The arithmetic mean of the following frequency distribution is 25.25. Determine the value of p:
Class
0-10
10-20
20-30
30-40
40-50
Frequency
7
8
p
15
4

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Solution

Class Interval $f_{i}$ Class Mark $x_{i}$ $x_{i} F_{i}$
0-10 7 5 35
10-20 8 15 120
20-30 P 25 25p
30-40 15 35 525
40-50 4 45 180
$Σ f_{i} = 34 + p$ $Σ x_{i} f_{i} = 860 + 25 p$

$F o r g r o u p e d d a t a w h e n C l a s s I n t e r v a l i s g i v e n, w e f i n d t h e C l a s s M a r k . C l a s s M a r k = m i d p o i n t o f a n i n t e r v a l = \frac{(L o w e r l i m i t + U p p e r l i m i t)}{2} C l a s s M a r k i s t a k e n a s x_{i} f o r e a c h c l a s s i n t e r v a l . F i n d x_{i} f_{i} . F r o m t h e t a b l e, w e h a v e Σ x_{i} f_{i} = 860 + 25 p & Σ f_{i} = 34 + p G i v e n M e a n = 25.25 M e a n = \frac{Σ x_{i} f_{i}}{Σ f_{i}} \Rightarrow 25.25 = \frac{860 + 25 p}{34 + p} \Rightarrow 860 + 25 p = (34 + p) \times 25.25 \Rightarrow 860 + 25 p = 858.5 + 25.25 p \Rightarrow 25.25 p - 25 p = 860 - 858.5 \Rightarrow 0.25 p = 1.5 \Rightarrow p = \frac{1.5}{0.25} \Rightarrow p = \frac{15 \times 100}{25 \times 10} \Rightarrow p = \frac{15 \times 4}{10} \Rightarrow p = \frac{60}{10} \Rightarrow p = 6$

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