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Exercises · Q9
Q.

The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.

Production yield (kg. per hectare)50-5353-5656-5959-6262-6565-6868-7171-7474-77
Number of farms381430362816105

(Ans. mean = 63.82 kg. per hectare, median = 63.67 kg. per hectare, mode = 63.29 kg. per hectare)

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Compute all three averages for the same continuous series (N = 150). Using the step-deviation method the mean is 63.82; the median (from the 62–65 class) is 63.67; the mode (also the 62–65 class) is 63.29 kg per hectare — all clustered together, showing a fairly symmetrical distribution.

Set-up (mid-points, c.f., step deviations A=63.5, c=3A=63.5,\ c=3)

Yield (kg/ha)ffmid mmc.f.d′=m−63.53d'=\frac{m-63.5}{3}fd′fd'
50–53351.53−4−12
53–56854.511−3−24
56–591457.525−2−28
59–623060.555−1−30
62–653663.59100
65–682866.5119128
68–711669.5135232
71–741072.5145330
74–77575.5150420
Total15016

Mean (step-deviation method)

Xˉ=A+∑fd′N×c=63.5+16150×3=63.5+0.32=63.82\bar{X} = A + \frac{\sum fd'}{N}\times c = 63.5 + \frac{16}{150}\times 3 = 63.5 + 0.32 = 63.82

Median

N2=75\dfrac{N}{2}=75 → first c.f. ≥75\geq 75 is 91, so the median class is 62–65 (L=62, c.f.=55, f=36, h=3L=62,\ c.f.=55,\ f=36,\ h=3): …

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