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A-Level · Statistics

S4 Statistics: Statistical Distributions (Binomial/Normal)

EDEXCEL 9MA0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Daniel is a quality inspector at a factory that produces steel bolts. From long-term records, 8% of bolts produced are defective. Daniel selects a random sample of 10 bolts for inspection. Let X be the number of defective bolts in the sample.

(a)State two conditions needed for X to be modelled by a binomial distribution.(2)

(b)Given that X ~ B(10, 0.08), calculate P(X = 2).(2)

(c)Calculate P(X ≤ 1).(2)

(Total for Question 1 is 6 marks)

Fatima is an engineer testing a large batch of light bulbs. It is known that 15% of bulbs in the batch are defective. She takes a random sample of 14 bulbs. Let X be the number of defective bulbs in the sample, so X ~ B(14, 0.15).

(a)Write down the values of E(X) and Var(X).(2)

(b)Calculate P(X ≤ 2).(2)

(c)Calculate the probability that at least 4 bulbs in the sample are defective.(2)

(Total for Question 2 is 6 marks)

Aaliyah is analysing height data. Heights of adult women in the UK, X cm, are modelled by X ~ N(163, 6.5²).

(a)Calculate P(X > 170).(3)

(b)Calculate P(155 < X < 170).(3)

(c)Give one criticism of using a normal distribution to model adult female height.(1)

(Total for Question 3 is 7 marks)

Grace is the quality control manager at a factory in Leeds that fills bags of sugar. The mass of sugar in a bag, W grams, is modelled by W ~ N(1010, σ²). It is known that 10% of bags weigh less than 1000 g.

(a)State the value of z such that P(Z < z) = 0.10.(1)

(b)Find the value of σ, giving your answer to 3 significant figures.(3)

(c)Using this value of σ, calculate the probability that a randomly selected bag weighs more than 1020 g.(2)

(d)Calculate P(995 < W < 1020).(3)

(Total for Question 4 is 9 marks)

A manufacturing engineer, Mr Whitfield, times how long a robot arm takes to complete an assembly task. The time, X seconds, is modelled by X ~ N(72, 5²).

(a)Find the value of a such that P(X > a) = 0.2005.(4)

(b)Find the value of b such that P(X < b) = 0.2005.(2)

(c)Hence, or otherwise, calculate P(b < X < a).(2)

(Total for Question 5 is 8 marks)

Dr Okafor, a lecturer, models the scores on a Sociology module test as X ~ N(μ, σ²). It is known that P(X < 60) = 0.2 and P(X > 85) = 0.1.

(a)State the two z-values used, giving each to 4 decimal places.(2)

(b)Form two equations in μ and σ.(2)

(c)Solve the equations to find the values of μ and σ, giving each to 3 significant figures.(4)

(Total for Question 6 is 8 marks)

Priya is a marketing analyst. A company claims that the probability a customer makes a purchase within a week of receiving a promotional email is 0.3. After a new campaign, Priya suspects this probability has increased. She takes a random sample of 20 customers who receive the email and finds that 10 make a purchase within a week. Let X be the number of customers, out of 20, who make a purchase, and let p be the true probability of a purchase under the new campaign.

(a)State suitable null and alternative hypotheses for Priya's test.(2)

(b)Find the critical region for a test at the 5% significance level, in the form X ≥ c, showing that your value of c is correct.(4)

(c)Using the observed value X = 10, state the conclusion of the test in context.(2)

(d)State the actual significance level of this test.(1)

(Total for Question 7 is 9 marks)

Liam works for a polling company, Meridian Polling. National figures suggest 55% of voters support a proposed policy. Liam surveys a random sample of 200 voters. Let X be the number, out of 200, who support the policy, so X ~ B(200, 0.55).

(a)Give a reason why a normal approximation to X is appropriate here.(1)

(b)Using a suitable approximation with a continuity correction, calculate P(X > 110).(4)

(c)Using a suitable approximation with a continuity correction, calculate P(100 ≤ X ≤ 120).(4)

(Total for Question 8 is 9 marks)

Mrs Chen grows pea plants on her allotment in Bristol. The probability that a planted seed germinates is 0.85, independently of other seeds. She plants 16 seeds. Let X be the number of seeds that germinate, and let L be the pod length, in cm, of a germinated plant, modelled by L ~ N(9.2, 1.1²).

(a)State the distribution of X, including its parameters.(1)

(b)Calculate the probability that all 16 seeds germinate.(2)

(c)Calculate the probability that at least 14 seeds germinate.(3)

(d)For a randomly selected germinated plant, calculate P(L > 10.5).(3)

(e)State one assumption made in modelling L using a normal distribution that may not hold in practice.(1)

(Total for Question 9 is 10 marks)

Tom works at a call centre and makes 25 sales calls each day. The probability that any call results in a sale is 0.24, independently of other calls. Let X be the number of sales Tom makes in a day, so X ~ B(25, 0.24). Tom's daily bonus, in pounds, is L = 8X + 15.

(a)Find E(X) and Var(X).(2)

(b)Find E(L) and Var(L).(4)

(c)Find the standard deviation of Tom's daily bonus, giving your answer to 3 significant figures.(2)

(Total for Question 10 is 8 marks)

Zara, a statistics student, is investigating a binomial model X ~ B(n, p) where n = 48. She is given that Var(X) = 0.75 x E(X).

(a)Show that p = 0.25.(4)

(b)State the values of np and n(1-p), and explain why a normal approximation to X would be appropriate.(2)

(c)Using the normal approximation to X, with a continuity correction, estimate P(X ≤ 8).(4)

(Total for Question 11 is 10 marks)

Mr Hussain, a maths tutor, asks his class to investigate the shape of the binomial distribution X ~ B(25, 0.4).

(a)Show that, for r = 1, 2, ..., 25, P(X=r) / P(X=r-1) = [(26-r)/r] x (2/3).(3)

(b)Hence determine the mode of X, justifying your answer.(3)

(c)Using your answer to part (a), explain, without further calculation, why P(X=25) is much smaller than P(X=10).(2)

(Total for Question 12 is 8 marks)

Mark scheme · S4 Statistics: Statistical Distributions (Binomial/Normal)

Question 1

(a) B1 there is a fixed number of trials (n=10, known in advance), oe
(a) B1 each bolt is independent with only two outcomes (defective/not defective) and a constant probability of being defective (p=0.08), oe
(a) Answer: Fixed number of trials; independent trials with constant probability of defect
(b) M1 uses P(X=2) = C(10,2) x 0.08² x 0.92⁸
(b) A1 P(X=2) = awrt 0.148
(b) Answer: P(X=2) = awrt 0.148
(c) M1 finds P(X=0) + P(X=1) = 0.92¹⁰ + 10x0.08x0.92⁹
(c) A1 P(X≤1) = awrt 0.812
(c) Answer: P(X≤1) = awrt 0.812

Question 2

(a) B1 E(X) = 14 x 0.15 = 2.1, cao
(a) B1 Var(X) = 14 x 0.15 x 0.85 = 1.785, cao
(a) Answer: E(X) = 2.1, Var(X) = 1.785
(b) M1 sums P(X=0)+P(X=1)+P(X=2) using the binomial formula or cumulative tables
(b) A1 P(X≤2) = awrt 0.648
(b) Answer: P(X≤2) = awrt 0.648
(c) M1 uses P(X≥4) = 1 - P(X≤3), with P(X=3) = C(14,3) x 0.15³ x 0.85¹¹ found and added to part (b)
(c) A1 P(X≥4) = awrt 0.147
(c) Answer: P(X≥4) = awrt 0.147

Question 3

(a) M1 standardises: z = (170-163)/6.5, oe
(a) A1 z = 1.08 (or better) and uses the normal distribution correctly
(a) A1 P(X>170) = awrt 0.140
(a) Answer: P(X>170) = awrt 0.140
(b) M1 finds both z-values: z1=(155-163)/6.5=-1.23, z2=(170-163)/6.5=1.08
(b) A1 uses P(155<X<170) = Phi(1.08) - Phi(-1.23), oe
(b) A1 P(155<X<170) = awrt 0.751
(b) Answer: P(155<X<170) = awrt 0.751
(c) B1 the normal distribution is defined for all real values (including negative values), but height cannot be negative, oe; or the true distribution may have a natural upper/lower limit not reflected by the normal model
(c) Answer: The normal model allows impossible negative heights, so it can only be an approximation

Question 4

(a) B1 z = -1.2816 (or -1.28), using the fact that P(Z<1.2816)=0.90
(a) Answer: z = -1.2816
(b) M1 forms the equation (1000-1010)/σ = -1.2816, oe
(b) M1 rearranges to σ = 10/1.2816
(b) A1 σ = awrt 7.80
(b) Answer: σ = awrt 7.80
(c) M1ft standardises: z = (1020-1010)/7.80 = 1.28 (ft their σ)
(c) A1ft P(W>1020) = awrt 0.100
(c) Answer: P(W>1020) = awrt 0.100
(d) M1ft finds both z-values: z1=(995-1010)/7.80=-1.92, z2=(1020-1010)/7.80=1.28
(d) A1ft uses P(995<W<1020) = Phi(1.28) - Phi(-1.92), oe
(d) A1ft P(995<W<1020) = awrt 0.872
(d) Answer: P(995<W<1020) = awrt 0.872

Question 5

(a) M1 recognises P(Z<z) = 1-0.2005 = 0.7995
(a) A1 z = 0.84
(a) M1 forms a = 72 + 0.84x5
(a) A1 a = 76.2
(a) Answer: a = 76.2
(b) B1 uses symmetry: z = -0.84
(b) A1 b = 72 - 0.84x5 = 67.8
(b) Answer: b = 67.8
(c) M1 uses P(b<X<a) = 1 - 2(0.2005), oe
(c) A1 P(67.8<X<76.2) = 0.599, cao
(c) Answer: P(67.8<X<76.2) = 0.599

Question 6

(a) B1 z1 = -0.8416, since P(Z< 0.8416)=0.80
(a) B1 z2 = 1.2816, since P(Z<1.2816)=0.90
(a) Answer: z1 = -0.8416, z2 = 1.2816
(b) M1 60 = μ - 0.8416 σ, oe
(b) M1 85 = μ + 1.2816 σ, oe
(b) Answer: μ - 0.8416sigma = 60 and μ + 1.2816sigma = 85
(c) M1 subtracts the equations to eliminate μ: 25 = 2.1232 σ
(c) dM1 solves for σ = 25/2.1232
(c) A1 σ = awrt 11.8
(c) A1ft μ = awrt 69.9 (ft their σ substituted into either equation)
(c) Answer: μ = awrt 69.9, σ = awrt 11.8

Question 7

(a) B1 H0: p = 0.3
(a) B1 H1: p > 0.3
(a) Answer: H0: p=0.3, H1: p>0.3
(b) M1 under H0, X ~ B(20, 0.3), and attempts P(X≥k) for values of k near the upper tail
(b) A1 P(X≥10) = awrt 0.0479 (< 0.05)
(b) A1 P(X≥9) = awrt 0.1133 (> 0.05)
(b) A1 critical region is X ≥ 10, since this is the smallest region with probability ≤ 0.05, cso
(b) Answer: Critical region: X ≥ 10
(c) B1 X=10 lies in the critical region (X≥10), so H0 is rejected
(c) B1 there is sufficient evidence, at the 5% level, that the proportion of customers making a purchase has increased since the new campaign, cao
(c) Answer: Reject H0; evidence that the purchase proportion has increased
(d) B1 actual significance level = P(X≥10) = awrt 0.0479 (4.79%)
(d) Answer: 0.0479 (4.79%)

Question 8

(a) B1 n is large and both np=110>5 and n(1-p)=90>5, so a normal approximation is valid, oe
(a) Answer: np=110>5 and n(1-p)=90>5, so normal approximation is valid
(b) B1 states Y ~ N(110, 49.5), since np=110, npq=200x0.55x0.45=49.5
(b) M1 applies continuity correction: P(X>110) approx P(Y>110.5)
(b) M1 standardises: z = (110.5-110)/√49.5 = 0.07
(b) A1 P(X>110) = awrt 0.472
(b) Answer: P(X>110) = awrt 0.472
(c) M1 applies continuity correction at both ends: P(99.5 < Y < 120.5)
(c) M1 standardises both values: z1=(99.5-110)/7.036=-1.49, z2=(120.5-110)/7.036=1.49
(c) A1 uses P(100≤X≤120) approx 2Phi(1.49)-1, oe
(c) A1 P(100≤X≤120) = awrt 0.864
(c) Answer: P(100≤X≤120) = awrt 0.864

Question 9

(a) B1 X ~ B(16, 0.85)
(a) Answer: X ~ B(16, 0.85)
(b) M1 uses P(X=16) = 0.85¹⁶
(b) A1 P(X=16) = awrt 0.0743
(b) Answer: P(X=16) = awrt 0.0743
(c) M1 identifies P(X≥14) = P(X=14)+P(X=15)+P(X=16)
(c) M1 computes each term correctly using the binomial formula
(c) A1 P(X≥14) = awrt 0.561
(c) Answer: P(X≥14) = awrt 0.561
(d) M1 standardises: z = (10.5-9.2)/1.1
(d) A1 z = 1.18
(d) A1 P(L>10.5) = awrt 0.119
(d) Answer: P(L>10.5) = awrt 0.119
(e) B1 assumes pod length is symmetric about the mean and can theoretically take any value (including implausibly large or negative lengths), which may not be realistic, oe
(e) Answer: Pod length cannot realistically be negative or unbounded, but the normal model allows this

Question 10

(a) B1 E(X) = 25 x 0.24 = 6, cao
(a) B1 Var(X) = 25 x 0.24 x 0.76 = 4.56, cao
(a) Answer: E(X) = 6, Var(X) = 4.56
(b) M1 uses E(L) = 8E(X) + 15
(b) A1 E(L) = 63, cao
(b) M1 uses Var(L) = 8² Var(X)
(b) A1 Var(L) = 291.84, cao
(b) Answer: E(L) = 63, Var(L) = 291.84
(c) M1 uses SD(L) = √Var(L)
(c) A1 SD(L) = awrt 17.1 (pounds)
(c) Answer: SD(L) = awrt GBP17.1

Question 11

(a) M1 states E(X) = 48p
(a) M1 states Var(X) = 48p(1-p)
(a) M1 forms 48p(1-p) = 0.75 x 48p and divides both sides by 48p (p not 0) to get 1-p = 0.75
(a) A1 p = 0.25, cso
(a) Answer: p = 0.25
(b) B1 np = 12 and n(1-p) = 36
(b) B1 both values are greater than 5, so X can be well approximated by a normal distribution, oe
(b) Answer: np=12, n(1-p)=36; both exceed 5, so normal approximation is appropriate
(c) B1 states Y ~ N(12, 9), since Var(X)=9 from part (a)
(c) M1 applies continuity correction: P(X≤8) approx P(Y<8.5)
(c) M1 standardises: z = (8.5-12)/3 = -1.17
(c) A1 P(X≤8) = awrt 0.121
(c) Answer: P(X≤8) = awrt 0.121

Question 12

(a) M1 writes the ratio of binomial coefficients C(25,r)/C(25,r-1) = (25-r+1)/r = (26-r)/r
(a) M1 writes the ratio of probability terms as (p/(1-p)) = 0.4/0.6 = 2/3
(a) A1 combines to give P(X=r)/P(X=r-1) = [(26-r)/r] x (2/3), cso
(a) Answer: P(X=r)/P(X=r-1) = [(26-r)/r] x (2/3)
(b) M1 sets [(26-r)/r] x (2/3) ≥ 1 and solves to find r ≤ 10.4
(b) A1 deduces P(X=r) > P(X=r-1) for r ≤ 10, and P(X=r) < P(X=r-1) for r ≥ 11
(b) A1 mode of X = 10, cao
(b) Answer: Mode = 10
(c) B1 for r>10.4 the ratio [(26-r)/r]x(2/3) is less than 1, so probabilities decrease steadily as r increases from 11 to 25
(c) B1 at r=25 the ratio [(26-25)/25]x(2/3) = 2/75, a very small factor, so after many successive decreasing multiplications P(X=25) is far smaller than the peak value P(X=10), oe
(c) Answer: The ratio [(26-r)/r]x(2/3) falls well below 1 for large r (e.g. 2/75 at r=25), so repeated multiplication by small factors makes P(X=25) far smaller than P(X=10)

Statistics: Statistical Distributions (Binomial/Normal) - Worksheets, Questions and Revision