Review of Term 1
Answers are for parents and teachers.
1
A juice stall records the price x (LE) of a bottle and the bottles sold per hour y (in tens). The pairs are (1, 9), (2, 8), (3, 6), (4, 5), (5, 3), (6, 2). Given Σx = 21, Σy = 33, Σxy = 90, Σx² = 91, Σy² = 219, find r and the regression line. Predict y at x = 4.5, and say why x = 10 should not be used.
nΣxy − ΣxΣy = 540 − 693 = −153; nΣx² − (Σx)² = 546 − 441 = 105; nΣy² − (Σy)² = 1314 − 1089 = 225. r = −153 ÷ √(105 × 225) ≈ −0.995. b = −153 ÷ 105 ≈ −1.457 and a = (33 + 1.457 × 21) ÷ 6 = 10.6, so y = 10.6 − 1.457x. At x = 4.5: y ≈ 4.04. At x = 10 the line gives about −4.0, impossible; 10 is outside the data (1 to 6).
2
Six pupils’ marks in two tests are: test 1: 70, 85, 60, 85, 90, 75; test 2: 65, 80, 70, 55, 90, 75. Rank each test from the lowest mark (ties take the mean of their places), find ΣD² and Spearman’s r.
Ranks of test 1: 2, 4.5, 1, 4.5, 6, 3. Ranks of test 2: 2, 5, 3, 1, 6, 4. D = 0, −0.5, −2, 3.5, 0, −1 (sum 0), D² = 0, 0.25, 4, 12.25, 0, 1, so ΣD² = 17.5. r = 1 − 6 × 17.5 ÷ (6 × 35) = 1 − 0.5 = 0.5.
3
Twelve pupils’ scores are 34, 41, 28, 52, 37, 45, 31, 39, 47, 36, 43, 50. Find Q₁, Q₂, Q₃ and Q. Then a thirteenth pupil scores 95: find the new Q and compare how Q and the range change.
Ordered: 28, 31, 34, 36, 37, 39, 41, 43, 45, 47, 50, 52; places 3.25, 6.5, 9.75: Q₁ = 34.5, Q₂ = 40, Q₃ = 46.5 and Q = 6. With 95 added (n = 13, places 3.5, 7, 10.5): Q₁ = 35, Q₃ = 48.5, Q = 6.75. Q moves from 6 to 6.75, while the range jumps from 24 to 67.
4
Six cards are numbered 1 to 6 and two are drawn without replacement. Find P(both even), P(the sum is even) and P(the second is even | the first is even).
Both even: 3/6 × 2/5 = 1/5. Both odd is also 1/5, so the sum is even with probability 2/5. Given that the first is even, 5 cards remain, 2 of them even: 2/5.
5
Of 80 pupils, 40 wear a watch, 20 walk to school, and 10 both wear a watch and walk. Are “wears a watch” and “walks to school” independent? Compare P(watch | walks) with P(watch).
P(watch) = 40/80 = 0.5, P(walks) = 20/80 = 0.25 and P(both) = 10/80 = 0.125 = 0.5 × 0.25. So they are independent. Also P(watch | walks) = 10/20 = 0.5 = P(watch).