JAMB - Mathematics (1994 - No. 37)
| Class interval | 1 - 5 | 6 - 10 | 11 - 15 | 16 - 20 | 21 - 25 |
| Frequency | 6 | 15 | 20 | 7 | 2 |
Estimate the median of the frequency distribution above
10\(\frac{1}{2}\)
11\(\frac{1}{2}\)
12
13
Explanation
Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5
= 11 + (\(\frac{25 - 21}{20}\))5
= 11 + (\(\frac{(4)}{20}\))
11 + 1 = 12
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5
= 11 + (\(\frac{25 - 21}{20}\))5
= 11 + (\(\frac{(4)}{20}\))
11 + 1 = 12
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