Statistics — Test Yourself

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 each of 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 each of 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.