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Mathematics

Calculate the mean and the median for the following distribution :

NumberFrequency
51
102
155
206
253
302
351

Measures of Central Tendency

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Answer

The given numbers are already in ascending order. We construct the cumulative frequency table as under :

Number (fi)Frequency (xi)Cumulative frequencyfixi
5115
102320
155875
20614120
2531775
3021960
3512035
Total20390

Here, n (no. of observations) = 20, which is even.

Median=n2th observation+(n2+1) th observation2=202th observation+(202+1) th observation2= 10th observation + 11th observation2\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{20}{2} \text{th observation} + \big(\dfrac{20}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 10th observation + 11th observation}}{2} \\[1em]

All observations from 9th to 14th are equal, each = 20.

Hence, median

=20+202=402=20.= \dfrac{20 + 20}{2} \\[1em] = \dfrac{40}{2} \\[1em] = 20.

Now calculating mean,

 Mean=fixifi=39020=19.5\text{ Mean} = \dfrac{∑fixi}{∑f_i} \\[1em] = \dfrac{390}{20} \\[1em] = 19.5

Hence, mean = 19.5 and median = 20.

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