Measures of Central Tendency (Mean) — Analytical & Application Based Questions

The class mark and frequency of a data is given in the graph. From the graph, Find:

(a) the table showing the class interval and frequency.

(b) the mean

Answer

(a) Class marks : 13, 15, 17, 19, 21, 23.

Difference between two consecutive class marks = 15 - 13 = 2.

Lower limit of class = Class mark - 1

Upper limit of class = Class mark + 1

(b)

Mean = $\dfrac{Σfx}{Σf} = \dfrac{504}{28}$ = 18.

Hence, mean = 18.

The mean of 5, 7, 8, 4 and n is o and the mean of 5, 7, 8, 4, n and o is n. Find the values of n and o.

Answer

Given,

Mean of 5, 7, 8, 4 and n is o.

By formula,

$\Rightarrow \text{Mean} = \dfrac{\sum x_i}{n} \\[1em] \Rightarrow o = \dfrac{5 + 7 + 8 + 4 + n}{5} \\[1em] \Rightarrow 5o = 24 + n \\[1em] \Rightarrow o = \dfrac{24 + n}{5} \text{ .....(1)}$

Mean of 5, 7, 8, 4, n and o is n.

$\Rightarrow n = \dfrac{5 + 7 + 8 + 4 + n + o}{6} \\[1em] \Rightarrow 6n = 24 + n + o \\[1em] \Rightarrow 5n = 24 + o \text{ .....(2)}$

From equation (1), o = $\dfrac{24 + n}{5}$.

Substituting in equation (2), we get:

$\Rightarrow 5n = 24 + \dfrac{24 + n}{5} \\[1em] \Rightarrow 5n = \dfrac{120 + 24 + n}{5} \\[1em] \Rightarrow 25n = 120 + 24 + n \\[1em] \Rightarrow 24n = 144 \\[1em] \Rightarrow n = 6.$

Substituting n = 6 in equation (1), we get:

$\Rightarrow o = \dfrac{24 + 6}{5} \\[1em] \Rightarrow o = \dfrac{30}{5} = 6.$

Hence, n = 6 and o = 6.