Frequency Distribution

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.

If in a grouped frequency distribution, the classes are 20 - 30, 30 - 40, 40 - 50, ..., then the observation 40 is included in the class :

1. 30 - 40

2. 40 - 50

3. 50 - 60

4. 20 - 30

Answer

Since the given class-intervals are in exclusive form.

So, in exclusive form lower limit is included but the upper limit is excluded.

So 40 is included in the class-interval, 40-50.

Hence, option 2 is the correct option.

If in a grouped frequency distribution, the class intervals are 20 - 29, 30 - 39, 40 - 49, ...., then the observation 29.4 is included in the class :

1. 20 - 29

2. 30 - 39

3. 40 - 49

4. 29 - 30

Answer

Given class intervals are in inclusive form but the observation is in decimal form (29.4).

So convert the inclusive form into exclusive form.

Gap between the classes = 30 - 29 = 1

Adjustment factor = 0.5

So new class boundaries become

19.5 - 29.5, 29.5 - 39.5, 39.5 - 49.5, ...,

29.4 lies between the class-interval 19.5 - 29.5.

19.5 - 29.5 is the exclusive form of 20 - 29.

Hence, option 1 is the correct option.

The classes of a frequency distribution are 30 -34, 35 - 39, ..., 50 - 54. The lower boundary of the class 35 - 39 is :

1. 30.5

2. 35

3. 34.5

4. 39

Answer

Gap between the classes = 35 - 34 = 1

Adjustment factor = $\dfrac{1}{2}$ = 0.5

New class boundaries :

29.5 - 34.5, 34.5 - 39.5, ..., 49.5 - 54.5.

Thus lower limit of class 35 - 39 is 34.5

Hence, option 3 is the correct option.

In a grouped frequency distribution, the class interval are 40 - 49, 50 - 59, ..., 80 - 89. The upper boundary of the class (50 - 59) is :

1. 59

2. 59.5

3. 49.5

4. 60

Answer

Gap between the classes = 50 - 49 = 1

Adjustment factor = 0.5

Upper boundary for the class interval (50 - 59) = Upper limit + 0.5

= 59 + 0.5

= 59.5

Hence, option 2 is the correct option.

If 0 - 9, 10 - 19, 20 - 29, ...., are the classes of a grouped frequency distribution, then the width of each class is :

1. 9

2. 9.5

3. 10

4. 9 or 10

Answer

Gap between classes = 10 - 9 = 1

Adjustment factor = 0.5

So the new class boundaries become

-0.5 - 9.5, 9.5 - 19.5, 19.5 - 29.5, ...,

Class width = Upper boundary - Lower boundary

Considering second class :

Class width = 19.5 - 9.5 = 10

Similarly, for rest of the class intervals.

Hence, option 3 is the correct option.

If 10, 15, 20, ... are respectively the mid-value of the classes of a grouped frequency distribution, then the class whole mid-value is 25 is :

1. 20 - 27

2. 22.5 - 27.5

3. 21.5 - 28.5

4. 21 - 28

Answer

Difference between the mid values = 15-10 = 5

So, the class width = 5

For the class whose mid-value 25 is

Lower limit = 25 - $\dfrac{\text{Class width}}{2}$ = 22.5

Upper limit = 25 + $\dfrac{\text{Class width}}{2}$ = 27.5

So the Class is 22.5 - 27.5.

Hence, option 2 is the correct option.

The width of each class interval of a grouped frequency distribution is 8. The lower boundary of the class whose mid-value is 10 is :

1. 2

2. 4.5

3. 5.5

4. 6

Answer

Class width = 8

Half of class width = $\dfrac{8}{2}$ = 4

Mid-value = 10

So, Lower boundary = 10 - 4

= 6

Hence, option 4 is the correct option.

The width of each class of a frequency distribution is 5. The upper boundary of the class whose class mark is 25 is :

1. 22

2. 22.5

3. 27

4. 27.5

Answer

Class width = 5

Class mark = 25

Half of class width = $\dfrac{5}{2}$ = 2.5

Upper boundary = 25 + 2.5 = 27.5

Hence, option 4 is the correct option.

Class marks of a distribution are: 47, 52, 57, 62, 67, 72, 77, 82. Class size of the given distribution is :

1. 5

2. 6

3. 7

4. 2

Answer

Class size = Difference between the consecutive class marks.

Class size = 52 - 47 = 5

57 - 52 = 5

Therefore class size = 5

Hence, option 1 is the correct option.

Mid value for the class interval 19.5 - 29.5 is :

1. 14.5

2. 24.5

3. 15.5

4. 25.5

Answer

Class interval = 19.5-29.5

Mid value = $\dfrac{\text{Lower limit} + \text{Upper limit}}{2}$

= $\dfrac{19.5 + 29.5}{2}$

= $\dfrac{49.5}{2}$

= 24.5

Hence, option 2 is the correct option.

Case Study

To maintain the health report card in a hospital, the weights of new born babies are recorded on regular basis.

During a week, the weights (in kg) of new born were recorded as below :

2.7, 2.6, 3.1, 2.2, 2.4, 2.9, 3.0, 2.5, 2.8, 2.5, 2.8, 2.9, 2.2, 2.9, 2.7, 2.8, 2.3, 2.8, 2.1, 2.2

Based on the above information, answer the following questions:

1. Range of the data is :
   (a) 1 kg
   (b) 0.4 kg
   (c) 0.5 kg
   (d) 1.1 kg

2. How many babies are born during the week?
   (a) 10
   (b) 16
   (c) 17
   (d) 20

3. How many babies weigh less than 2.4 kg?
   (a) 7
   (b) 6
   (c) 5
   (d) 4

4. How many babies weigh more than 2.7 kg?
   (a) 10
   (b) 9
   (c) 8
   (d) 7

5. How many babies weigh 2.8 kg?
   (a) 4
   (b) 5
   (c) 6
   (d) 3

Answer

1. Range = Highest value - Lowest value

= 3.1 - 2.1

= 1 kg

Hence, option (a) is the correct option.

2. Total babies born during the week = 20

Hence, option (d) is the correct option.

3. Babies weighing less than 2.4 kg are

2.2, 2.2, 2.3, 2.1, 2.2

Total = 5

Hence, option (c) is the correct option.

4. Babies weighing more than 2.7 kg are

2.8, 2.9, 3.0, 2.8, 2.9, 2.9, 2.8, 2.8, 3.1

Total = 9

Hence, option (b) is the correct option.

5. From data,

Babies weighing 2.8 kg = 4.

Hence, option (a) is the correct answer

Case Study

The following data gives the weights (in grams) of some marbles picked up randomly from 3 boxes. The data was presented in the tabular form as below:

Based on the above information, answer the following questions:

1. Class size of the given distribution is :
   (a) 5
   (b) 10
   (c) 50
   (d) 85

2. Class mark of the 3rd class is :
   (a) 60
   (b) 65
   (c) 125
   (d) 62.5

3. The number of marbles whose weight is more than or equal to 75 g is :
   (a) 0
   (b) 1
   (c) 2
   (d) 28

4. The number of marbles whose weight is less than 70 g is :
   (a) 26
   (b) 28
   (c) 4
   (d) 2

5. The class boundaries of the class 7th class interval are :
   (a) 80, 85
   (b) 80.5, 85.5
   (c) 79.5, 85.5
   (d) 80, 85.5

Answer

1. Class size = Upper limit - lower limit

= 55 - 50

= 5.

Hence, option (a) is the correct option.

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

= $\dfrac{60 + 65}{2}$

= $\dfrac{125}{2}$

= 62.5

Hence, option (d) is the correct option.

3. Marbles weighing more than or equal to 75 g are

75-80 ⟶ 1

80-85 ⟶ 1

Total = 1 + 1 = 2

Hence, option (c) is the correct option.

4. Marbles weighing less than 70 g are

50-55 ⟶ 5

55-60 ⟶ 7

60-65 ⟶ 8

65-70 ⟶ 6

Total = 5 + 7 + 8 + 6 = 26

Hence, option (a) is the correct option.

5. Class interval of 7th class = 80 - 85

Hence, option (a) is the correct option.

Assertion (A) : Classes of the form : 1-10, 11-20, 21-30 etc. are said to be in inclusive form.

Reason (R) : A frequency distribution in which each upper limit as well as lower limit is included is called an inclusive form.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given class intervals 1-10, 11-20, 21-30 etc shows that gap between the upper limit of one class (10) and the lower limit of the next (11).

Since gap shows that these classes don't overlap, thus they are in inclusive form.

∴ Assertion (A) is true.

We know that,

A frequency distribution in which each upper limit as well as lower limit is included is known as inclusive form.

∴ Reason (R) is true.

Both Assertion and Reason are True.

Hence, option 3 is the correct option.

Assertion (A) : In the exclusive form, the upper and lower limits of a class are respectively known as the true upper limit and true lower limit.

Reason (R) : Class size = $\dfrac{\text{ True lower limit} + \text{ True upper limit}}{2}$

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

In the exclusive form (e.g., 10 - 20, 20 - 3)there are no gaps between the classes. Because the classes are continuous, therefore, these limits are indeed the true lower limit and true upper limit.

∴ Assertion (A) is true.

We know that,

Class size = Upper limit - Lower limit

∴ Reason (R) is false.

Hence, option 1 is the correct option.

Age of teachers of a school varies from 24 years to 72 years. Rishabh has to make a grouped frequency distribution table of age of teachers. Which among the following class intervals of age will be most appropriate to make meaningful conclusion in less time and effort?

1. 24 - 26, 26 - 28, 28 - 30, and so on

2. 24 - 30, 30 - 36, 36 - 42, and so on

3. 24 - 48, 48 - 72

4. none of these

Answer

Age of teachers varies from 24 years to 72 years

So, Range = 72 - 24 = 48

By checking the option (ii)

Class size = 6

Total classes = $\dfrac{48}{6}$

= 8 classes

∴ 8 class intervals are suitable to check the age of teachers.

Hence, option 2 is the correct option.

The distribution of marks of students of a class in a test is shown in the table below.

Which of the following statements are correct?

(i) We can definitely say that no one scored 60 marks.

(ii) The number of students who got at least 30 marks is 12.

1. only (i)

2. only (ii)

3. both (i) and (ii)

4. none of these

Answer

(i) We know that given table is in exclusive form. So in exclusive form generally the upper limit of class is excluded from that group and moved to next one.

However, last class interval of the given table is 45 - 60.

So, we can strongly say that no one scored 60 marks.

∴ Statement (i) is correct.

(ii) The number of students who got at least 30 marks are

30 – 45 ⟶ 8 students

45 – 60 ⟶ 4 students

Total = 8 + 4 = 12.

∴ Statement (ii) is Correct.

Hence, option 3 is the correct option.

Range of 14, 12, 17, 18, 16 and p is 20. If p>0, then p= ?

1. 10

2. 26

3. 32

4. 35

Answer

Given numbers 14, 12, 17, 18, 16 and p

Maximum Value = 18

Minimum value = 12

Range = Maximum value - Minimum value = 18 - 12 = 6 (excluding p)

But the range given is 20, So p must be either maximum value or minimum value

If p is the maximum value, then

p - 12 = 20

P = 32

If p is the minimum value, then

18 - p = 20

p = -2

Since p > 0

∴ p = 32

Hence, option 3 is the correct option.

The class mark of a class-interval is 'a'. If its lower limit is a-b, then its upper limit is :

1. a + b

2. ab

3. b-a

4. 2b

Answer

Class mark = a

Lower limit = a − b

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

$a = \dfrac{(a - b) + \text{Upper limit}}{2}$

2a = (a - b) + Upper limit

Upper limit = 2a − (a − b)

= 2a - a + b

= a + b.

Hence, option 1 is the correct option.