Parcels handled in a day by 25 workers at an Indore sorting centre: 60-80: 4, 80-100: 6, 100-120: 8, 120-140: 5, 140-160: 2. Rohit used the direct method and got a mean of 106 parcels. Sneha used the assumed mean method with a = 110. What value of f_i d_i must she have obtained?

Question

Accepted Answer

-100. Class marks 70, 90, 110, 130, 150 give f_i x_i = 2650 and f_i = 25 , so the mean is 106. Since d_i = x_i - a , f_i d_i = f_i x_i - a f_i = 2650 - 110(25) = -100 . Both methods must agree: 110 + (-100/25) = 110 - 4 = 106. "100" comes from writing d_i = a - x_i (sign reversed); "-4" is the ratio f_i d_i / f_i , not the sum itself; "-5" comes from also dividing by the class size h = 20 as if this were the step-deviation method.

Parcels handled in a day by 25 workers at an Indore sorting centre:
60-80: 4, 80-100: 6, 100-120: 8, 120-140: 5, 140-160: 2.
Rohit used the direct method and got a mean of 106 parcels. Sneha used the assumed mean method with $a = 110$ . What value of $\sum f_{i} d_{i}$ must she have obtained?

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More Class 10 Mean of Grouped Data practice

Parcels handled in a day by 25 workers at an Indore sorting centre:60-80: 4, 80-100: 6, 100-120: 8, 120-140: 5, 140-160: 2.Rohit used the direct method and got a mean of 106 parcels. Sneha used the assumed mean method with a=110. What value of ∑fi​di​ must she have obtained?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Mean of Grouped Data practice

Parcels handled in a day by 25 workers at an Indore sorting centre:
60-80: 4, 80-100: 6, 100-120: 8, 120-140: 5, 140-160: 2.
Rohit used the direct method and got a mean of 106 parcels. Sneha used the assumed mean method with $a = 110$ . What value of $\sum f_{i} d_{i}$ must she have obtained?