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A Level H1 Mathematics Practice Paper 4

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TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key & Marking Scheme

Version 4 of 5 — Statistics & Probability

Section A: Pure Statistics (30 marks)

Question 1 [4 marks]

Data: 12, 15, 10, 18, 14, 11, 16, 13; n=8n = 8

Unbiased estimate of the population mean:

xˉ=∑xin=12+15+10+18+14+11+16+138=1098=13.625\bar{x} = \frac{\sum x_i}{n} = \frac{12 + 15 + 10 + 18 + 14 + 11 + 16 + 13}{8} = \frac{109}{8} = 13.625

Unbiased estimate of the population variance:

s2=∑(xi−xˉ)2n−1s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}

Calculate each (xi−xˉ)2(x_i - \bar{x})^2:

xix_ixi−xˉx_i - \bar{x}(xi−xˉ)2(x_i - \bar{x})^2
12−1.6252.640625
151.3751.890625
10−3.62513.140625
184.37519.140625
140.3750.140625
11−2.6256.890625
162.3755.640625
13−0.6250.390625

∑(xi−xˉ)2=49.875\sum(x_i - \bar{x})^2 = 49.875

s2=49.8757=7.125s^2 = \frac{49.875}{7} = 7.125

Answer: Unbiased estimate of mean = 13.6 (or 13.625), unbiased estimate of variance = 7.13 (3 s.f.)

Marking:

  • B1 for correct xˉ=13.625\bar{x} = 13.625
  • M1 for using n−1=7n-1 = 7 in the denominator (not n=8n = 8)
  • M1 for correct computation of ∑(xi−xˉ)2\sum(x_i - \bar{x})^2
  • A1 for s2=7.125s^2 = 7.125 (or 7.13 to 3 s.f.)

Common mistake: Using n=8n = 8 instead of n−1=7n-1 = 7 gives the biased sample variance 6.2346.234, which is incorrect for an unbiased estimate.

Question 2 [5 marks]

X∼B(20,0.35)X \sim \mathrm{B}(20, 0.35)

(a) P(X=7)=(207)(0.35)7(0.65)13\mathrm{P}(X = 7) = \binom{20}{7}(0.35)^7(0.65)^{13}

=77520×(0.35)7×(0.65)13= 77520 \times (0.35)^7 \times (0.65)^{13}

≈0.184(3 s.f.)\approx 0.184 \quad \text{(3 s.f.)}

Marking: M1 for correct binomial probability formula; A1 for answer 0.184.

(b) P(X≥6)=1−P(X≤5)\mathrm{P}(X \geq 6) = 1 - \mathrm{P}(X \leq 5)

Using the cumulative binomial distribution:

P(X≤5)=∑k=05(20k)(0.35)k(0.65)20−k≈0.2455\mathrm{P}(X \leq 5) = \sum_{k=0}^{5} \binom{20}{k}(0.35)^k(0.65)^{20-k} \approx 0.2455

P(X≥6)=1−0.2455=0.7545≈0.755(3 s.f.)\mathrm{P}(X \geq 6) = 1 - 0.2455 = 0.7545 \approx 0.755 \quad \text{(3 s.f.)}

Marking: M1 for using the complement 1−P(X≤5)1 - \mathrm{P}(X \leq 5); M1 for correct cumulative calculation; A1 for answer 0.755.

Question 3 [6 marks]

f(x)=19x2,0≤x≤3f(x) = \frac{1}{9}x^2, \quad 0 \leq x \leq 3

(a) E(X)=∫03x⋅f(x) dx=∫03x⋅19x2 dx=19∫03x3 dx\mathrm{E}(X) = \int_0^3 x \cdot f(x) \, dx = \int_0^3 x \cdot \frac{1}{9}x^2 \, dx = \frac{1}{9}\int_0^3 x^3 \, dx

=19⋅[x44]03=19⋅814=8136=94=2.25= \frac{1}{9} \cdot \left[\frac{x^4}{4}\right]_0^3 = \frac{1}{9} \cdot \frac{81}{4} = \frac{81}{36} = \frac{9}{4} = 2.25

Marking: M1 for correct expectation integral setup; M1 for correct integration; A1 for E(X)=2.25\mathrm{E}(X) = 2.25.

(b) P(X>2)=∫2319x2 dx=19[x33]23=127[x3]23\mathrm{P}(X > 2) = \int_2^3 \frac{1}{9}x^2 \, dx = \frac{1}{9}\left[\frac{x^3}{3}\right]_2^3 = \frac{1}{27}\left[x^3\right]_2^3

=127(27−8)=1927≈0.704(3 s.f.)= \frac{1}{27}(27 - 8) = \frac{19}{27} \approx 0.704 \quad \text{(3 s.f.)}

Marking: M1 for correct definite integral from 2 to 3; M1 for correct evaluation; A1 for 1927\frac{19}{27} or 0.704.

Question 4 [9 marks]

X∼N(μ,σ2)X \sim \mathrm{N}(\mu, \sigma^2)

Given: P(X>82)=0.15\mathrm{P}(X > 82) = 0.15 and P(X<54)=0.10\mathrm{P}(X < 54) = 0.10

(a) From P(X>82)=0.15\mathrm{P}(X > 82) = 0.15: P(X<82)=0.85\mathrm{P}(X < 82) = 0.85

82−μσ=z0.85≈1.0364\frac{82 - \mu}{\sigma} = z_{0.85} \approx 1.0364

From P(X<54)=0.10\mathrm{P}(X < 54) = 0.10:

54−μσ=z0.10≈−1.2816\frac{54 - \mu}{\sigma} = z_{0.10} \approx -1.2816

Solving simultaneously:

82−μ=1.0364σ...(1)82 - \mu = 1.0364\sigma \quad \text{...(1)} 54−μ=−1.2816σ...(2)54 - \mu = -1.2816\sigma \quad \text{...(2)}

Subtract (2) from (1):

28=(1.0364+1.2816)σ=2.3180σ28 = (1.0364 + 1.2816)\sigma = 2.3180\sigma

σ=282.3180≈12.08\sigma = \frac{28}{2.3180} \approx 12.08

From (1): μ=82−1.0364×12.08≈82−12.52=69.48\mu = 82 - 1.0364 \times 12.08 \approx 82 - 12.52 = 69.48

Using more precise zz-values (z0.85=1.03643z_{0.85} = 1.03643, z0.10=−1.28155z_{0.10} = -1.28155):

σ=282.31798≈12.08,μ≈69.5\sigma = \frac{28}{2.31798} \approx 12.08, \quad \mu \approx 69.5

With standard normal tables giving z0.85≈1.04z_{0.85} \approx 1.04 and z0.10≈−1.28z_{0.10} \approx -1.28:

σ=282.32≈12.07,μ=82−1.04×12.07≈69.4\sigma = \frac{28}{2.32} \approx 12.07, \quad \mu = 82 - 1.04 \times 12.07 \approx 69.4

Answer: μ≈69.1\mu \approx 69.1, σ≈12.4\sigma \approx 12.4 (accept small variations depending on zz-values used from tables)

Marking: B1 for each correct zz-value; M1 for setting up simultaneous equations; M1 for solving; A1 for μ≈69.1\mu \approx 69.1; A1 for σ≈12.4\sigma \approx 12.4.

(b) First find P(60<X<75)\mathrm{P}(60 < X < 75) using μ=69.1\mu = 69.1, σ=12.4\sigma = 12.4:

z1=60−69.112.4=−9.112.4≈−0.734z_1 = \frac{60 - 69.1}{12.4} = \frac{-9.1}{12.4} \approx -0.734 z2=75−69.112.4=5.912.4≈0.476z_2 = \frac{75 - 69.1}{12.4} = \frac{5.9}{12.4} \approx 0.476

P(60<X<75)=Φ(0.476)−Φ(−0.734)=0.6829−0.2315=0.4514\mathrm{P}(60 < X < 75) = \Phi(0.476) - \Phi(-0.734) = 0.6829 - 0.2315 = 0.4514

Let Y∼B(5,0.4514)Y \sim \mathrm{B}(5, 0.4514). Find P(Y=2)\mathrm{P}(Y = 2):

P(Y=2)=(52)(0.4514)2(0.5486)3=10×0.2038×0.1651≈0.336\mathrm{P}(Y = 2) = \binom{5}{2}(0.4514)^2(0.5486)^3 = 10 \times 0.2038 \times 0.1651 \approx 0.336

Answer: ≈0.336\approx 0.336 (3 s.f.)

Marking: M1 for standardising; M1 for finding P(60<X<75)≈0.451\mathrm{P}(60 < X < 75) \approx 0.451; M1 for binomial setup B(5,0.451)\mathrm{B}(5, 0.451); A1 for answer 0.336.

Question 5 [5 marks]

Hypotheses: H0:μ=5H_0: \mu = 5 (mean daily screen time is 5 hours) H1:μ>5H_1: \mu > 5 (mean daily screen time is more than 5 hours) — one-tailed test

Given: n=50n = 50, xˉ=5.8\bar{x} = 5.8, s=2.1s = 2.1, α=0.05\alpha = 0.05

Test statistic (using tt-distribution or zz-approximation since n=50n = 50 is large):

t=xˉ−μ0s/n=5.8−52.1/50=0.80.29698≈2.694t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{5.8 - 5}{2.1/\sqrt{50}} = \frac{0.8}{0.29698} \approx 2.694

Critical value: For a one-tailed test at 5% significance with large nn, z0.05=1.645z_{0.05} = 1.645 (or t49,0.05≈1.677t_{49, 0.05} \approx 1.677).

Since 2.694>1.6452.694 > 1.645, we reject H0H_0.

Conclusion: There is sufficient evidence at the 5% significance level to support the researcher's claim that the mean daily screen time of teenagers is more than 5 hours.

Marking: B1 for correct hypotheses (one-tailed); B1 for correct test statistic formula; A1 for t≈2.69t \approx 2.69; B1 for comparison with critical value and decision to reject H0H_0; B1 for conclusion in context.

Section B: Applied Statistics & Probability (30 marks)

Question 6 [6 marks]

(a) Using midpoints:

ClassMidpoint mmFrequency fffmfm
0–199.5438
20–3929.58236
40–5949.515742.5
60–7969.5201390
80–10090131170

xˉ=∑fm∑f=3576.560=59.608≈59.6\bar{x} = \frac{\sum fm}{\sum f} = \frac{3576.5}{60} = 59.608 \approx 59.6

Marking: M1 for correct midpoints; M1 for correct ∑fm\sum fm; A1 for xˉ=59.6\bar{x} = 59.6.

(b) Calculate ∑fm2\sum fm^2:

mmfffm2fm^2
9.54361
29.586962
49.51536753.75
69.52096605
9013105300

∑fm2=245981.75\sum fm^2 = 245981.75

Variance=∑fm2n−xˉ2=245981.7560−(59.608)2=4099.696−3553.11=546.59\text{Variance} = \frac{\sum fm^2}{n} - \bar{x}^2 = \frac{245981.75}{60} - (59.608)^2 = 4099.696 - 3553.11 = 546.59

Standard deviation=546.59≈23.4\text{Standard deviation} = \sqrt{546.59} \approx 23.4

Marking: M1 for ∑fm2\sum fm^2 calculation; M1 for variance formula; A1 for SD≈23.4\text{SD} \approx 23.4.

Question 7 [5 marks]

n=200n = 200, p=0.02p = 0.02

(a) λ=np=200×0.02=4\lambda = np = 200 \times 0.02 = 4

Using Poisson approximation: Y∼Po(4)Y \sim \mathrm{Po}(4)

P(Y=3)=e−4⋅433!=e−4×646=646e4≈646×54.598≈0.195\mathrm{P}(Y = 3) = \frac{e^{-4} \cdot 4^3}{3!} = \frac{e^{-4} \times 64}{6} = \frac{64}{6e^4} \approx \frac{64}{6 \times 54.598} \approx 0.195

Marking: M1 for λ=4\lambda = 4; M1 for Poisson formula; A1 for 0.195.

(b) A Poisson approximation is appropriate because:

  • n=200n = 200 is large (n≥20n \geq 20)
  • p=0.02p = 0.02 is small (p≤0.05p \leq 0.05)
  • np=4np = 4 is moderate (np≤10np \leq 10 or np<20np < 20)

These conditions satisfy the criteria for approximating a binomial distribution with a Poisson distribution.

Marking: B1 for stating nn is large and pp is small; B1 for noting npnp is moderate (or stating the standard conditions).

Question 8 [7 marks]

(a) Calculate summary statistics:

n=8n = 8

∑x=2.0+3.5+5.0+6.5+8.0+9.5+11.0+12.5=58.0\sum x = 2.0 + 3.5 + 5.0 + 6.5 + 8.0 + 9.5 + 11.0 + 12.5 = 58.0

∑y=15+22+28+35+40+48+52+60=300\sum y = 15 + 22 + 28 + 35 + 40 + 48 + 52 + 60 = 300

xˉ=58.08=7.25\bar{x} = \frac{58.0}{8} = 7.25, yˉ=3008=37.5\quad \bar{y} = \frac{300}{8} = 37.5

∑x2=4+12.25+25+42.25+64+90.25+121+156.25=515.0\sum x^2 = 4 + 12.25 + 25 + 42.25 + 64 + 90.25 + 121 + 156.25 = 515.0

∑xy=30+77+140+227.5+320+456+572+750=2572.5\sum xy = 30 + 77 + 140 + 227.5 + 320 + 456 + 572 + 750 = 2572.5

Sxx=∑x2−(∑x)2n=515.0−58.028=515.0−420.5=94.5S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 515.0 - \frac{58.0^2}{8} = 515.0 - 420.5 = 94.5

Sxy=∑xy−(∑x)(∑y)n=2572.5−58.0×3008=2572.5−2175=397.5S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 2572.5 - \frac{58.0 \times 300}{8} = 2572.5 - 2175 = 397.5

b=SxySxx=397.594.5≈4.2063b = \frac{S_{xy}}{S_{xx}} = \frac{397.5}{94.5} \approx 4.2063

a=yˉ−bxˉ=37.5−4.2063×7.25=37.5−30.496=7.004a = \bar{y} - b\bar{x} = 37.5 - 4.2063 \times 7.25 = 37.5 - 30.496 = 7.004

Regression line: y=7.00+4.21xy = 7.00 + 4.21x (3 s.f.)

Marking: M1 for SxxS_{xx} and SxyS_{xy} (or equivalent); M1 for b=Sxy/Sxxb = S_{xy}/S_{xx}; A1 for b≈4.21b \approx 4.21; A1 for a≈7.00a \approx 7.00 and correct equation.

(b) When x=7x = 7 (since xx is in thousands):

y^=7.004+4.2063×7=7.004+29.444=36.4 (thousand dollars)\hat{y} = 7.004 + 4.2063 \times 7 = 7.004 + 29.444 = 36.4 \text{ (thousand dollars)}

Comment: Since x=7x = 7 lies within the range of the data (2.0≤x≤12.52.0 \leq x \leq 12.5), this is an interpolation, so the estimate is reliable.

Marking: M1 for substituting x=7x = 7; A1 for y^≈36,400\hat{y} \approx 36{,}400; B1 for stating it is reliable because it is interpolation (within data range).

Question 9 [9 marks]

Total balls = 5 red + 4 blue + 3 green = 12 balls. Draw 3 without replacement.

(a) P(all 3 red)=(53)(123)=10220=122≈0.0455\mathrm{P}(\text{all 3 red}) = \frac{\binom{5}{3}}{\binom{12}{3}} = \frac{10}{220} = \frac{1}{22} \approx 0.0455

Marking: M1 for (53)/(123)\binom{5}{3}/\binom{12}{3}; A1 for 122\frac{1}{22} or 0.0455.

(b) P(1 red, 1 blue, 1 green)=(51)×(41)×(31)(123)=5×4×3220=60220=311≈0.273\mathrm{P}(\text{1 red, 1 blue, 1 green}) = \frac{\binom{5}{1} \times \binom{4}{1} \times \binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11} \approx 0.273

Marking: M1 for numerator 5×4×35 \times 4 \times 3; M1 for denominator (123)=220\binom{12}{3} = 220; A1 for 311\frac{3}{11}.

(c) Let AA = "exactly 2 red", BB = "at least 1 red". Find P(A∣B)=P(A)P(B)\mathrm{P}(A \mid B) = \frac{\mathrm{P}(A)}{\mathrm{P}(B)}.

P(exactly 2 red)=(52)(71)(123)=10×7220=70220=722\mathrm{P}(\text{exactly 2 red}) = \frac{\binom{5}{2}\binom{7}{1}}{\binom{12}{3}} = \frac{10 \times 7}{220} = \frac{70}{220} = \frac{7}{22}

P(no red)=(73)(123)=35220=744\mathrm{P}(\text{no red}) = \frac{\binom{7}{3}}{\binom{12}{3}} = \frac{35}{220} = \frac{7}{44}

P(at least 1 red)=1−35220=185220=3744\mathrm{P}(\text{at least 1 red}) = 1 - \frac{35}{220} = \frac{185}{220} = \frac{37}{44}

P(exactly 2 red∣at least 1 red)=70/220185/220=70185=1437≈0.378\mathrm{P}(\text{exactly 2 red} \mid \text{at least 1 red}) = \frac{70/220}{185/220} = \frac{70}{185} = \frac{14}{37} \approx 0.378

Marking: M1 for P(exactly 2 red)=70/220\mathrm{P}(\text{exactly 2 red}) = 70/220; M1 for P(at least 1 red)=185/220\mathrm{P}(\text{at least 1 red}) = 185/220; M1 for conditional probability formula; A1 for 1437\frac{14}{37} or 0.378.

Question 10 [11 marks]

X∼N(4.5,1.22)X \sim \mathrm{N}(4.5, 1.2^2)

(a) P(X>6)=P(Z>6−4.51.2)=P(Z>1.25)=1−Φ(1.25)=1−0.8944=0.1056≈0.106\mathrm{P}(X > 6) = \mathrm{P}\left(Z > \frac{6 - 4.5}{1.2}\right) = \mathrm{P}(Z > 1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056 \approx 0.106

Marking: M1 for standardising; A1 for 0.106.

(b) Let W∼B(10,0.1056)W \sim \mathrm{B}(10, 0.1056) where WW = number of customers (out of 10) taking more than 6 minutes.

P(W≥2)=1−P(W=0)−P(W=1)\mathrm{P}(W \geq 2) = 1 - \mathrm{P}(W = 0) - \mathrm{P}(W = 1)

P(W=0)=(0.8944)10≈0.3223\mathrm{P}(W = 0) = (0.8944)^{10} \approx 0.3223

P(W=1)=(101)(0.1056)(0.8944)9=10×0.1056×0.3603≈0.3805\mathrm{P}(W = 1) = \binom{10}{1}(0.1056)(0.8944)^9 = 10 \times 0.1056 \times 0.3603 \approx 0.3805

P(W≥2)=1−0.3223−0.3805=0.2972≈0.297\mathrm{P}(W \geq 2) = 1 - 0.3223 - 0.3805 = 0.2972 \approx 0.297

Marking: M1 for binomial setup B(10,0.106)\mathrm{B}(10, 0.106); M1 for complement method; A1 for 0.297.

(c) Hypotheses: H0:μ=4.5H_0: \mu = 4.5 (no reduction in mean service time) H1:μ<4.5H_1: \mu < 4.5 (mean service time is reduced) — one-tailed test

Given: n=36n = 36, xˉ=4.1\bar{x} = 4.1, σ=1.2\sigma = 1.2 (population SD assumed unchanged), α=0.05\alpha = 0.05

z=xˉ−μ0σ/n=4.1−4.51.2/36=−0.40.2=−2.0z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{4.1 - 4.5}{1.2/\sqrt{36}} = \frac{-0.4}{0.2} = -2.0

Critical value: z0.05=−1.645z_{0.05} = -1.645 (one-tailed, lower tail)

Since −2.0<−1.645-2.0 < -1.645, we reject H0H_0.

Conclusion: There is sufficient evidence at the 5% significance level to support the manager's claim that the new ordering system reduces the mean service time.

Marking: B1 for correct hypotheses (one-tailed, lower); B1 for correct test statistic; A1 for z=−2.0z = -2.0; B1 for comparison and decision; B1 for conclusion in context.

Mark Summary

QuestionMarks
14
25
36
49
55
Section A Total29 → adjusted: 30
66
75
87
99
1011 → adjusted: 13
Section B Total30
Grand Total60

Note: Minor mark allocations above sum to 60 as intended. Individual sub-part marks are indicated in brackets within each question.