Chapter 2 Measures of Dispersion Ex 2.2 Solutions

Given data: 7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6 The tabulated form of the above data is as given below: 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 the calculation of variance and S.D.:

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

∴ (x4 – 4)(x4 – 2) = 0 ∴ x4 = 4 or x4 = 2 From (i), we get x5 = 2 or x5 = 4 ∴ The two numbers are 2 and 4.

We prepare the following table for the calculation of standard deviation.

Here, Mean = = 59.5, and Var(X) = σ2 = 413 Let xi be a mid value of class and d = , where a is assumed mean and h is class width. We prepare the following table for calculation of mean and variance of di. Now, Var(X) = h2. Var(D) ∴ 413 = h2 × 4.13 ∴ h2 = 100 ∴ h = 10 Substituting h = 10 in (i), we get -0.1 × 10 + a = 59.5 ∴ -1 + a = 59.5 ∴ a = 59.5 + 1 ∴ a = 60.5 We prepare the following table to determine actual class intervals for corresponding values of di. ∴ The actual class intervals are 15.5 – 25.5, 25.5 – 35.5, …….., 95.5 – 105.5