Question : 250 apples of a box were weighed and the distribution of masses of the apples is given in the following table :
Mass (in grams) |
80-100 | 100-120 | 120-140 | 140-160 | 160-180 |
Number of apples |
20 | 60 | 70 | x | 60 |
i) Find the value of x and the mean mass of the apples. [3 Marks]
ii) Find the modal mass of the apples. [2 Marks]
CBSE 2023
Solution :
i) Total apples = 250
20+60+70+x+60=250
210+x=250
x=250-210
x=40
Mass (C.I.) | No. of Apples (fi) | Class Marks (xi) | di=xi-a | ui=di/a | fiui |
80-100 | 20 | 90 | -40 | -2 | -40 |
100-120 | 60 | 110 | -20 | -1 | -60 |
120-140 | 70 | 130 = a | 0 | 0 | 0 |
140-160 | 40 | 150 | 20 | 1 | 40 |
160-180 | 60 | 170 | 40 | 2 | 120 |
Σfi=250 |
Σfiui=60 |
Using Step Deviation Method
Mean (x̄)
= a+(Σfiui/Σfi)×h
= 130+(60/250)×20
=130+(1200/250)
=130+4.8
=134.8 grams
Hence, the mean mass of the apple is 134.8 grams
ii) Modal Mass
Mass (C.I.) | No. of Apples (fi) |
80-100 | 20 |
100-120 | 60 - f0 |
120-140 | 70 - f1 |
140-160 | 40 - f2 |
160-180 | 60 |
Σfi=250 |
Here
Maximum Frequency = 70
Modal Class = 120-140
Lower Limit of the model class (l) = 120
Class size (h) = 140-120
=20
f0=60
f1=70
f2=40
Mode = l+[(f1-f0)/2f1-f0-f2)]×h
= 120+[(70-60)/(140-60-40)]×20
= 120+[10/(140-100)]×20
= 120+[10/40]×20
= 120+10/2
= 120+5
= 125 grams
Hence, the modal mass of the apple is 125 grams.