F71SM STATISTICAL METHODS
Tutorial on Section 1 DATA SUMMARY: SOLUTIONS
1.
(a) n=7,x ̄=£3562.7,s=£5067.9
(b) n = 6, x ̄ = £1649.5, s = £$270.3
The effect of the outlier (£15,042) on both x ̄ and s is dramatic —- including it increases both measures hugely. In fact, x ̄ and s are inappropriate as summary measures for the data set which includes the observation £15,042.
2. (a) x ̄ = 25.24, median is 50.5th sorted value = (25.1 + 25.2)/2 = 25.15
3.
4. 5.
6.
7.
8.
(b) Q1 is 25.25th value = 22.9 + 0.25 × (23.0 − 22.9) = 22.925 Q3 is 75.75th value = 27.3 + 0.75 × (27.4 − 27.3) = 27.375 so IQR = 4.45
Standard deviation s = 3.00
8i=1 xi = 8 × 41.825 = 334.6, 8i=1 x2i = 7 × 27.2362 + 334.62/8 = 19187.243 Adding new observation (60.4) gives: 9i=1 xi = 395.0, 9i=1 x2i = 22835.403 ⇒ x ̄ = 43.889, ie £43,890, and s = 26.219, ie £26,220
Invoice amount y = 70 + 30x, where x is length of job ⇒y ̄=70+30×3.5=£175,sy =30×0.5=£15
Lety=a+bx,thensy =|b|sx ⇒|b|=2
Taking b = 2 and adjusting the mean ⇒ a = −74 so y = −74 + 2x (We can also have b = −2, then y = 174 − 2x)
x ̄ = 32.16, sx = 5.6338. Assuming all gross incomes in the sample are over £10,000, then net income y = x−0.3(x−10) = 0.7x+3 ⇒ y ̄ = 0.7×32.16+3 = 25.512 ie £25,510; sy = 0.7 × 5.6338 = 3.944, ie £3,940.
Looking at the relative frequencies is instructive:
x 1234 Relative frequency A 0.4 0.3 0.2 0.1 Relative frequency B 0.55 0.25 0.15 0.05
B is more concentrated at x = 1 and less concentrated at higher values. So (a) A (b) A (c) B (a) Letz=100(x−2.08)thenx=z/100+2.08.
x z f fz fz2 2.08 0 1 0 0 2.09 1 3 3 3 2.10 2 3 6 12 2.11 3 7 21 63 2.12 4 10 40 160 2.13 5 23 115 575 2.14 6 56 336 2016 2.15 7 97 679 4753
200 1200 7582 1
x ̄ =
s2 = 1
1 100
1200 + 2.08 = 2.14kg 200
1 7582− 12002 = 1.91960×10−4 ⇒ s = 0.014kg. 199 200
x 1002
(b) Using a weighted average of the two means:
0.08 × 2.14 + 0.92(mean) = 2.342 ⇒ mean = 2.360kg
2