Find the value of p, if the mean of the following distribution is 18.

Variate (x)	Frequency (f)
13	8
15	2
17	3
19	4
20 + p	5p
23	6

We construct the table as under :

$\text{Mean} = \dfrac{Σfx}{Σf} \\[1em] \therefore 18 = \dfrac{5p^2 + 100p + 399}{5p + 23} \\[1em] \Rightarrow 90p + 414= 5p^2 + 100p + 399 \\[1em] \Rightarrow 5p^2 + 100p - 90p + 399 - 414 = 0 \\[1em] \Rightarrow 5p^2 + 10p - 15 = 0 \\[1em] \Rightarrow 5p^2 + 15p - 5p - 15 = 0 \\[1em] \Rightarrow 5p(p + 3) - 5(p + 3) = 0 \\[1em] \Rightarrow (5p - 5)(p + 3) = 0 \\[1em]$

⇒ 5p - 5 = 0 or p + 3 = 0
⇒ 5p = 5 or p = -3
⇒ p = 1 or p = -3.

Since, frequency cannot be negative so p = 1.

Hence, the value of p = 1.