Question 1 [8 marks]
(a) P(exactly 4 cars in 1 minute). [2 marks]
$X \sim \mathrm{Po}(3.6)$
$$\mathrm{P}(X = 4) = \frac{\mathrm{e}^{-3.6} \times 3.6^4}{4!} = 0.1912$$
$\mathrm{P}(X = 4) = 0.191$ (3 s.f.)
(b) P(at least 3 cars in 1 minute). [2 marks]
$$\mathrm{P}(X \geqslant 3) = 1 - \mathrm{P}(X \leqslant 2)$$
Using tables or calculator: $\mathrm{P}(X \leqslant 2) = 0.3027$
$\mathrm{P}(X \geqslant 3) = 1 - 0.3027 = 0.697$ (3 s.f.)
(c) Normal approximation for P(more than 20 in 5 minutes). [3 marks]
For a 5-minute period: $\lambda = 3.6 \times 5 = 18$
Since $\lambda$ is large, use a normal approximation:
$$X \sim \mathrm{Po}(18) \approx \mathrm{N}(18, 18)$$
Continuity correction: $\mathrm{P}(X > 20) \approx \mathrm{P}(X > 20.5)$
Standardise: $z = \dfrac{20.5 - 18}{\sqrt{18}} = 0.5893$
From tables: $\Phi(0.5893) = 0.7222$
$\mathrm{P}(Z > 0.5893) = 1 - 0.7222 = 0.2778$
$\mathrm{P}(X > 20) \approx 0.278$ (3 s.f.)
Without continuity correction: $z = \frac{20-18}{\sqrt{18}} = 0.471$, giving 0.319 (wrong). The continuity correction makes a significant difference here.
(d) Conditional: 5 in first 2 minutes given 12 in 4 minutes. [1 mark]
Under a Poisson process, given a total of $N$ events in time $[0, T]$, the number in $[0, t]$ follows $\mathrm{B}(N, t/T)$.
Here $N = 12$, $t = 2$, $T = 4$, so $p = 2/4 = 0.5$.
$$\mathrm{P}(\text{5 in first 2 min}) = \binom{12}{5}(0.5)^{12} = 0.1934$$
$0.193$ (3 s.f.)
Question 2 [13 marks]
(a) Show that $k = \frac{3}{16}$. [3 marks]
For a valid PDF: $\int_0^2 kx(4-x)\,\mathrm{d}x = 1$
$$\int_0^2 kx(4-x)\,\mathrm{d}x = k\int_0^2(4x - x^2)\,\mathrm{d}x = k\Bigl[2x^2 - \frac{x^3}{3}\Bigr]_0^2$$
$$= k\Bigl(8 - \frac{8}{3}\Bigr) = k \cdot \frac{16}{3}$$
$$\frac{16k}{3} = 1 \quad \Rightarrow \quad k = \frac{3}{16}$$
$k = \frac{3}{16}$ as required.
(b) Find $\mathrm{P}(0.5 < X < 1.5)$. [2 marks]
$$\mathrm{P}(0.5 < X < 1.5) = \int_{0.5}^{1.5} \frac{3}{16}x(4-x)\,\mathrm{d}x = \frac{3}{16}\Bigl[2x^2 - \frac{x^3}{3}\Bigr]_{0.5}^{1.5}$$
$$= \frac{3}{16}\Bigl[\Bigl(4.5 - 1.125\Bigr) - \Bigl(0.5 - 0.04167\Bigr)\Bigr]$$
$$= \frac{3}{16}(3.375 - 0.45833) = 0.5469$$
$\mathrm{P}(0.5 < X < 1.5) = 0.547$ (3 s.f.)
(c) Find $\mathrm{E}(X)$. [2 marks]
$$\mathrm{E}(X) = \int_0^2 x \cdot \frac{3}{16}x(4-x)\,\mathrm{d}x = \frac{3}{16}\int_0^2(4x^2 - x^3)\,\mathrm{d}x$$
$$= \frac{3}{16}\Bigl[\frac{4x^3}{3} - \frac{x^4}{4}\Bigr]_0^2 = \frac{3}{16}\Bigl(\frac{32}{3} - 4\Bigr) = \frac{3}{16} \cdot \frac{20}{3} = \frac{20}{16} = \frac{5}{4}$$
$\mathrm{E}(X) = \frac{5}{4} = 1.25$
(d) Explain why the mode is 2. [2 marks]
$$\mathrm{f}'(x) = \frac{3}{16}(4 - 2x)$$
$\mathrm{f}'(x) = 0$ only at $x = 2$. For $0 \leqslant x < 2$, $\mathrm{f}'(x) > 0$, so $\mathrm{f}$ is increasing on $[0, 2]$.
The maximum therefore occurs at the endpoint $x = 2$, with $\mathrm{f}(2) = \frac{3}{4}$.
The mode is 2 because $\mathrm{f}$ increases throughout $[0, 2]$ and attains its maximum at the right endpoint.
This is the "mode at endpoint" scenario that examiners like to test. The stationary point at $x = 2$ is found by setting $\mathrm{f}'(x)=0$, but it is the boundary of the domain, and the function reaches its maximum there.
(e) Find $\mathrm{P}(X > \mathrm{E}(X))$. [2 marks]
$\mathrm{P}(X > 1.25) = 1 - \mathrm{F}(1.25)$ where $\mathrm{F}(x) = \frac{3}{16}\Bigl(2x^2 - \frac{x^3}{3}\Bigr)$
$$\mathrm{F}(1.25) = \frac{3}{16}\Bigl(3.125 - 0.6510\Bigr) = 0.4639$$
$\mathrm{P}(X > 1.25) = 0.536$ (3 s.f.)
(f) Show median between 1.30 and 1.31. [1 mark]
$$\mathrm{F}(1.30) = \frac{3}{16}\Bigl(3.38 - 0.7323\Bigr) = 0.4964$$
$$\mathrm{F}(1.31) = \frac{3}{16}\Bigl(3.4322 - 0.7485\Bigr) = 0.5030$$
$\mathrm{F}(1.30) = 0.4964 < 0.5 < 0.5030 = \mathrm{F}(1.31)$, so the median lies between 1.30 and 1.31.
(g) Sketch the PDF. [1 mark]
The graph is an upside-down parabola segment: passes through the origin, concave down, increasing on $[0, 2]$, with maximum at $(2, \frac{3}{4})$.
Sketch: increasing curve from $(0, 0)$ to $(2, 0.75)$, concave down throughout.
Question 3 [11 marks]
(a) State the distribution. [1 mark]
Fixed $n = 30$, constant $p = 0.08$, independent trials.
$X \sim \mathrm{B}(30, 0.08)$
(b) $\mathrm{P}(X = 2)$. [2 marks]
$\mathrm{P}(X = 2) = \binom{30}{2}(0.08)^2(0.92)^{28} = 0.2696$
$0.270$ (3 s.f.)
(c) $\mathrm{P}(X \geqslant 4)$. [2 marks]
$\mathrm{P}(X \geqslant 4) = 1 - \sum_{r=0}^{3}\binom{30}{r}(0.08)^r(0.92)^{30-r} = 1 - 0.7838 = 0.2162$
$0.216$ (3 s.f.)
(d) Hypothesis test at 5% significance. [4 marks]
Step 1: $\mathrm{H}_0: p = 0.08$, $\mathrm{H}_1: p > 0.08$
Step 2: Under $\mathrm{H}_0$, $X \sim \mathrm{B}(30, 0.08)$. Observed $x = 6$.
$$\mathrm{P}(X \geqslant 6 \mid p = 0.08) = 1 - \mathrm{P}(X \leqslant 5) = 1 - 0.9707 = 0.0293$$
Step 3: $0.0293 < 0.05$, so reject $\mathrm{H}_0$.
There is sufficient evidence at the 5% significance level to support the inspector's suspicion that the proportion of defective components is greater than 8%.
(e) Critical region. [2 marks]
Since $\mathrm{P}(X \geqslant 6) = 0.0293 < 0.05$ and $\mathrm{P}(X \geqslant 5) = 0.081 > 0.05$, the critical region consists of values 6 and above.
Critical region: $X \geqslant 6$
Question 4 [12 marks]
(a) Test whether the mean number of complaints has decreased. [7 marks]
Step 1 — Hypotheses:
$\mathrm{H}_0: \lambda = 3$ (mean unchanged); $\mathrm{H}_1: \lambda < 3$ (mean decreased)
Step 2 — Test statistic: Under $\mathrm{H}_0$, over a 4-week period:
$$X \sim \mathrm{Po}(4 \times 3) = \mathrm{Po}(12)$$
Step 3 — p-value: Observed $x = 6$.
$$\mathrm{P}(X \leqslant 6 \mid \lambda = 12) = \sum_{r=0}^6 \frac{\mathrm{e}^{-12} \times 12^r}{r!} = 0.0458$$
Step 4 — Comparison: $0.0458 < 0.05$, reject $\mathrm{H}_0$.
Step 5 — Conclusion in context: Significant evidence that the mean number of complaints per week has decreased after staff training.
Critical region: $\mathrm{P}(X \leqslant 6) = 0.0458 < 0.05$, $\mathrm{P}(X \leqslant 7) = 0.0895 > 0.05$, so critical region is $X \leqslant 6$.
There is significant evidence at the 5% level that the mean number of complaints per week has decreased. Critical region: $X \leqslant 6$.
(b) Critical region. [1 mark]
$X \leqslant 6$
(c) Condition for Poisson model. [2 marks]
Complaints occur independently of each other and at a constant average rate over time.
(d) At 1% level. [2 marks]
At 1% significance: $\mathrm{P}(X \leqslant 4) = 0.0076 < 0.01$, $\mathrm{P}(X \leqslant 5) = 0.0203 > 0.01$.
Critical region at 1%: $X \leqslant 4$. Observed value 6 is not in this region.
Yes, the conclusion would change. At the 1% level we would not reject $\mathrm{H}_0$; there would be insufficient evidence of a decrease in complaints.
Question 5 [11 marks]
(a) P(exactly 6 in 1 hour). [2 marks]
$X \sim \mathrm{Po}(8.4)$
$$\mathrm{P}(X = 6) = \frac{\mathrm{e}^{-8.4} \times 8.4^6}{6!} = 0.10972$$
$0.110$ (3 s.f.)
(b) P(at most 5 in 1 hour). [2 marks]
$\mathrm{P}(X \leqslant 5) = \sum_{r=0}^{5}\mathrm{e}^{-8.4}\frac{8.4^r}{r!} = 0.1573$
$0.157$ (3 s.f.)
(c) Normal approx: P(more than 55 in 6 hours). [5 marks]
For 6 hours: $\lambda = 8.4 \times 6 = 50.4$. Since $\lambda$ is large, use normal approximation:
$$X \sim \mathrm{Po}(50.4) \approx \mathrm{N}(50.4, 50.4)$$
Continuity correction: $\mathrm{P}(X > 55) \approx \mathrm{P}(X > 55.5)$
Standardise: $z = \dfrac{55.5 - 50.4}{\sqrt{50.4}} = \dfrac{5.1}{7.099} = 0.7183$
From tables: $\Phi(0.7183) = 0.7638$
$$\mathrm{P}(Z > 0.7183) = 1 - 0.7638 = 0.2362$$
$\mathrm{P}(X > 55) \approx 0.236$ (3 s.f.)
(d) Conditional: 10 in first hour given 24 in 3 hours. [2 marks]
Under a Poisson process, given 24 customers in 3 hours, the number in the first hour follows $\mathrm{B}(24, \frac{1}{3})$.
$$\mathrm{P}(\text{10 in first hour}) = \binom{24}{10}\Bigl(\frac{1}{3}\Bigr)^{10}\Bigl(\frac{2}{3}\Bigr)^{14} = 0.11377$$
$0.114$ (3 s.f.)
Question 6 [11 marks]
(a) Show $\mathrm{f}(x)$ is a valid PDF. [2 marks]
Substituting $u = x - 2$, so $\mathrm{d}u = \mathrm{d}x$, $x = 2 \to u = 0$, $x = 6 \to u = 4$:
$$\int_2^6 \frac{3}{64}(x-2)^2(6-x)\,\mathrm{d}x = \frac{3}{64}\int_0^4 u^2(4-u)\,\mathrm{d}u$$
$$= \frac{3}{64}\int_0^4(4u^2 - u^3)\,\mathrm{d}u = \frac{3}{64}\Bigl[\frac{4u^3}{3} - \frac{u^4}{4}\Bigr]_0^4$$
$$= \frac{3}{64}\Bigl(\frac{256}{3} - \frac{256}{4}\Bigr) = \frac{3}{64}\Bigl(\frac{256}{3} - 64\Bigr) = \frac{3}{64} \cdot \frac{64}{3} = 1$$
Also $\mathrm{f}(x) \geqslant 0$ for $2 \leqslant x \leqslant 6$ because $(x-2)^2 \geqslant 0$ and $(6-x) \geqslant 0$.
$\int \mathrm{f}(x)\,\mathrm{d}x = 1$ and $\mathrm{f}(x) \geqslant 0$ everywhere, so $\mathrm{f}(x)$ is a valid PDF.
(b) Find $\mathrm{E}(X)$. [2 marks]
$\mathrm{E}(X) = \int_2^6 x \cdot \frac{3}{64}(x-2)^2(6-x)\,\mathrm{d}x$. Let $u = x - 2$, $x = u + 2$:
$$= \frac{3}{64}\int_0^4(u+2)u^2(4-u)\,\mathrm{d}u = \frac{3}{64}\int_0^4(2u^3 - u^4 + 8u^2)\,\mathrm{d}u$$
$$= \frac{3}{64}\Bigl[\frac{u^4}{2} - \frac{u^5}{5} + \frac{8u^3}{3}\Bigr]_0^4 = \frac{3}{64}\Bigl(128 - \frac{1024}{5} + \frac{512}{3}\Bigr) = 4.4$$
$\mathrm{E}(X) = 4.4 = \frac{22}{5}$
(c) Find $\operatorname{Var}(X)$. [2 marks]
$\mathrm{E}(X^2) = \int_2^6 x^2 \cdot \frac{3}{64}(x-2)^2(6-x)\,\mathrm{d}x = 20$
$$\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 20 - 4.4^2 = 20 - 19.36 = 0.64$$
$\operatorname{Var}(X) = 0.64$
(d) Find the mode. [2 marks]
$$\mathrm{f}'(x) = \frac{3}{64}\Bigl[2(x-2)(6-x) + (x-2)^2(-1)\Bigr]$$
$$= \frac{3}{64}(x-2)(12 - 2x - x + 2) = \frac{3}{64}(x-2)(14 - 3x)$$
$\mathrm{f}'(x) = 0$ at $x = 2$ and $x = \frac{14}{3}$.
$\mathrm{f}(2) = 0$, $\mathrm{f}(\frac{14}{3}) > 0$, and $\mathrm{f}''(\frac{14}{3}) < 0$, so maximum at $\frac{14}{3}$.
Mode $= \frac{14}{3} \approx 4.67$
(e) Skewness. [2 marks]
$\mathrm{E}(X)=4.4$ and mode $=\frac{14}{3}\approx 4.67$. Since mean $<$ mode, the distribution is negatively skewed.
The distribution is negatively skewed because $\mathrm{E}(X) < \text{mode}$.
For a negatively skewed distribution: mean $<$ median $<$ mode. Here mean $=4.4$, median $\approx 4.457$, mode $\approx 4.667$, confirming negative skew.
(f) Median between 4.45 and 4.46. [1 mark]
$\mathrm{F}(x) = \frac{3}{64}\bigl(\frac{4u^3}{3} - \frac{u^4}{4}\bigr) = \frac{u^3(16-3u)}{256}$ where $u = x - 2$.
$$\mathrm{F}(4.45) = \frac{3}{64}\bigl(\frac{4 \cdot 2.45^3}{3} - \frac{2.45^4}{4}\bigr) = 0.4969$$
$$\mathrm{F}(4.46) = \frac{3}{64}\bigl(\frac{4 \cdot 2.46^3}{3} - \frac{2.46^4}{4}\bigr) = 0.5013$$
$\mathrm{F}(4.45) = 0.4969 < 0.5 < 0.5013 = \mathrm{F}(4.46)$, so the median lies between 4.45 and 4.46.
Question 7 [9 marks]
(a) $\mathrm{P}(10 \leqslant X \leqslant 20)$ where $X \sim \mathrm{B}(50, 0.28)$. [2 marks]
$$X \sim \mathrm{B}(50, 0.28)$$
$$\mathrm{P}(10 \leqslant X \leqslant 20) = \sum_{r=10}^{20} \binom{50}{r}(0.28)^r(0.72)^{50-r} = 0.9028$$
$0.903$ (3 s.f.)
Using binomial tables: $\mathrm{P}(X \leqslant 20) - \mathrm{P}(X \leqslant 9) = 0.9736 - 0.0708 = 0.9028$.
(b) Normal approximation for $\mathrm{P}(40 \leqslant Y \leqslant 60)$ with $Y \sim \mathrm{B}(180, 0.28)$. [3 marks]
Parameters: $\mu = 180 \times 0.28 = 50.4$; $\sigma^2 = 180 \times 0.28 \times 0.72 = 36.288$; $\sigma = 6.024$
Continuity correction:
$$\mathrm{P}(40 \leqslant Y \leqslant 60) \approx \mathrm{P}(39.5 < Y < 60.5)$$
Standardise:
$$z_1 = \frac{39.5 - 50.4}{6.024} = -1.810$$
$$z_2 = \frac{60.5 - 50.4}{6.024} = 1.677$$
From tables:
$$\Phi(1.677) \approx 0.9535,\quad \Phi(-1.810) \approx 0.0351$$
$\mathrm{P}(40 \leqslant Y \leqslant 60) \approx 0.9535 - 0.0351 = 0.918$ (3 s.f.)
Without continuity correction: $\mathrm{P}(40 \leqslant Y \leqslant 60) \approx \Phi(1.594) - \Phi(-1.727) = 0.9445 - 0.0422 = 0.902$ (slight underestimate of the true binomial probability).
(c) Why continuity correction? [1 mark]
A continuity correction is needed because a discrete distribution (binomial) is being approximated by a continuous distribution (normal).
(d) Conditions for normal approximation to binomial. [1 mark]
$np > 5$ and $n(1-p) > 5$.
Here $np = 50.4$ and $n(1-p) = 129.6$, both well above 5, so the approximation is valid.
(e) Normal approximation for $\mathrm{P}(C \leqslant 35)$ where $C \sim \mathrm{Po}(40)$. [2 marks]
$\lambda = 40 > 15$, so normal approximation is appropriate:
$$C \sim \mathrm{Po}(40) \approx \mathrm{N}(40, 40)$$
Continuity correction: $\mathrm{P}(C \leqslant 35) \approx \mathrm{P}(C < 35.5)$
Standardise: $z = \dfrac{35.5 - 40}{\sqrt{40}} = \dfrac{-4.5}{6.325} = -0.7115$
From tables: $\Phi(-0.7115) = 0.2384$
$\mathrm{P}(C \leqslant 35) \approx 0.238$ (3 s.f.)
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