Mean Questions

In mathematics and statistics, the mean is one of the important concepts that can be understood easily without much prior knowledge. Different types of the mean are defined based on the data provided. In this article, you will get questions and solutions for finding the mean of a given data set and grouped data (frequency distribution). These questions will help the students of Classes 9 and 10 to get enough practice for this concept.

What is Mean in Statistics?

Mean is one of the measures of central tendency in statistics. The mean is the average of the given data set, which means it can be calculated by dividing the sum of the given data values by the total number of data values.

Mean for ungrouped data:

Mean (x̄) = ∑x_i/n

x_i = x₁, x₂, x₃,…, x_n such that i = 1, 2, 3,…n.

Number of observations = n

Mean for grouped data:

Mean (x̄) = ∑f_ix_i/ ∑f_i

Here, f_i’s are frequencies of x_i’s.

Mean Questions and Answers

1. Calculate the mean from the data showing marks of students in a class in a test: 40, 50, 55, 78, 58.

Solution:

Given marks:

40, 50, 55, 78, 58

Here, the number of data values = 5

We know that:

Mean = Sum of data values/Total number of data values

= (40 + 50 + 55 + 78 + 58)/5

= 281/5

= 56.2

Therefore, the mean for the given data is 56.2.

2. A class consists of 50 students, out of which 30 are girls. The mean of marks scored by girls in a test is 73 (out of 100), and that of boys is 71. Determine the mean score of the whole class.

Solution:

Given,

Total number of students in a class = 50

Number of girls in the class = 30

Number of boys in the class = 50 – 30 = 20

Mean marks scored by girls = 73

Mean marks scored by boys = 71

Thus, the total marks scored by girls = 73 × 30 = 2190

Also, the total marks scored by boys = 71 × 20 = 1420

Mean score of the class = (Total marks scored by girls and boys)/Total number of students

= (2190 + 1420)/50

= 3610/50

= 72.2

3. The mean of the following distribution is 50.

Find the value of a and hence the frequencies of 30 and 70.

Solution:

We know that,

Mean for grouped data (x̄) = ∑f_ix_i/n

(640a + 2800)/ (12a + 60) = 50 {given that mean = 50}

640a + 2800 = 50(12a + 60)

640a + 2800 = 600a + 3000

640a – 600a = 3000 – 2800

40a = 200

a = 200/40

a = 5

Therefore, the frequency for 30 = 5a + 3 = 5(5) + 3 = 28

And the frequency for 70 = 7a – 11 = 7(5) – 11 = 24

4. Find the mean salary of 60 workers of a factory from the following table:

Solution:

Mean (x̄) = ∑f_ix_i/ ∑f_i

Mean = (305000)/60 = 5083.33.

Therefore, the mean salary = Rs. 5083.33

5. A total of 25 patients admitted to a hospital are tested for levels of blood sugar, (mg/dl) and the results obtained were as follows:

87, 71, 83, 67, 85, 77, 69, 76, 65, 85, 85, 54, 70, 68, 80, 73, 78, 68, 85, 73, 81, 78, 81, 77, 75

Find the mean (mg/dl) of the above data.

Solution:

Sum of data values = 87 + 71 + 83 + 67 + 85 + 77 + 69 + 76 + 65 + 85 + 85 + 54 + 70 + 68 + 80 + 73 + 78 + 68 + 85 + 73 + 81 + 78 + 81 + 77 + 75

= 1891

Mean = 1891/25

= 75.64

6. Calculate the mean for the following distribution.

Solution:

Mean = a + (∑f_id_i/∑f_i)

= 47.5 + (435/30)

= 47.5 + 14.5

= 62

7. The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Solution:

The given distribution has unequal class heights. So, we can use the direct method for calculating the mean.

Mean = ∑f_ix_i/ ∑f_i

= 6880/45

= 152.89

Here the mean signifies that, on average, 45 bowlers take 152.89 wickets.

8. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

Solution:

Here, the values of fi and xi are small. Thus, we can use the direct method to calculate the mean.

So, Mean = ∑f_ix_i/ ∑f_i

= 162/20

= 8.1

Therefore, the mean number of plants = 8.1

9. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency x.

Solution:

Mean pocket allowance = a + (∑f_id_i/ ∑f_i)

18 + [(2x – 40)/(44 + x)] = 18 [from the given]

(2x – 40)/(44 + x) = 18 – 18

(2x – 40)/(44 + x) = 0

2x – 40 = 0

2x = 40

x = 20

Hence, the missing frequency is 20.

10. The following is the cumulative frequency distribution (of less than type) of 1000 persons, each one of age 20 years and above. Determine the mean age.

Solution:

For the given frequency distribution, we need to write the class intervals and frequencies and then proceed with the necessary calculations for estimating the mean age.

Mean = a + h(∑f_iu_i/∑f_i)

= 55 + 10(-370/1000)

= 55 – 3.7

= 51.3

Therefore, the mean age is 51.3 years.

Practice Questions on Mean

1. The points scored by a basketball team in a series of matches are as follows:
   
   17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28

Find the mean for the data.

2. 5 people were asked about the time in a week they spent doing social work in their community. They said 10, 7, 13, 20 and 15 hours, respectively. Find the mean (or average) time in a week devoted by them to social work.
3. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers	A	B	C	D	E	F	G	H	I	J
Daily income (in Rs)	120	150	180	200	250	300	220	350	370	260

4. The frequency distribution table of agricultural holdings in a village is given below.

Area of land (in hectares)	1 – 3	3 – 5	5 – 7	7 – 9	9 – 11	11 – 13
Number of families	20	45	80	55	40	12

Find the mean agricultural holdings of the village.

5. Calculate the mean for the following distribution.

CI	14 – 15	16 – 17	18 – 20	21 – 24	25 – 29	30 – 34	35 – 39
Frequency	60	140	150	110	110	100	90