Chapter 13 Solutions — Statistics

n = 8, Sum = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72

Mean x̄ = 72/8 = 9

Deviations (xi − x̄) and (xi − x̄)²:

6 − 9 = −3, (−3)² = 9

7 − 9 = −2, (−2)² = 4

10 − 9 = 1, 1² = 1

12 − 9 = 3, 3² = 9

13 − 9 = 4, 4² = 16

4 − 9 = −5, (−5)² = 25

8 − 9 = −1, (−1)² = 1

12 − 9 = 3, 3² = 9

Σ(xi − x̄)² = 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9 = 74

Variance σ² = Σ(xi − x̄)²/n = 74/8 = 9.25.

Standard deviation σ = √9.25 ≈ 3.04.

Let the original observations be x1, x2, …, x6 with mean x̄ = 8 and σ = 4.

New observations: yi = 3xi.

New mean ȳ = (1/6)Σyi = (1/6)Σ3xi = 3 × (1/6)Σxi = 3 × 8 = 24.

New variance:

σ²_y = (1/6)Σ(yi − ȳ)² = (1/6)Σ(3xi − 3x̄)² = (1/6) × 9 × Σ(xi − x̄)² = 9σ²ₓ = 9 × 16 = 144.

New standard deviation σ_y = √144 = 12.

The first n natural numbers are 1, 2, 3, …, n.

Mean x̄ = (1 + 2 + … + n)/n = [n(n+1)/2]/n = (n+1)/2.

To find the variance, use:

Σxi² = 1² + 2² + … + n² = n(n+1)(2n+1)/6

Variance σ² = Σxi²/n − (x̄)²

= [n(n+1)(2n+1)/6]/n − [(n+1)/2]²

= (n+1)(2n+1)/6 − (n+1)²/4

Factor out (n+1):

= (n+1)[(2n+1)/6 − (n+1)/4]

= (n+1)[2(2n+1) − 3(n+1)] / 12

= (n+1)[4n+2 − 3n − 3] / 12

= (n+1)(n−1)/12

= (n² − 1)/12.