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Cambridge AS & A Level Mathematics (9709), Paper 7, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s19_qp_729709_s19_ms_729709/72/M/J/19
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
The random variable X has the distribution Po 5. (i) Find P X = 2. [1] It is given that P X = n = P X = n + 1. (ii) Write down an equation in n. [1] (iii) Hence or otherwise find the value of n. [1]
Mark scheme
1(i) 0.0842 (3 sf) B1 1 1(ii) 5n 5n +1 B1 5n 5n +1 e-5 x = e-5 x or = or better n! ( n + 1)! n! ( n + 1)! ISW 1 1(iii) 5 B1 1= n +1 n=4 1
Question
The random variable X has mean 372 and standard deviation 54. (i) Describe fully the distribution of the mean of a random sample of 36 values of X . [3] (ii) The distribution in part (i) might be either exact or approximate. State a condition under which the distribution is exact. [1]
Mark scheme
2(i) Normal with mean 372 B1 54 M1 542 sd = or variance = M1 36 36 (= 9) A1 (= 81) A1 3 2(ii) Pop normal B1 Allow X is normal 1
Question
It is claimed that, on average, a particular train journey takes less than 1.9 hours. The times, t hours, taken for this journey on a random sample of 50 days were recorded. The results are summarised below. n = 50 Σ t = 92.5 Σ t2 = 175.25 (i) Calculate unbiased estimates of the population mean and variance. [3] (ii) Test the claim at the 5% significance level. [5]
Mark scheme
3(i) Est(µ) = 1.85 B1 50 175.25 M1 50 175.25 Est(σ2) = − '1.85'2 Allow − '1.85'2 or 0.0290 for M1 49 50 49 150 33 A1 Cao = 0.0842 (3 sf) or 50 392 If omitted (giving var = 0.0825 or sd = 0.287) M0A0 49 3 3(ii) H0: Pop mean time = 1.9 (h) B1 H1: Pop mean time < 1.9 (h) Allow ‘µ’ but not just ‘mean’ 1.85 − 1.9 M1 1.85 − 1.9 ± ± Accept totals method (92.5–95) / 4.21 ' 0.0842 ' ' 0.290 ' 50 50 = –1.22 A1 = –1.22 comp z = –1.645 M1 Or other valid comparison 0.888 or 0.889 < 0.95 OR 0.111 or 0.112>0.05 No evidence that mean time < 1.9 h A1 FT their z. Correct conclusion. No contradictions If 50 49 not used in (1): var = 0.8225, sd = 0.907, cr = 1.17 can score all marks in (ii) Note- 2 tail test can score B0 M1 A1 M1 (comparison with 1.96) A0 (no ft) max3/5 5
Question
The heights of a certain variety of plant are normally distributed with mean 110 cm and variance 1050 cm2 . Two plants of this variety are chosen at random. Find the probability that the height of one of these plants is at least 1.5 times the height of the other. [7]
Mark scheme
4 Use of 1.5X1 – X2 or similar B1 E(1.5X1–X2) =1.5(110) –110 (= 55) B1 or E(X1–1.5X2) =110 – 1.5(110) (= –55) Var(1.5X1–X2) =1.52×1050+1050 (or 3412.5) M1 Correct expression or result 0−55 or 0−( −55) (= ± 0.942) M1 Their ‘55’. Allow incorrect var (dep > 0 and ≠ 1050) '3412.5' '3412.5' 1 – Φ(‘0.942’) M1 Area consistent with their working = 0.173 A1 Ans 0.346 (3 sf) B1 FT double their prob (must be <1) 7
Question
The manufacturer of a certain type of biscuit claims that 10% of packets include a free offer printed on the packet. Jyothi suspects that the true proportion is less than 10%. He plans to test the claim by looking at 40 randomly selected packets and, if the number which include the offer is less than 2, he will reject the manufacturer’s claim. (i) State suitable hypotheses for the test. [1] (ii) Find the probability of a Type I error. [3] On another occasion Jyothi looks at 80 randomly selected packets and finds that exactly 6 include the free offer. (iii) Calculate an approximate 90% confidence interval for the proportion of packets that include the offer. [3] (iv) Use your confidence interval to comment on the manufacturer’s claim. [1]
Mark scheme
5(i) H0: p = 0.1 B1 H1: p < 0.1 1 5(ii) B(40, 0.1) stated or implied by use of B1 e.g. by 40Cx or 0.9p×0.1q (p + q = 40) 0.940 + 40×0.939×0.1 M1 Correct working (if seen). If working not seen, M1 may be implied by 0.0805 = 0.0805 A1 3 5(iii) z = 1.645 B1 seen 6 ×(80−6) M1 Formula of correct form. Must be a ‘z’ 6 ± z 80 80 80 80 = 0.0266 to 0.123 (3 sfs) A1 Allow 0.03 to 0.12 or better Must be an interval 3 5(iv) 10% (or manufacturer’s claim) is within CI B1 FT Allow ‘10% is within CI, accept claim’ oe Hence no reason to question claim Must include both parts. No contradictions. FT their CI Note if CI is centred on 0.1 allow ft 0.075 is within CI, accept claim 1
Question
X is a random variable with probability density function given by d a 1 ≤ x ≤ b, f x = x2 0 otherwise, where a and b are constants. a (i) Show that b = . [3] a−1 (ii) Given that the median of X is 23 , find the values of a and b. [3] (iii) Use your values of a and b from part (ii) to find E X . [3]
Mark scheme
6(i) b M1 a ∫1 1 dx = 1 x2 Attempt int f(x) and = 1, ignore limits b A1 correct integ and limits = 1 a[ − 1x ] =1 1 a[1 – b1 ] = 1 or a× bb−1 = 1 A1 No errors seen a AG b = a−1 3 6(ii) 3 2 M1 a ∫1 x1 dx = 12 2 Attempt int f(x) with limits 1 to 32 and = 12 3 a [ − 1x ] 2 = 12 1 a [1 – 23 ] = 12 A1 oe correct equn in a a = 32 , b = 3 A1 Both 3 6(iii) 3 M1 Attempt int xf(x), ignore limits – condone missing a 3 2 ∫1 1 dx x 3 A1 FT Correct integ and their limits 1 to b – condone missing a = 32 [ln x] 1 = 32 ln 3 or 1.65 (3 sf) A1 FT their a and b (valid b i.e. >1) 3
Question
All the seats on a certain daily flight are always sold. The number of passengers who have bought seats but fail to arrive for this flight on a particular day is modelled by the distribution B 320, 0.005. (i) Explain what the number 320 represents in this context. [1] (ii) The total number of passengers who have bought seats but fail to arrive for this flight on 2 randomly chosen days is denoted by X . Use a suitable approximating distribution to find P 2 < X < 6. [3] (iii) Justify the use of your approximating distribution. [2] After some changes, the airline wishes to test whether the mean number of passengers per day who fail to arrive for this flight has decreased. (iv) During 5 randomly chosen days, a total of 2 passengers failed to arrive. Carry out the test at the 2.5% significance level. [5] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(i) Max no. of passengers plane can take oe B1 oe e.g. No of passengers who bought tickets 1 7(ii) λ = 3.2 B1 e-3.2( 3.2 3 + 3.2 4 + 3.2 5 ) M1 Any λ. Allow one end error 3! 4! 5! = 0.5146 = 0.515 (3 sfs) A1 SR Use of Bin(640,0.005) scores B1 (only) for 0.516 3 7(iii) n > 50 B1 Accept n is large np = 1.6, which is < 5 or p=0.005 which is <0.1 B1 Allow np = 3.2 2 7(iv) H0: Pop mean (for 5 days) = 8 B1 or Pop mean (for 1 day) = 1.6 H1: Pop mean (for 5 days) < 8 Pop mean (for 1 day) < 1.6 Allow λ or µ but not just ‘mean’ 2 e-8(1 + 8 + 82! ) M1 Any λ (≠ 1.6) No end errors. Accept use of Bin(1600,0.005) P(0,1,2)=0.0136 = 0.0138 A1 Comp 0.025 M1 Valid comparison Evidence that mean no. failing to arrive has A1 FT their ‘0.0138’ or ‘0.0136’. No contradictions decreased 5
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Mathematics 9709/72 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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