Problems on Mean of Ungrouped Data

Here we will learn how to solve the different types of problems on mean of ungrouped data.

1. (i) Find the mean of 6, 10, 0, 7, 9.

(ii) Find the mean of the first four odd natural numbers.

Solution:

(i) We know that the mean of five variates x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), x\(_{5}\) is given by

A = \(\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5}\)

   = \(\frac{6 + 10 + 0 + 7 + 9}{5}\)

   = \(\frac{32}{5}\)

   = 6.4

(ii) The first four odd natural numbers are 1, 3, 5, 7.

Therefore, mean A = \(\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4}\)

= \(\frac{1 + 3 + 5 + 7}{4}\)

= \(\frac{16}{4}\)

= 4.


2. Find the mean of the following data:

                10, 15, 12, 16, 15, 10, 14, 15, 12, 10.

Solution:

There are ten variates. So,

mean = A = \(\frac{10 + 15 + 12 + 16 + 15 + 10 + 14 + 15 + 12 + 10}{10}\)

                 = \(\frac{129}{10}\)

                = 12.9

Alternatively,

As variates are repeated in the collection, we take note of their frequencies.

Variate

(x\(_{1}\))

10

12

14

15

16

Total

Frequency

(f\(_{1}\))

3

2

1

3

1

10

Therefore, mean = A = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3}  + x_{4}f_{4} + x_{5}f_{5}}{ f_{1} +  f_{2} + f_{3} + f_{4} + f_{5}}\)

                             = \(\frac{10 × 3 + 12 × 2 + 14 × 1 + 15 × 3 + 16 × 1}{3 + 2 + 1 + 3 + 1}\)

= \(\frac{30 + 24 + 14 + 45 + 16}{10}\)

= \(\frac{129}{10}\)

= 12.9


3. The mean age of five boys is 16 years. If the ages of four of them be 15 years, 18 years, 14 years and 19 years then find the age of the fifth boy.

Solution:

Let the age of the fifth boy be x years.

Then the mean age of the five boys = \(\frac{15 + 18 + 14 + 19 + x}{5}\) years.

Therefore, from the question, 16 = \(\frac{15 + 18 + 14 + 19 + x}{5}\)

⟹ 80 = 66 + x

Therefore, x = 80 – 66

x = 14.

Therefore, the age of the fifth boy is 14 years.


4. The mean of five data is 10. If a new variate is included, the mean of the six data becomes 11. Find the sixth data.

Solution:

Let the first five data be x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), x\(_{5}\) and the sixth data be x\(_{6}\).

The mean of the first five data = \(\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5}\)

From the question, 10 = \(\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{6}\)

Therefore, x\(_{1}\) + x\(_{2}\) + x\(_{3}\) + x\(_{4}\) + x\(_{5}\) = 50 ........................ (i)

Again, from the question, 11 = \(\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{6}\)

Therefore, x\(_{1}\) + x\(_{2}\) + x\(_{3}\) + x\(_{4}\) + x\(_{5}\) + x\(_{6}\) = 66

Therefore, 50 + x\(_{6}\) = 66, [Using the equation (i)]

Therefore, x\(_{6}\) = 66 - 50

                x\(_{6}\) = 16

Therefore, sixth data is 16.

`





9th Grade Math

From Problems on Mean of Ungrouped Data to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.