Statistics

Which of the following variables are discrete?

1. daily temperature of your city

2. sizes of shoes

3. distance travelled by a man

4. time

Answer

A discrete variable is a variable which is incapable of taking all possible numerical values.

From the given options, only shoe sizes are considered discrete variables, whereas daily temperature in your city, distance traveled by a person, and time are continuous variables.

Hence, option 2 is the correct option.

The marks obtained by 15 students in a test (out of hundred) are given below :

81, 72, 90, 90, 80, 55, 72, 66, 69, 80, 36, 54, 62, 56 and 58

The range of data is :

1. 46

2. 54

3. 90

4. 100

Answer

Range is defined as the difference between the highest and lowest values in the data.

From the given data, highest value = 90.

Lowest value = 36

Range = 90 - 36

= 54

Hence, option 2 is the correct option.

The class-mark of the class 35-45 is :

1. 35

2. 45

3. 40

4. 42

Answer

Class-mark is the value midway between its actual lower limit and actual upper limit.

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

Lower class limit = 35

Upper class limit = 45

$= \dfrac{35 + 45}{2}\\[1em] = \dfrac{80}{2}\\[1em] = 40$

Hence, option 3 is the correct option.

In the class-intervals 1-10, 11-20, 21-30; the class 11-20 after adjustment is :

1. 0.5-10.5

2. 20.5-30.5

3. 10.5-30.5

4. 10.5-20.5

Answer

The adjustment factor is $\dfrac{11-10}{2} = 0.5$

The adjusted class would then be as follows:

The class 11-20 after adjustment is 10.5-20.5.

Hence, option 4 is the correct option.

The class marks of a frequency distribution are 10, 15, 20, 25, ............... The class corresponding to the class mark 15 is :

1. 12.5-17.5

2. 10-20

3. 10.25-17

4. 14-16

Answer

The given class marks are 10, 15, 20, 25, ........

The class width is the difference between any two consecutive class marks.

Class width = 15 - 10 = 5

If a is the class mark of a class interval of width h, then the lower and upper limits of the class interval are:

Lower limit = a - $\dfrac{h}{2}$ upper limit = a + $\dfrac{h}{2}$

So, a = 15 and h = 5

Class interval = $\Big(15 - \dfrac{5}{2}\Big) to \Big(15 + \dfrac{5}{2}\Big)$

= (15 - 2.5) to (15 + 2.5)

= 12.5 to 17.5

Hence, the class corresponding to the class mark 15 is 12.5-17.5.

Hence, option 1 is the correct option.

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

(a) number of children in your class.

(b) distance travelled by a car.

(c) sizes of shoes.

(d) time.

(e) number of patients in a hospital.

Answer

(a) Discrete variable

Reason

A discrete variable is a variable which is incapable of taking all possible numerical values. In this case, the number of children is a finite, whole number.

(b) Continuous variable

Reason

Continuous variable is a variable which can take any numerical value within a certain range. The distance can be measured in infinitely small increments.

(c) Discrete variable

Reason

A discrete variable is a variable which is incapable of taking all possible numerical values. Sizes of shoes are represented by specific values.

(d) Continuous variable

Reason

Continuous variable is a variable which can take any numerical value within a certain range. Time can take on an infinite number of values within a range.

(e) Discrete variable

Reason

A discrete variable is a variable which is incapable of taking all possible numerical values. The number of patients is countable and can only be whole numbers

Given below are the marks obtained by 30 students in an examination :

08 17 33 41 47 23 20 34

09 18 42 14 30 19 29 11

36 48 40 24 22 02 16 21

15 32 47 44 33 01

Taking class intervals 1-10, 11-20, ..............., 41-50; make a frequency table for the above distribution.

Answer

The frequency table for the given distribution is :

The marks of 24 candidates in the subject mathematics are given below :

45 48 15 23 30 35 40 11

29 0 3 12 48 50 18 30

15 30 11 42 23 2 3 44

The maximum marks are 50. Make a frequency distribution taking class intervals 0-10, 10-20, ............... .

Answer

The frequency table for the given distribution is :

Fill in the blanks :

(a) A quantity which can vary from one individual to another is called a ...............

(b) Sizes of shoes are ............... variables.

(c) Daily temperature is ............... variable.

(d) The range of the data 7, 13, 6, 25, 18, 20, 16 is ...............

Range of the data is the difference between the highest and the lowest values.

(e) In the class interval 35-46; the lower limit is ............... and upper limit is ...............

(f) The class mark of class interval 22 - 29 is ...............

Answer

(i) A quantity which can vary from one individual to another is called a variable.

(ii) Sizes of shoes are distinct variables.

(iii) Daily temperature is continuous variable.

(iv) The range of the data 7, 13, 6, 25, 18, 20, 16 is (25-6) = 19.

Range of the data is the difference between the highest and the lowest values.

(v) In the class interval 35 - 46; the lower limit is 35 and upper limit is 46.

(vi) The class mark of class interval 22 - 29 is 25.5.

Find the actual lower class limits, upper class limits and the mid-values of the classes :

10-19, 20-29, 30-39 and 40-49.

Answer

In case of frequency 10-19 the lower class limit is 9.5, the upper class limit is 19.5 and the mid value is

$= \dfrac{9.5 + 19.5}{2} \\[1em] = \dfrac{29}{2} \\[1em] = 14.5$

In case of frequency 20-29 the lower class limit is 19.5, the upper class limit is 29.5 and the mid value is

$= \dfrac{19.5 + 29.5}{2} \\[1em] = \dfrac{49}{2} \\[1em] = 24.5$

In case of frequency 30-39 the lower class limit is 29.5, the upper class limit is 39.5 and the mid value is

$= \dfrac{29.5 + 39.5}{2} \\[1em] = \dfrac{69}{2} \\[1em] = 34.5$

In case of frequency 40-49 the lower class limit is 39.5, the upper class limit is 49.5 and the mid value is

$= \dfrac{39.5 + 49.5}{2} \\[1em] = \dfrac{89}{2} \\[1em] = 44.5$

Find the actual lower and upper class limits and also the class marks of the classes :

1.1-2.0, 2.1-3.0 and 3.1-4.0.

Answer

In case of frequency 1.1-2.0 the lower class limit is 1.05, the upper class limit is 2.05 and the class mark is

$= \dfrac{1.05 + 2.05}{2} \\[1em] = \dfrac{3.1}{2} \\[1em] = 1.55$

In case of frequency 2.1-3.0 the lower class limit is 2.05, the upper class limit is 3.05 and the class mark is

$= \dfrac{2.05 + 3.05}{2} \\[1em] = \dfrac{5.1}{2} \\[1em] = 2.55$

In case of frequency 3.1-4.0 the lower class limit is 3.05, the upper class limit is 4.05 and the class mark is

$= \dfrac{3.05 + 4.05}{2} \\[1em] = \dfrac{7.1}{2} \\[1em] = 3.55$

Use the table given below to find :

(a) The actual class limits of the fourth class.

(b) The class boundaries of the sixth class.

(c) The class mark of the third class.

(d) The upper and lower limits of the fifth class.

(e) The size of the third class.

Answer

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

(ii) The class boundaries of the sixth class is 54.5-59.5.

(iii) The class mark of the third class

$= \dfrac{40 + 44}{2}\\[1em] = \dfrac{84}{2}\\[1em] = 42$

(iv) The upper and lower limits of the fifth class (50-54) is 54 and 50, respectively.

(v) Interval of third class = 40-44

The size of class = Upper limit - lower limit + 1

The size of third class = 44 - 40 + 1

= 4 + 1

= 5

Hence, the size of third class is 5.

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

(i)

(ii)

Answer

(i) The cumulative frequency distribution table is:

(ii) The cumulative frequency distribution table is -

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

(i)

(ii)

Answer

(i) The frequency distribution table is

(ii) The frequency distribution table is

Construct a frequency polygon for the following distribution :

Answer

Steps:

1. Find the class-mark (mid-value) of each given class-interval.

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

2. On a graph paper, mark class-marks along x-axis and frequencies along y-axis.

3. On this graph paper, mark points taking values of class-marks along x-axis and the values of their corresponding frequencies along y-axis.

4. Draw line segments joining the consecutive points marked in step (3) above.

Construct a combined histogram and frequency polygon for the following frequency distribution :

Answer

Steps:

1. Draw a histogram for the given data.

2. Mark the mid-point at the top of each rectangle of the histogram drawn.

3. Also, mark the mid-point of the immediately lower class-interval (in the given question, the immediately lower class-interval is 0 - 10) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 60 - 70).

4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Construct a frequency polygon for the following data :

Answer

The given class interval are inclusive. For constructing polygon we will first convert them into the exclusive form.

Steps:

1. Find the class-mark (mid-value) of each given class-interval.

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

2. On a graph paper, mark class-marks along x-axis and frequencies along y-axis.

3. On this graph paper, mark points taking values of class-marks along x-axis and the values of their corresponding frequencies along y-axis.

4. Draw line segments joining the consecutive points marked in step (3) above.

The daily wages in a factory are distributed as follows :

Draw a frequency polygon for this distribution.

Answer

Steps:

1. Find the class-mark (mid-value) of each given class-interval.

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

2. On a graph paper, mark class-marks along x-axis and frequencies along y-axis.

3. On this graph paper, mark points taking values of class-marks along x-axis and the values of their corresponding frequencies along y-axis.

4. Draw line segments joining the consecutive points marked in step (3) above.

The class-mark of class 19 - 30 is:

1. $\dfrac{30 + 19}{2}$

2. $\dfrac{29.5 - 19.5}{2}$

3. $\dfrac{30.5 - 19}{2}$

4. $\dfrac{29.5 - 18.5}{2}$

Answer

By formula, class-mark = $\dfrac{\text{Lower limit + Upper limit}}{2}$

Substituting the values, we get :

Class-mark of class 19 - 30 = $\dfrac{19 + 30}{2}$

Hence, option 1 is the correct option.

In a frequency distribution, mid-values of the class is 10 and width of the class is 6, the lower limit of the class is :

1. 6

2. 7

3. 8

4. 9

Answer

Lower limit of class = Mid value - $\dfrac{\text{Width}}{2}$

= 10 - $\dfrac{6}{2}$

= 10 - 3

= 7.

Hence, option 2 is the correct option.

Statement 1: Let m be the mid-value and x be the upper limit of a class in the continuous frequency distribution, then the lower limit of this class is 2m - x.

Statement 2: For a given class: $\dfrac{\text{lower limit + upper limit}}{2}$ = mid-value of the class

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, mid-value = m

Upper limit of a class = x

As we know,

Mid-value = $\dfrac{\text{lower class-limit + upper class-limit}}{2}$

So, statement 2 is true.

⇒ m = $\dfrac{\text{lower class-limit} + x}{2}$

⇒ 2m = lower class-limit + x

⇒ Lower class-limit = 2m - x

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): 30 children were asked about the number of hours they watched TV program everyday. The results are recorded as under.

Then the number of children who watched TV for 10 or more hours a day is 22.

Reason (R): The assertion is not correct as the required number is 4 + 2 = 6.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

10 or more hours covers the 10 – 15 and 15 – 20 classes.

4 students are there in 10 - 15 class.

2 students are there in 15 - 20 class.

Total students = 4 + 2 = 6

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Construct a frequency table from the following data :

Answer`

The frequency table for the given distribution is :

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

(i) State the number of students in the age group 10-13.

(ii) State the age-group which has the least number of students.

Answer

The frequency table for the given distribution is :

(i) There are 103 students in the age group 10-13.

(ii) The age group 7-10 has the least number of students.

Fill in the blanks in the following table:

Answer

The value of π upto 50 decimal places is :

3.14159265358979323846264338327950288419716939937510

(i) Make a frequency distribution table of the digits from 0 to 9 after the decimal place.

(ii) Which are the most and the least occurring digits ?

Answer

(i) The frequency distribution table of the digits from 0 to 9 after the decimal place is :

(ii) From the frequency distribution table, the frequency of digit 9 and 3 is maximum (8) and the frequency of digit 0 is minimum (2).

Hence, the most occurring digits are 9 and 3 and the least occurring digit is 0.

Draw frequency polygons for the following frequency distribution :

(a) using histogram

(b) without using histogram.

Answer

(a) Using histogram

Steps:

1. Draw a histogram for the given data.

2. Mark the mid-point at the top of each rectangle of the histogram drawn.

3. Also, mark the mid-point of the immediately lower class-interval (in the given question, the immediately lower class-interval is -10 - 10) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 150 - 170).

4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

(b) Without using histogram

Steps:

1. Find the class-mark (mid-value) of each given class-interval.

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

2. On a graph paper, mark class-marks along x-axis and frequencies along y-axis

3. On this graph paper, mark points taking values of class-marks along x-axis and the values of their corresponding frequencies along y-axis.

4. Draw line segments joining the consecutive points marked in step (3) above.

Draw frequency polygons for the following frequency distribution :

(a) using histogram

(b) without using histogram.

Answer

(a) Using histogram

Steps:

1. Draw a histogram for the given data.

2. Mark the mid-point at the top of each rectangle of the histogram drawn.

3. Also, mark the mid-point of the immediately lower class-interval (in the given question, the immediately lower class-interval is 0 - 5) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 65 - 75).

4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

(b) Without using histogram

Steps:

1. Find the class-mark (mid-value) of each given class-interval.

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

2. On a graph paper, mark class-marks along x-axis and frequencies along y-axis

3. On this graph paper, mark points taking values of class-marks along x-axis and the values of their corresponding frequencies along y-axis.

4. Draw line segments joining the consecutive points marked in step (3) above.

Using the class intervals 0-9, 10-19, 20-29, ............... , construct the frequency distribution for :

15, 8, 12, 7, 13, 16, 22, 29, 35, 49, 37 and 48.

Answer

The frequency table for the given distribution is :

Construct the cumulative frequency table for :

Answer

The cumulative frequency table for the given distribution is :

Construct a combined histogram and frequency polygon for the following frequency distribution :

Answer

Steps:

1. Draw a histogram for the given data.

2. Mark the mid-point at the top of each rectangle of the histogram drawn.

3. Also, mark the mid-point of the immediately lower class-interval (in the given question, the immediately lower class-interval is 10 - 20) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 70 - 80).

4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.