1. The yield of soyabean per acre in quintal in Mukund’s field for 7 years was 10, 7, 5, 3, 9, 6, 9.
   Find the mean of yield per acre.

Solution:

Here, N = 7

Now, Mean

\(\displaystyle \overline{x} = \frac{10+7+5+3+9+6+9}{7}\)

\(\displaystyle \therefore \overline{x} = \frac{49}{7}\)

\(\displaystyle \therefore \overline{x} = 7\)

∴ The mean yield of soyabean per acre in Mukund’s farm is 7 quintal.

2. Find the median of the observations: 59, 75, 68, 70, 74, 75, 80.

Solution:

Let’s write the given data in ascending order:

59, 68, 70, 74, 75, 75, 80

Here, N = 7 (Odd number)

∴ Median is the value of the middle score.

The middle score is 74.

∴ Median is 74.

3. The marks (out of 100) obtained by 7 students in Mathematics examination are given below. Find the mode for these marks:
   99, 100, 95, 100, 100, 60, 90

Solution:

Let’s write the given data in ascending order:

60, 90, 95, 99, 100, 100, 100

Here, the maximum frequency is of the score 100.

∴ The mode is 100 marks.

4. The monthly salaries in rupees of 30 workers in a factory are given below:

5000	7000	3000	4000	4000	3000	3000	3000	8000	4000
4000	9000	3000	5000	5000	4000	4000	3000	5000	5000
6000	8000	3000	3000	6000	7000	7000	6000	6000	4000

From the above data find the mean of monthly salary.

Solution:

Let’s prepare a table as shown below:

Solution:

Now, \(\displaystyle \overline{x} = \frac{\sum{f_i}{x_i}}{\sum{f_i}}\)

\(\displaystyle \therefore \overline{x} = \frac{147000}{30}\)

\(\displaystyle \therefore \overline{x} = 4900\)

∴ The mean of monthly salary is ₹ 4,900/-

5. In a basket, there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows:
   60, 70, 90, 95, 50, 65, 70, 80, 85, 95:
   Find the median of the weight of these tomatoes.

Solution:

Let’s write the given data in ascending order:

50, 60, 65, 70, 70, 80, 85, 90, 95, 95

Here, N = 10 (Even number)

∴ Median is the average of the two middle scores.

∴ Median is the average of 70 and 80.

\(\displaystyle \therefore Median = \frac{70+80}{2}\)

\(\displaystyle \therefore Median = \frac{150}{2}\)

\(\displaystyle \therefore Median = 75\)

∴ The median weight of tomatoes is 75 grams.

6. A hockey player has scored following number of goals in 9 matches:
   5, 4, 0, 2, 2, 4, 4, 3, 3.
   Find the mean, median and mode of the data.

Solution:

Here, N = 7

Now, Mean

\(\displaystyle \overline{x} = \frac{5+4+0+2+2+4+4+3+3}{9}\)

\(\displaystyle \therefore \overline{x} = \frac{27}{9}\)

\(\displaystyle \therefore \overline{x} = 3\)

∴ Mean = 3 ... (i)

For finding the median and mode, let’s arrange the data in ascending order:

0, 2, 2, 3, 3, 4, 4, 4, 5

Here, N = 9 (Odd number)

∴ Median is the value of the middle score (5^th observation).

The middle score is 3.

∴ Median = 3 ... (ii)

For finding the mode:

The maximum frequency is of the score 4.

∴ Mode = 4 ... (iii)

7. The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What was the correct mean?

Solution:

Here, N = 50 and \(\displaystyle \overline{x} = 80\)

Now, \(\displaystyle \overline{x} = \frac{\sum{x_i}}{N}\)

\(\displaystyle \therefore 80 = \frac{\sum{x_i}}{50}\)

\(\displaystyle \therefore 80 \times 50 = \sum{x_i}\)

\(\displaystyle \therefore 4000 = \sum{x_i}\)

i.e. \(\displaystyle \sum{x_i} = 4000\)

∴ The sum of first 50 observations is 4000.

Now, one observation 19, was taken as 91.

∴ The increase in the sum
= 91 − 19
= 72

∴ Corrected sum
= 4000 − 72
= 3928

∴ Corrected mean = \(\displaystyle \frac{3928}{50}\)

∴ The correct mean was 78.56

8. Following 10 observations are arranged in ascending order as follows:
   2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20
   If the median of the data is 11, find the value of x.

Solution:

Here, N = 10 (Even number)

∴ Median is the average of middle two scores.

∴ Median is the average of x + 1 and x + 3.

∴ \(\displaystyle 11 = \frac{x + 1 + x + 3}{2}\)

∴ \(\displaystyle 11 = \frac{2x + 4}{2}\)

∴ \(\displaystyle 11 \times 2 = 2x + 4\)

∴ \(\displaystyle 22 = 2x + 4\)

i.e. \(\displaystyle 2x + 4 = 22\)

∴ \(\displaystyle 2x = 22 - 4\)

∴ \(\displaystyle 2x = 18\)

∴ \(\displaystyle x = \frac{18}{2}\)

∴ \(\displaystyle x = 9\)

∴ The value of x is 9.

9. The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observation is 25. Find the 18th observation.

Solution:

The mean of 35 observations is 20.

∴ The sum of 35 observations
= 36 × 20
= 700 ... (i)

Also, the mean of first 18 observations is 15.

∴ The sum of first 18 observations
= 18 × 15
= 270 ... (ii)

And, the mean of last 18 observations is 25.

∴ The sum of last 18 observations
= 18 × 25
= 450 ... (iii)

∴ 18^th observation = The sum of last 18 observations − The sum of first 18 observations − The sum of 35 observations
= 270 − 450 + 700 ... [From (ii), (iii) and (i)]
= 20

∴ 18^th observation is 20.

10. The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.

Solution:

The mean of 5 observations is 50.

∴ The sum of 5 observations
= 5 × 50
= 250 ... (i)

Now, one observation was removed from the data, hence the mean became 45.

∴ The sum of remaining 4 observations
= 4 × 45
= 180 ... (ii)

From (i) and (ii),

The removed observation
= 250 − 180
= 70

∴ The observation which was removed is 70.

11. There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.

Solution:

No. of girls = 40 − 25 = 15

Now, the mean of marks obtained by 15 boys is 33.

∴ The sum of the marks obtained by 15 boys
= 15 × 33
= 495 ... (i)

Also, the mean of marks obtained by 25 girls is 35.

∴ The sum of the marks obtained by 25 girls
= 25 × 35
= 875 ... (ii)

∴ Total marke scored by 40 students
= 495 + 875
= 1370 ... (iii)

\(\displaystyle \therefore \overline{x} = 1370\) and N = 40

\(\displaystyle \therefore \overline{x} = \frac{1370}{40}\)

\(\displaystyle \therefore \overline{x} = 34.25\)

∴ The mean of the marks scored by all students in the class is 34.25.

12. The weights of 10 students (in kg) are given below:
    40, 35, 42, 43, 37, 35, 37, 37, 42, 37.
    Find the mode of the data.

Solution:

For finding the mode, let’s arrange the data in ascending order:

35, 35, 37, 37, 37, 37, 40, 42, 42, 43

Here, the maximum frequency is 4 and it is of 37 kg.

∴ The mode is 37 kg.

13. In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.

No. of Siblings	1	2	3	4
Families	15	25	5	5

Solution:

Here, the maximum frequency is 25 and it is of 2 siblings.

∴ The mode is 2.

14. Find the mode of the following data:

Marks	35	36	37	38	39	40
No. of Students	9	7	9	4	4	2

Solution:

Here, the maximum frequncey is 9 and it is of 35 and 37 marks.

∴ The mode is 35 marks and 37.