Find the median from the following data:
Class groups
Frequency
110-120
6
120-130
25
130-140
48
140-150
72
150-160
116
160-170
60
170-180
38
180-190
22
190-200
3
mean is 153.8
If true then enter $1$ and if false then enter $0$

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Solution

$F o r c o n t i n u o u s d a t a, m e d i a n = L + (\frac{\frac{N}{2} - C}{f}) \times H W h e r e N = T o t a l c u m u l a t i v e f r e q u e n c y L = l o w e r l i m i t o f t h e m e d i a n c l a s s I n t e r v a l; M e d i a n C l a s s = t h e c l a s s$
$i n t e r v a l h a v i n g c u m u l a t i v e f r e q \geq \frac{N}{2} l i e s C = c u m u l a t i v e f r e q u e n c y o f t h e c l a s s$
$i n t e r v a l j u s t b e f o r e t h e m e d i a n c l a s s i n t e r v a l f = f r e q u e n c y o f t h e m e d i a n c l a s s i n t e r v a l; H = w i d t h o f t h e m e d i a n c l a s s$
$i n t e r v a l$
$H e r e N = 390 \frac{N}{2} = 195, t h i s c u m u l a t i v e f r e q l i e s w i t h i n i n t e r v a l (150 - 160); L = 150; C = 151; f = 116; H = 10 H e n c e m e d i a n = 150 + (\frac{\frac{390}{2} - 151}{116} \times 10) = 150 + (\frac{195 - 151}{116} \times 10) = 150 + (\frac{44}{116} \times 10)$
$= 150 + (\frac{22}{58} \times 10) = 150 + (\frac{11}{29} \times 10) = 150 + (\frac{110}{29}) = 150 + 3.79 = 150 + 3.8 = 153.8$

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