjustment factor

= $\dfrac{\text{Lower limit of one class - Upper limit of previous class}}{2} = \dfrac{35 - 34}{2} = \dfrac{1}{2}$ = 0.5

Subtract the lower limit of each class by 0.5 and add 0.5 to upper limit of each class.

(i) The true class limits of the fourth class interval is 44.5 - 49.5.

(ii) The class-boundaries of the fifth class interval is 49.5 - 54.5.

(iii) The class-mark of the third class interval

= $\dfrac{39.5 + 44.5}{2}$

= $\dfrac{84}{2}$

= 42.

Hence, the class mark of third class interval = 42.

(iv) Sixth class interval : 54.5 - 59.5

Class size = Upper limit - lower limit

Class size = 59.5 - 54.5 = 5.

Hence, the class size of sixth class interval is 5.