Basic Data & Probability

Question 1 (8)

(a) Primary data (1) — because the student collected it directly/first-hand via their own survey, rather than reusing someone else's published data. (1)

(b) Counting the raw list:

Total $= 5+8+10+4+2+1 = 30$ ✓ (4) — 1 mark per 2 correct rows; deduct if total ≠ 30.

(c) A frequency table groups identical values with their counts, so instead of adding 30 separate numbers you compute $\sum fx$ with only 6 products — fewer operations, less error, and it makes the $\sum f$ (total) immediately visible. (2)

Question 2 (10)

(a) Formula: $\bar{x}=\dfrac{\sum fx}{\sum f}$. (1)

$\sum fx = (0)(5)+(1)(8)+(2)(10)+(3)(4)+(4)(2)+(5)(1)$ $= 0+8+20+12+8+5 = 53$ (2)

$\bar{x} = \dfrac{53}{30} = 1.7\overline{6} \approx 1.77$ siblings (1)

(b) $n=30$ (even), so median = mean of the $15$th and $16$th values (position rule $\frac{n}{2}$ and $\frac{n}{2}+1$). (1) Cumulative: 0→5, 1→13, 2→23. Both 15th and 16th fall in the value-2 group. (1) Median $= 2$. (1)

(c) Range = largest value − smallest value (the spread of the data). (1) Mode = 2 (highest frequency, 10). (1) Range $= 5 - 0 = 5$. (1)

Question 3 (9)

(a) Original values are lost once grouped, so we assume every item lies at the midpoint of its class (a uniform/central-representative assumption); this lets us estimate $\sum fx$. (2)

(b) Midpoints: 2.5, 7.5, 12.5, 17.5. (1)

$\sum fx = 6(2.5)+14(7.5)+12(12.5)+8(17.5)$ $= 15+105+150+140 = 410$ (3) — 1 mark midpoints, 2 marks products/sum.

$\bar{x} = \dfrac{410}{40} = 10.25$ kg (1)

(c) Modal class $= 5 \le m < 10$ (highest frequency 14). (1) Median position $= \frac{40}{2}=20$th value; cumulative 6, 20 — the 20th value is the last of the second class, so median class $= 5 \le m < 10$. (1)

Question 4 (7)

(a) Total = 60 people over $360°$ ⇒ $\frac{360}{60}=6°$ per person. (1)

• Car: $25 \times 6 = 150°$
• Bus: $15 \times 6 = 90°$
• Walk: $12 \times 6 = 72°$
• Cycle: $8 \times 6 = 48°$

(3) — 1 mark per two correct angles.

(b) $150+90+72+48 = 360°$ ✓ (1) — matters because sectors must fill the whole circle; a wrong total signals an arithmetic error.

(c) $15 \div 4 = 3.75$ ⇒ 4 symbols (3 full + one three-quarter symbol). (2)

Question 5 (9)

(a) All ordered pairs $(a,b)$ with $a,b \in \{1,...,6\}$; size $= 6 \times 6 = 36$. (2)

(b) Sum 7 outcomes: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ = 6. (2) $P(\text{sum}=7)=\dfrac{6}{36}=\dfrac{1}{6}$. (1)

(c) Sum ≥ 10: sum 10 = 3 ways, sum 11 = 2, sum 12 = 1 ⇒ 6 outcomes. $P=\dfrac{6}{36}=\dfrac{1}{6}$. (2)

(d) $P(\text{sum}\le 9) = 1 - P(\text{sum}\ge 10) = 1 - \dfrac{1}{6} = \dfrac{5}{6}$. (1) Faster because counting all outcomes ≤ 9 means summing many rows; the complement is just one small count subtracted from 1. (1)

Question 6 (7)

(a) Positive correlation — as hours revised increases, exam score tends to increase (upward left-to-right trend). (2)

(b) (2, 90%) is an outlier (anomalous point). (1) It does not fit the trend. (1) A line of best fit should generally ignore/downweight it so it isn't dragged off the true trend. (1)

(c) Gradient $= \dfrac{80-30}{10-0} = \dfrac{50}{10} = 5$. (1) Interpretation: each extra hour of revision is associated with about 5 extra percentage points on the exam. (1)

[
  {"claim":"Q2 mean of siblings = 53/30", "code":"fx=0*5+1*8+2*10+3*4+4*2+5*1; total=5+8+10+4+2+1; result = (fx==53 and total==30 and Rational(fx,total)==Rational(53,30))"},
  {"claim":"Q3 grouped estimated mean = 10.25", "code":"s=6*Rational(5,2)+14*Rational(15,2)+12*Rational(25,2)+8*Rational(35,2); result = (s==410 and Rational(s,40)==Rational(41,4))"},
  {"claim":"Q4 pie angles sum to 360", "code":"deg=6; total=25*deg+15*deg+12*deg+8*deg; result = (total==360 and 25*deg==150 and 8*deg==48)"},
  {"claim":"Q5 P(sum<=9)=5/6 via complement", "code":"pge10=Rational(6,36); result = (1-pge10==Rational(5,6) and Rational(6,36)==Rational(1,6))"},
  {"claim":"Q6 gradient of line of best fit = 5", "code":"m=Rational(80-30,10-0); result = (m==5)"}
]