Basic Data & Probability

Question 1

(a) Primary data — the student collected it themselves directly from their own experiment (not from a published/existing source). [1 statement + 1 reason = 2]

(b) Outlier is $5.9$ s. Cause: e.g. a mistimed stopwatch start/stop, the ball caught on the ramp, or a reaction-time delay. [1 identify + 1 valid physical cause = 2]

(c) Full data (10 values), sum: $4.1+4.3+4.2+4.6+4.3+4.2+4.3+5.9+4.4+4.2 = 44.5$

• Mean $= 44.5/10 = 4.45$ s → to 3 s.f. $4.45$ s. [2]
• Median: sort → $4.1,4.2,4.2,4.2,4.3,4.3,4.3,4.4,4.6,5.9$; average of 5th & 6th $=(4.3+4.3)/2 = 4.3$ s. [2]
• Mode: most frequent values $4.2$ and $4.3$ each appear 3× → bimodal: $4.2$ and $4.3$ s (accept stating both). [1]
• Range $= 5.9 - 4.1 = 1.8$ s. [1]

(d) Remove $5.9$: sum $=44.5-5.9=38.6$, over 9 values → mean $=38.6/9 = 4.2\overline{8} \approx 4.29$ s. New range $=4.6-4.1=0.5$ s. [2 for two correct values] Comment: mean dropped from $4.45$ to $4.29$ (change $0.16$), median stayed at $4.3$ → the median is more resistant to the outlier. [2]

(e) Midpoints: $4.1, 4.3, 4.5, 4.7$. $\text{mean}=\frac{6(4.1)+18(4.3)+12(4.5)+4(4.7)}{40}=\frac{24.6+77.4+54.0+18.8}{40}=\frac{174.8}{40}=4.37$ → $4.37$ s (3 s.f.). [2]

Question 2

(a) Two independent rolls, each 6 outcomes → $6\times6 = 36$ ordered pairs. [2]

(b) Sum $=8$: $(2,6),(3,5),(4,4),(5,3),(6,2)$ → 5 outcomes. $P(S)=\frac{5}{36}$ (already lowest terms). [list 1 + count 1 + answer 1 = 3]

(c) $D'$ = "both rolls the same" = $\{(1,1),(2,2),\dots,(6,6)\}$ → 6 outcomes, $P(D')=6/36=1/6$. $P(D)=1-P(D')=1-\tfrac16=\tfrac56.$ [complement 1 + $P(D')$ 1 + answer 1 = 3]

(d) (i) estimate → $P(S)=5/36\approx 0.139$. [1] (ii) Single = performs assignment not comparison (equality test needs ==). The if would evaluate the assigned/truthy value rather than testing equality, so the condition no longer checks "sum equals 8" — the count is corrupted and the estimate does not converge to $5/36$. It's a syntax/semantics confusion between assignment and equality. [3]

(e) $S$ and $S'$ are complementary and mutually exclusive covering the whole sample space, so their probabilities sum to 1 (certainty). $P(S')=1-\frac{5}{36}=\frac{31}{36}$. [reason 1 + value 1 = 2]

Question 3

(a) Negative correlation, strong (points lie close to a curve/line). [type 1 + strength 1 = 2]

(b) Since this is a controlled experiment where $h$ is deliberately varied and other factors held fixed, a causal claim is more justifiable than in mere observational correlation. However, correlation alone still doesn't guarantee causation; here the physics (potential energy → kinetic energy) supports genuine causation. Accept: reasonable to infer causation because it's controlled + mechanism exists. [3]

(c) Range $=5.0-2.2=2.8$ s. [1] A line graph is appropriate because ramp height is a continuous quantitative variable and joining points shows the trend/relationship between $h$ and $t$; bar charts suit discrete categories. [2]

(d) Range would decrease: reducing random error clusters repeated times more tightly, shrinking max − min. [direction 1 + justification 1 = 2]

[
  {"claim":"Q1c mean of 10 values is 4.45","code":"data=[4.1,4.3,4.2,4.6,4.3,4.2,4.3,5.9,4.4,4.2]; result = Rational(sum([Rational(str(x)) for x in data]),10)==Rational('4.45')"},
  {"claim":"Q1c range is 1.8","code":"data=[4.1,4.3,4.2,4.6,4.3,4.2,4.3,5.9,4.4,4.2]; result = (Rational('5.9')-Rational('4.1'))==Rational('1.8')"},
  {"claim":"Q1e grouped mean is 4.37","code":"num=6*Rational('4.1')+18*Rational('4.3')+12*Rational('4.5')+4*Rational('4.7'); result = num/40==Rational('4.37')"},
  {"claim":"Q2b P(S)=5/36","code":"result = Rational(5,36)==Rational(5,36)"},
  {"claim":"Q2c P(D)=5/6 via complement","code":"result = 1-Rational(6,36)==Rational(5,6)"},
  {"claim":"Q2e P(S')=31/36","code":"result = 1-Rational(5,36)==Rational(31,36)"}
]