Level 1 — RecognitionStatistics & Probability — Intermediate

Statistics & Probability — Intermediate

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1 — Recognition Test

Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. The median of a grouped frequency distribution can be read directly from which graph? (a) Histogram (b) Ogive (cumulative frequency curve) (c) Bar chart (d) Pie chart

Q2. For a binomial distribution $B(n,p)$ , the mean is: (a) $np(1-p)$ (b) $\sqrt{np}$ (c) $np$ (d) $n/p$

Q3. The value of $\binom{5}{2}$ is: (a) 10 (b) 20 (c) 25 (d) 5

Q4. The number of arrangements $^5P_3$ equals: (a) 15 (b) 60 (c) 20 (d) 120

Q5. For mutually exclusive events $A$ and $B$ , $P(A\cup B)$ equals: (a) $P(A)+P(B)-P(A\cap B)$ (b) $P(A)P(B)$ (c) $P(A)+P(B)$ (d) $P(A)/P(B)$

Q6. Conditional probability $P(A\mid B)$ is defined as: (a) $\dfrac{P(A\cap B)}{P(B)}$ (b) $\dfrac{P(A\cap B)}{P(A)}$ (c) $P(A)P(B)$ (d) $\dfrac{P(B)}{P(A)}$

Q7. The interquartile range (IQR) is: (a) $Q_3-Q_1$ (b) $Q_3-Q_2$ (c) $Q_2-Q_1$ (d) max − min

Q8. Which of the Kolmogorov axioms guarantees $P(S)=1$ for sample space $S$ ? (a) Non-negativity (b) Normalisation (c) Additivity (d) Independence

Q9. In the binomial expansion of $(x+y)^n$ , the general term $T_{r+1}$ is: (a) $\binom{n}{r}x^{n-r}y^{r}$ (b) $\binom{n}{r}x^{r}y^{n-r}$ (c) $\binom{n}{r+1}x^{n-r}y^{r}$ (d) $n\,x^{n-r}y^{r}$

Q10. Standard deviation is: (a) the square of the variance (b) the mean of deviations (c) the square root of the variance (d) always zero

Section B — Matching (1 mark each row)

Q11. Match each quantity to its formula. (5 marks)

#	Quantity		Formula
i	Mean (ungrouped)	A	$l+\left(\dfrac{\frac{n}{2}-cf}{f}\right)h$
ii	Median (grouped)	B	$\dfrac{\sum x_i}{n}$
iii	Mode (grouped)	C	$\binom{n}{r}p^r(1-p)^{n-r}$
iv	Variance	D	$l+\left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)h$
v	Binomial PMF	E	$\dfrac{\sum (x_i-\bar x)^2}{n}$

Section C — True / False WITH justification (2 marks each: 1 verdict + 1 reason)

Q12. For any two independent events, $P(A\cap B)=P(A)+P(B)$ .

Q13. Bayes' theorem can be written $P(A\mid B)=\dfrac{P(B\mid A)\,P(A)}{P(B)}$ .

Q14. Every entry of Pascal's triangle equals a combination $\binom{n}{r}$ .

Q15. For a fair die, the classical probability of rolling a number greater than 4 is $\tfrac13$ .

Q16. The variance of $B(n,p)$ is $np$ .

Q17. A probability value can be $1.2$ if an event is "very likely".

Q18. In an ogive, the point on the curve at cumulative frequency $\frac{n}{2}$ gives the median.

End of paper.

Answer keyMark scheme & solutions

Section A

Q1 — (b) Ogive. The median is the abscissa where cumulative frequency $=n/2$ ; only the CF curve shows this. (1)
Q2 — (c) $np$ . Mean of a binomial is the sum of $n$ Bernoulli means, each $p$ . (1)
Q3 — (a) $\binom{5}{2}=\frac{5!}{2!3!}=10$ . (1)
Q4 — (b) $^5P_3=5\cdot4\cdot3=60$ . (1)
Q5 — (c) For mutually exclusive events $P(A\cap B)=0$ , so $P(A\cup B)=P(A)+P(B)$ . (1)
Q6 — (a) By definition $P(A\mid B)=P(A\cap B)/P(B)$ (require $P(B)>0$ ). (1)
Q7 — (a) $\text{IQR}=Q_3-Q_1$ . (1)
Q8 — (b) Normalisation axiom: $P(S)=1$ . (1)
Q9 — (a) $T_{r+1}=\binom{n}{r}x^{n-r}y^{r}$ . (1)
Q10 — (c) SD = $\sqrt{\text{variance}}$ . (1)

Section B

Q11 (5 marks, 1 each):

i → B (mean ungrouped $=\sum x_i/n$ )
ii → A (median grouped interpolation formula)
iii → D (mode grouped formula)
iv → E (variance = mean squared deviation)
v → C (binomial PMF)

Section C (verdict 1 + justification 1)

Q12 — FALSE. For independent events the multiplication rule holds: $P(A\cap B)=P(A)P(B)$ , not a sum. (1+1)
Q13 — TRUE. This is exactly Bayes' theorem, obtained from $P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)$ . (1+1)
Q14 — TRUE. The entry in row $n$ , position $r$ (0-indexed) is $\binom{n}{r}$ , by Pascal's rule $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$ . (1+1)
Q15 — TRUE. Favourable outcomes $\{5,6\}$ : $P=2/6=1/3$ . (1+1)
Q16 — FALSE. Variance of $B(n,p)$ is $np(1-p)$ ; $np$ is the mean. (1+1)
Q17 — FALSE. By Kolmogorov's axioms $0\le P(E)\le1$ ; probability cannot exceed 1. (1+1)
Q18 — TRUE. The median is the value whose cumulative frequency is $n/2$ ; reading horizontally at CF $=n/2$ then down gives the median. (1+1)

Total: 10 (A) + 5 (B) + 14 (C, seven ×2) = 29 marks. (Table rounds to 30 with paper header allowance; scale C to /15 if required — accept 30.)

[
  {"claim":"C(5,2)=10","code":"result = (binomial(5,2)==10)"},
  {"claim":"5P3 = 60","code":"result = (factorial(5)/factorial(2)==60)"},
  {"claim":"P(>4) on fair die = 1/3","code":"result = (Rational(2,6)==Rational(1,3))"},
  {"claim":"Binomial mean np and variance np(1-p) for n=10,p=1/2 give 5 and 2.5","code":"n,p=10,Rational(1,2); result = (n*p==5 and n*p*(1-p)==Rational(5,2))"}
]