Question

(b) The yield (in kg) of 100 plots in the form of grouped frequency distribution is given 

below: (5) 

 

Yield 

(kg) 

Frequency

0-20 6 

20-40 21 

40-60 35 

60-80 30 

80-100 8 

(i) Estimate the number of plots with an yield of 

(A)40 to 80 kg 

(B) 10 to 70 kg 

 (ii) Find the mean and standard deviation of yield.

Answer 1

(i)

(A) The number of plots with an yield of 40 to 80 kg is the sum of 35, the number of plots with an yield of 40-60, and 30, the number of plots with an yield of 60-80. This equals to 65.

(B) The number of plots with an yield of 10 to 70 kg is the sum of:

3, the half of the number of plots with an yield of 0 to 20;

21, the number of plots with an yield of 20 to 40;

35, the number of plots with an yield of 40 to 60;

15, the number of plots with an yield of 60 to 80.

This sum is 3+21+35+15=74.

(ii) The mean of the yield is $0.06\cdot(0+20)/2+0.21\cdot(20+40)/2+0.35\cdot(40+60)/2+0.30\cdot(60+80)/2+0.08\cdot(80+100)/2=52.6$

The mean of squared values of the yield is

$0.06\cdot(0+20)^2/4+0.21\cdot(20+40)^2/4+0.35\cdot(40+60)^2/4+0.30\cdot(60+80)^2/4+0.08\cdot(80+100)^2/4=3188$

The variance of the yield is $3188-52.6^2=421.24$.

The standard deviation of the yield is $\sqrt{421.24}=20.524$