Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Solutions

Here, largest value (L) = 203, smallest value (S) = 114 ∴ Range = L – S = 203 – 114 = 89

Here, largest value (L) = 40, smallest value (S) = 10 ∴ Range = L – S = 40 – 10 = 30

Here, upper limit of the highest class (L) = 175, lower limit of the lowest class (S) = 115 ∴ Range = L – S = 175 – 115 = 60

The given data can be arranged in ascending order as follows: 24, 36, 40, 58, 62, 80 Here, n = 6 Q1 = value of observation = value of observation = value of (1.75)th observation = value of 1st observation + 0.75(value of 2nd observation – value of 1st observation) = 24 + 0.75(36 – 24) = 24 + 0.75(12) = 24 + 9 ∴ Q1 = 33 Q3 = value of 3 observation = value of 3 observation = value of (3 × 1.75)th observation = value of (5.25)th observation = value of 5th observation + 0.25(value of 6th observation – value of 5th observation) = 62 + 0.25(80 – 62) = 62 + 0.25(18) = 62 + 4.5 = 66.5 ∴ Q.D. = = 16.75

We construct the less than cumulative frequency table as follows: Here, N = 35 Q1 class = class containing observation ∴ = 8.75 Cumulative frequency which is just greater than (or equal to) 8.75 is 15. ∴ Q1 lies in the class 20-30. ∴ L = 20, c.f. = 8, f = 7, h = 10

Since open-ended classes are given, the appropriate measure of dispersion that we can compute is the quartile deviation. We construct the less than cumulative frequency table as follows: Here N = 250 Q1 class class containing observation ∴ = 62.5 Cumulative frequency which is just greater than (or equal to) 62.5 is 65. ∴ Q1 lies in the class 35-40. ∴ L = 35, c.f. = 15, f = 50, h = 5 The cumulative frequency which is just greater than (or equal to) 187.5 is 190. ∴ Q3 lies in the class 45-50. ∴ L = 45, c.f. = 150, f = 40, h = 5

We construct the less than cumulative frequency table as follows: Here, N = 120 Q1 class = class containing observation ∴ = 30 Cumulative frequency which is just greater than (or equal to) 30 is 35. ∴ Q1 lies in the class 4-6. ∴ L = 4, c.f. = 15, f = 20, h = 2 Cumulative frequency which is just greater than (or equal to) 90 is 90. ∴ Q3 lies in the class 10-12. ∴ L = 10, c.f. = 72, f = 18, h = 2

We prepare the following table for the calculation of variance and S.D.:

We prepare the following table for the calculation of variance and S.D:

We prepare the following table for calculation of arithmetic mean and S.D.:

Here, n = 200, = Mean = 60, S.D. = 20 Wrongly taken items are 3 and 67. Correct items are 13 and 17. Now, = 60 Correct value of (sum of wrongly taken items) + (sum of correct items) = 12000 – (3 + 67) + (13 + 17) = 12000 – 70 + 30 = 11960 Correct value of mean = × correct value of = × 11960 = 59.8 Now, S.D. = 20 Variance = (S.D.)2 = 202 ∴ Variance = 400 ∴ Correct value of = – (Sum of squares of wrongly taken items) + (Sum of squares of correct items) = 800000 – (32 + 672) + (132 + 172) = 800000 – (9 + 4489) + (169 + 289) = 800000 – 4498 + 458 = 795960 ∴ Correct value of Variance = ( × ) – (correct value of )2 = × 795960 – (59.8)2 = 3979.8 – 3576.04 = 403.76 ∴ The correct mean is 59.8 and correct variance is 403.76.

Let n1 and n2 be the number of boys and girls respectively. Let n = 200, = 65, = 70, = 62, σ1 = 8, σ2 = 10 Here, n1 + n2 = n ∴ n1 + n2 = 200 …….(i) Combined mean is given by ∴ 70n1 + 62n2 = 13000 ∴ 35n1 + 31n2 = 6500 ……..(ii) Solving (i) and (ii), we get n1 = 75, n2 = 125 Combined standard deviation is given by,

Since C.V. (I) > C.V. (II) ∴ the brand I is more variable.