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 |
Total |
Total Number of girls |
2 |
8 |
12 |
20 |
8 |
50 |
Find the mean, median and mode of the above data. [CBSE 2008]
To find mean, we will solve by direct method:
HEIGHT (cm.) |
MID - POINT(xi) |
TOTAL NUMBER OF GIRLS(fi) |
fixi |
120 - 130 |
125 |
2 |
250 |
130 - 140 |
135 |
8 |
1080 |
140 - 150 |
145 |
12 |
1740 |
150 - 160 |
155 |
20 |
3100 |
160 - 170 |
165 |
8 |
1320 |
TOTAL |
50 |
7490 |
We have got
Σfi = 50 & Σfixi = 7490
∵ mean is given by
⇒ 
⇒ 
To find median,
Assume Σfi = N = Sum of frequencies,
h = length of median class,
l = lower boundary of the median class,
f = frequency of median class
and Cf = cumulative frequency
Lets form a table.
HEIGHT (cm.) |
TOTAL NUMBER OF GIRLS(fi) |
Cf |
120 - 130 |
2 |
2 |
130 - 140 |
8 |
2 + 8 = 10 |
140 - 150 |
12 |
10 + 12 = 22 |
150 - 160 |
20 |
22 + 20 = 42 |
160 - 170 |
8 |
42 + 8 = 50 |
TOTAL |
50 |
So, N = 50
⇒ N/2 = 50/2 = 25
The cumulative frequency just greater than (N/2 = ) 25 is 42, so the corresponding median class is 150 - 160 and accordingly we get Cf = 22(cumulative frequency before the median class).
Now, since median class is 150 - 160.
∴ l = 150, h = 10, f = 20, N/2 = 25 and Cf = 22
Median is given by,
⇒ 
= 150 + 1.5
= 151.5
And we know that,
Mode = 3(Median) – 2(Mean)
= 3(151.5) – 2(149.8)
= 454.5 – 299.6
= 154.9
Hence, mean is 149.8, median is 151.5 and mode is 154.9.