ICSE Class 9 Maths Chapter 19 Mean and Median

Mean and Median

For Ungrouped Data Only

Mean Of Ungrouped Data

If \(x_1, x_2, x_3, x_4, \ldots, x_n\) are n observations in an ungrouped data, then their mean, in general denoted by \(\bar{x}\), is given by

\[\bar{x} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} = \frac{1}{n}\sum_{i=1}^{n} x_i\]

The symbol \(\sum_{i=1}^{n} x_i = x_1 + x_2 + x_3 + x_4 + \ldots + x_n\)

Mean, arithmetic mean and average are the same.

To find the mean of any set of observations, divide their sum by the total number of observations.

Example 1

The heights of 6 boys are 146 cm, 154 cm, 153 cm, 160 cm, 157 cm and 160 cm. Find their mean height.

Solution

Mean height = \[\frac{\text{Sum of heights of all the boys}}{\text{Number of boys}}\]

\[\bar{x} = \frac{146 + 154 + 153 + 160 + 157 + 160}{6} \text{ cm} = 155 \text{ cm}\]

Alternative method

\(\sum_{i=1}^{6} x_i = 146 \text{ cm} + 154 \text{ cm} + 153 \text{ cm} + 160 \text{ cm} + 157 \text{ cm} + 160 \text{ cm} = 930 \text{ cm}\)

and, \(n = 6\)

Mean \((\bar{x}) = \frac{1}{n}\sum_{i=1}^{6} x_i = \frac{1}{6} \times 930 \text{ cm} = 155 \text{ cm}\)

\(\sum_{i=1}^{n} x_i = x_1 + x_2 + x_3 + \ldots + x_n\)

\(\sum_{i=1}^{6} x_i = x_1 + x_2 + x_3 + x_4 + x_5 + x_6\) and so on.

Example 2

Find the mean of all prime numbers between 20 and 50.

Solution

Prime numbers between 20 and 50 are 23, 29, 31, 37, 41, 43 and 47.

Mean = \[\frac{23 + 29 + 31 + 37 + 41 + 43 + 47}{7} = \frac{251}{7} = 35\frac{6}{7}\]

Alternative method

\(\Sigma x = 23 + 29 + 31 + 37 + 41 + 43 + 47 = 251\)

and \(n = 7\)

Mean \((\bar{x}) = \frac{251}{7} = 35\frac{6}{7}\) cm

For \(\sum_{i=1}^{n} x_i\), we can simply write \(\Sigma x\).

Teacher's Note

Understanding how to calculate the mean helps in everyday situations like determining average test scores, daily temperatures, or typical spending patterns over a week.

Properties Of Mean

Property 1

If \(\bar{x}\) is the mean of n number of observations \(x_1, x_2, x_3, x_4, \ldots, x_n\); then the sum of deviations of \(\bar{x}\) from the observations \(x_1, x_2, x_3, x_4, \ldots, x_n\), is zero i.e., \(\Sigma(x - \bar{x}) = 0\).

Example 3

(i) Find the mean \((\bar{x})\) of first 5 even natural numbers.

(ii) If x denotes the different even natural numbers, taken in part (i) above, show that \(\Sigma(x - \bar{x}) = 0\).

Solution

(i) Different values of x are: 2, 4, 6, 8 and 10

\(\Sigma x = 2 + 4 + 6 + 8 + 10 = 30\) and \(n = 5\)

Mean \((\bar{x}) = \frac{\Sigma x}{n} = \frac{30}{5} = 6\)

(ii)

\(\Sigma(x - \bar{x})\) = -4 - 2 + 0 + 2 + 4 = 0

Example 4

Find the sum of the deviations of the data 4, 5, 7, 9 and 15 from their mean.

Solution

Mean of given data = \[\frac{4 + 5 + 7 + 9 + 15}{5} = 8\]

Sum of deviations of the given data from their mean

= (4 - 8) + (5 - 8) + (7 - 8) + (9 - 8) + (15 - 8)

= -4 - 3 - 1 + 1 + 7 = 0

Teacher's Note

The property that deviations sum to zero explains why the mean is a point of balance - it's like the center point that equally distributes differences above and below it.

Property 2

If \(\bar{x}\) is the mean of n number of observations \(x_1, x_2, x_3, \ldots, x_n\), then the mean of observations \(x_1 + a, x_2 + a, x_3 + a, \ldots, x_n + a\) is \(\bar{x} + a\).

i.e., if each observation under consideration is increased by quantity a, then their mean is also increased by the same quantity a.

Property 3

If \(\bar{x}\) is the mean of n number of observations \(x_1, x_2, x_3, \ldots, x_n\), then the mean of observations \(x_1 - a, x_2 - a, x_3 - a, \ldots, x_n - a\) is \(\bar{x} - a\).

i.e., if each observation under consideration is decreased by quantity a, then their mean is also decreased by the same quantity a.

Property 4

If \(\bar{x}\) is the mean of n number of observations \(x_1, x_2, x_3, \ldots, x_n\), then mean of \(ax_1, ax_2, ax_3 \ldots, ax_n\) is \(a\bar{x}\).

i.e., if each observation under consideration is multiplied by quantity a, then mean is also multiplied by the same quantity a.

Property 5

If \(\bar{x}\) is the mean of n number of observations \(x_1, x_2, x_3, \ldots, x_n\), then mean of \(\frac{x_1}{a}, \frac{x_2}{a}, \frac{x_3}{a}, \ldots, \frac{x_n}{a}\) is \(\frac{\bar{x}}{a}\).

i.e., if each observation under consideration is divided by quantity a, the mean is also divided by the same quantity a.

Example 5

(a) Find the mean of 5, 6, 17, 8, 9, 15, 23, 18, 10 and 24.

(b) Find the resulting mean, if each observation, given above, is:

(i) increased by 3.

(ii) decreased by 2.

(iii) multiplied by 4.

(iv) divided by 5.

Solution

(a) Mean = \[\frac{5 + 6 + 17 + 8 + 9 + 15 + 23 + 18 + 10 + 24}{10} = \frac{135}{10} = 13.5\]

(b) (i) According to the property 2, resulting mean = 13.5 + 3 = 16.5

(ii) According to the property 3, resulting mean = 13.5 - 2 = 11.5

(iii) According to the property 4, resulting mean = 13.5 \(\times\) 4 = 54

(iv) According to the property 5, resulting mean = \(\frac{13.5}{5}\) = 2.7

Teacher's Note

These properties help calculate adjusted means quickly - like figuring out new average prices after a discount or new test scores after a curve adjustment.