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Question
A survey regarding the heights (in cm) of 50 girls of a class was conducted and the following data was obtained:
Height in cm |
120 – 130 | 130 – 140 | 140 – 150 | 150 – 160 | 160 – 170 |
No. of girls |
2 | 8 | 12 | 20 | 8 |
Find the mean, median and mode of the above data.
Solution
We have the following
Height in cm | Mid value `(x_i)` | Frequency `(f_i)` | Cumulative frequency |
`(f_i × x_i)` |
120 – 130 | 125 | 2 | 2 | 250 |
130 – 140 | 135 | 8 | 10 | 1080 |
140 – 150 | 145 | 12 | 22 | 1740 |
150 – 160 | 155 | 20 | 42 | 3100 |
160 – 170 | 165 | 8 | 50 | 1320 |
`Ʃ f_i` = 50 | `Ʃ f_i × x_i` = 7490 |
Mean, x = `(sum ( f_i xx x_i))/(sum f_i)`
=`7490/50`
= 149.8
Now, N = 50
⇒`N/2 =25`
The cumulative frequency just greater than 25 is 42 and the corresponding class is 150 –
160.
Thus, the median class is 150 – 160.
∴ l = 150, h = 10, f = 20, c = cf of preceding class = 22 and `N/2 =25`
Now,
Median, `M_e = l + { h xx ((N/2-c)/f)}`
= `150+ { 10xx ((25-22)/20) }`
=` (150+ 10 xx 3/20)`
= 151.5
Mode = 3(median) – 2(mean)
= 3 × 151.5 – 2 × 149.8
= 154.9
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