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) 50–53 53–56 56–59 59–62 62–65 65–68 68–71 71–74 74–77 Number of farms 3 8 14 30 36 28 16 10 5
Mean = 63.82 kg per hectare, Median = 63.67 kg per hectare, Mode = 63.29 kg per hectare
Explanation
This is a comprehensive numerical problem requiring calculation of all three measures of central tendency. The student needs to construct frequency tables, identify median and modal classes, and apply the appropriate formulas for continuous series.
Solution Steps
Step 1: Calculation of Mean
Class Interval Frequency (f) Mid-point (x) f × x 50–53 3 51.5 154.5 53–56 8 54.5 436.0 56–59 14 57.5 805.0 59–62 30 60.5 1815.0 62–65 36 63.5 2286.0 65–68 28 66.5 1862.0 68–71 16 69.5 1112.0 71–74 10 72.5 725.0 74–77 5 75.5 377.5 Total 150 9573.0 Mean = Σfx/Σf = 9573/150 = 63.82 kg per hectare
Step 2: Calculation of Median
Class Frequency (f) Cumulative Frequency (cf) 50–53 3 3 53–56 8 11 56–59 14 25 59–62 30 55 62–65 36 91 65–68 28 119 68–71 16 135 71–74 10 145 74–77 5 150 N = 150, N/2 = 75 Median class = 62–65 (75th item lies here as cf reaches 91)
Median = L + [(N/2 - cf)/f] × h = 62 + [(75 - 55)/36] × 3 = 62 + (20/36) × 3 = 62 + 1.67 = 63.67 kg per hectare
Step 3: Calculation of Mode
Modal class = 62–65 (highest frequency = 36)
Mode = L + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h Where: L = 62, f₁ = 36, f₀ = 30, f₂ = 28, h = 3
= 62 + [(36 - 30)/( - 30 - 28)] × 3 = 62 + [6/(72 - 58)] × 3 = 62 + (6/14) × 3 = 62 + 1.29 = 63.29 kg per hectare