| Part | Mark Scheme | Marks |
|---|---|---|
| (a) | $X \sim \mathrm{Po}(3.6)$ $\mathrm{P}(X=4) = \mathrm{e}^{-3.6}\frac{3.6^4}{4!} = 0.191$ (3 s.f.) |
M1 A1 |
| (b) | $\mathrm{P}(X \geqslant 3) = 1 - \mathrm{P}(X \leqslant 2)$ M1 $= 1 - 0.3027 = 0.697$ (3 s.f.) A1 |
M1 A1 |
| (c) | $\lambda = 3.6 \times 5 = 18$ B1 $X \sim \mathrm{Po}(18) \approx \mathrm{N}(18, 18)$ M1 $\mathrm{P}(X > 20) \approx \mathrm{P}(X > 20.5)$ (cc) $z = \frac{20.5-18}{\sqrt{18}} = 0.589$ $\mathrm{P}(Z > 0.589) = 1 - \Phi(0.589) = 0.278$ (3 s.f.) A1 |
B1 M1 A1 |
| (d) | Given 12 in 4 min, count in first 2 min $\sim \mathrm{B}(12, \frac{1}{2})$ B1 $\mathrm{P}(\text{5 in first 2 min}) = \binom{12}{5}(0.5)^{12} = 0.193$ (3 s.f.) |
B1 |
| Notes: (c) Must show continuity correction. Accept 0.2776–0.2782. (d) Award B1 for recognising B(12, 0.5). | ||
| Part | Mark Scheme | Marks |
|---|---|---|
| (a) | $\int_0^2 kx(4-x)\,\mathrm{d}x = k\bigl[2x^2 - \frac{x^3}{3}\bigr]_0^2$ M1 $= k(8 - \frac{8}{3}) = \frac{16k}{3}$ A1 $\frac{16k}{3} = 1 \Rightarrow k = \frac{3}{16}$ A1 |
M1 A1 A1 |
| (b) | $\mathrm{P}(0.5 < X < 1.5) = \int_{0.5}^{1.5}\frac{3}{16}x(4-x)\,\mathrm{d}x$ M1 $= \frac{3}{16}\bigl[2x^2 - \frac{x^3}{3}\bigr]_{0.5}^{1.5} = 0.547$ (3 s.f.) A1 |
M1 A1 |
| (c) | $\mathrm{E}(X) = \int_0^2 x \cdot \frac{3}{16}x(4-x)\,\mathrm{d}x$ M1 $= \frac{3}{16}\int_0^2(4x^2 - x^3)\,\mathrm{d}x = \frac{3}{16}\bigl[\frac{4x^3}{3} - \frac{x^4}{4}\bigr]_0^2 = \frac{5}{4}$ A1 |
M1 A1 |
| (d) | $\mathrm{f}'(x) = \frac{3}{16}(4 - 2x)$ M1 $\mathrm{f}'(x) = 0$ at $x = 2$; $\mathrm{f}'(x) > 0$ on $[0, 2)$, so $\mathrm{f}$ increases. Maximum at endpoint $x = 2$, so mode $= 2$. A1 |
M1 A1 |
| (e) | $\mathrm{P}(X > 1.25) = 1 - \mathrm{F}(1.25)$ M1 $= 1 - \frac{3}{16}\bigl(2(1.25)^2 - \frac{1.25^3}{3}\bigr) = 0.536$ (3 s.f.) A1 |
M1 A1 |
| (f) | $\mathrm{F}(1.30) = 0.4964$, $\mathrm{F}(1.31) = 0.5030$ $\mathrm{F}(1.30) < 0.5 < \mathrm{F}(1.31)$, so median lies between 1.30 and 1.31. B1 |
B1 |
| (g) | Correct shape: increasing from origin, concave down, maximum at $(2, \frac{3}{4})$. B1 | B1 |
| Notes: (c) Accept $\frac{5}{4}$ or 1.25. (d) Full explanation needed for the A mark. (f) Accept any valid numerical method. | ||
| Part | Mark Scheme | Marks |
|---|---|---|
| (a) | $X \sim \mathrm{B}(30, 0.08)$ B1 | B1 |
| (b) | $\mathrm{P}(X = 2) = \binom{30}{2}(0.08)^2(0.92)^{28}$ M1 $= 0.270$ (3 s.f.) A1 |
M1 A1 |
| (c) | $\mathrm{P}(X \geqslant 4) = 1 - \sum_{r=0}^{3}\binom{30}{r}(0.08)^r(0.92)^{30-r}$ M1 $= 0.216$ (3 s.f.) A1 |
M1 A1 |
| (d) | $\mathrm{H}_0: p = 0.08$, $\mathrm{H}_1: p > 0.08$ B1 $\mathrm{P}(X \geqslant 6 \mid p=0.08) = 0.0293$ M1 A1 $0.0293 < 0.05$, reject $\mathrm{H}_0$. Significant evidence that proportion of defective components > 8%. A1 |
B1 M1 A1 A1 |
| (e) | $X \geqslant 6$ B1 B1 | B1 B1 |
| Notes: (d) M1 for correct probability calculation. Final A1 must be in context. (e) Condone $\{6, 7, \dots, 30\}$. | ||
| Part | Mark Scheme | Marks |
|---|---|---|
| (a) | $\mathrm{H}_0: \lambda = 3$, $\mathrm{H}_1: \lambda < 3$ B1 Under $\mathrm{H}_0$, 4-week total $\sim \mathrm{Po}(12)$ M1 $\mathrm{P}(X \leqslant 6 \mid \lambda = 12) = \sum_{r=0}^{6}\mathrm{e}^{-12}\frac{12^r}{r!}$ M1 $= 0.0458$ A1 $0.0458 < 0.05$, reject $\mathrm{H}_0$ M1 Significant evidence that mean complaints per week has decreased. A1 Critical region: $X \leqslant 6$ A1 |
B1 M1 M1 A1 M1 A1 A1 |
| (b) | $X \leqslant 6$ B1 | B1 |
| (c) | Complaints occur independently and at a constant average rate. B1 B1 | B1 B1 |
| (d) | At 1%: $\mathrm{P}(X \leqslant 4) = 0.0076 < 0.01$, $\mathrm{P}(X \leqslant 5) = 0.0203 > 0.01$ M1 Critical region $X \leqslant 4$. Observed 6 not in CR, so would not reject $\mathrm{H}_0$. A1 |
M1 A1 |
| Notes: (a) Award all 7 marks for a fully correct test with stated hypotheses, correct test statistic, p-value/critical value, comparison, and contextual conclusion. | ||
| Part | Mark Scheme | Marks |
|---|---|---|
| (a) | $X \sim \mathrm{Po}(8.4)$ $\mathrm{P}(X = 6) = \mathrm{e}^{-8.4}\frac{8.4^6}{6!}$ M1 $= 0.110$ (3 s.f.) A1 |
M1 A1 |
| (b) | $\mathrm{P}(X \leqslant 5) = \sum_{r=0}^{5}\mathrm{e}^{-8.4}\frac{8.4^r}{r!}$ M1 $= 0.157$ (3 s.f.) A1 | M1 A1 |
| (c) | $\lambda = 8.4 \times 6 = 50.4$ B1 $X \sim \mathrm{Po}(50.4) \approx \mathrm{N}(50.4, 50.4)$ M1 $\mathrm{P}(X > 55) \approx \mathrm{P}(X > 55.5)$ (cc) B1 $z = \frac{55.5 - 50.4}{\sqrt{50.4}} = 0.718$ M1 $\mathrm{P}(Z > 0.718) = 1 - \Phi(0.718) = 0.236$ (3 s.f.) A1 |
B1 M1 B1 M1 A1 |
| (d) | Given 24 in 3 h, count in first 1 h $\sim \mathrm{B}(24, \frac{1}{3})$ M1 $\mathrm{P}(\text{10 in first hour}) = \binom{24}{10}(\frac{1}{3})^{10}(\frac{2}{3})^{14} = 0.114$ (3 s.f.) A1 |
M1 A1 |
| Notes: (c) Must show cc. Accept 0.2362–0.2368. (d) M1 for recognising B(24, 1/3). | ||
| Part | Mark Scheme | Marks |
|---|---|---|
| (a) | $\int_2^6 \frac{3}{64}(x-2)^2(6-x)\,\mathrm{d}x$ (let $u=x-2$) M1 $= \frac{3}{64}\int_0^4 u^2(4-u)\,\mathrm{d}u = \frac{3}{64}\bigl[\frac{4u^3}{3} - \frac{u^4}{4}\bigr]_0^4 = 1$ A1 Also $\mathrm{f}(x) \geqslant 0$ on $[2,6]$. |
M1 A1 |
| (b) | $\mathrm{E}(X) = \int_2^6 x \cdot \frac{3}{64}(x-2)^2(6-x)\,\mathrm{d}x$ M1 $= 4.4$ ($\frac{22}{5}$) A1 |
M1 A1 |
| (c) | $\mathrm{E}(X^2) = \int_2^6 x^2 \cdot \frac{3}{64}(x-2)^2(6-x)\,\mathrm{d}x = 20$ M1 $\operatorname{Var}(X) = 20 - 4.4^2 = 0.64$ A1 |
M1 A1 |
| (d) | $\mathrm{f}'(x) = \frac{3}{64}(x-2)(14-3x) = 0$ M1 $x = 2$ or $\frac{14}{3}$; maximum at $\frac{14}{3}$. Mode $= \frac{14}{3}$. A1 |
M1 A1 |
| (e) | $\mathrm{E}(X) = 4.4 < \frac{14}{3} = \text{mode}$ A1 So negatively skewed. B1 |
A1 B1 |
| (f) | $\mathrm{F}(4.45) = 0.4969$, $\mathrm{F}(4.46) = 0.5013$ B1 $\mathrm{F}(4.45) < 0.5 < \mathrm{F}(4.46)$, so median lies between 4.45 and 4.46. |
B1 |
| Notes: (f) Accept any valid numerical method confirming the median lies between those values. (f) is 1 mark only. | ||
| Part | Mark Scheme | Marks |
|---|---|---|
| (a) | $X \sim \mathrm{B}(50, 0.28)$ B1 $\mathrm{P}(10 \leqslant X \leqslant 20) = 0.903$ (3 s.f.) A1 |
B1 A1 |
| (b) | $Y \sim \mathrm{B}(180, 0.28)$, $\mu = 50.4$, $\sigma^2 = 36.288$; apply cc: $\mathrm{P}(40 \leqslant Y \leqslant 60) \approx \mathrm{P}(39.5 < Y < 60.5)$ B1 $z_1 = \frac{39.5-50.4}{\sqrt{36.288}} = -1.810$, $z_2 = \frac{60.5-50.4}{\sqrt{36.288}} = 1.677$ M1 $\Phi(1.677) - \Phi(-1.810) = 0.9535 - 0.0351 = 0.918$ (3 s.f.) A1 |
B1 M1 A1 |
| (c) | A discrete (binomial) distribution is approximated by a continuous (normal) distribution. B1 | B1 |
| (d) | $np > 5$ and $n(1-p) > 5$ (or $n$ large, $p$ not too close to 0 or 1). B1 | B1 |
| (e) | $C \sim \mathrm{Po}(40) \approx \mathrm{N}(40, 40)$ B1 $\mathrm{P}(C \leqslant 35) \approx \mathrm{P}(C < 35.5)$ (cc) $z = \frac{35.5-40}{\sqrt{40}} = -0.7115$ $\mathrm{P}(Z < -0.7115) = \Phi(-0.7115) = 0.238$ (3 s.f.) A1 |
B1 A1 |
| Notes: (b) Must apply continuity correction at both boundaries. Accept 0.9182–0.9188. (e) B1 for recognising Poisson with $\lambda > 15$ may be normal-approximated. | ||