# Thursday 22 May 2014 Morning

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2 2 Section A (36 marks) 1 The ages, x years, of the senior members of a running club are summarised in the table below. Age (x) 20 G x G x G x G x G x G x G x 1 90 Frequency (i) Draw a cumulative frequency diagram to illustrate the data. [5] (ii) Use your diagram to estimate the median and interquartile range of the data. [3] 2 Candidates applying for jobs in a large company take an aptitude test, as a result of which they are either accepted, rejected or retested, with probabilities 0.2, 0.5 and 0.3 respectively. When a candidate is retested for the first time, the three possible outcomes and their probabilities remain the same as for the original test. When a candidate is retested for the second time there are just two possible outcomes, accepted or rejected, with probabilities 0.4 and 0.6 respectively. (i) Draw a probability tree diagram to illustrate the outcomes. [3] (ii) Find the probability that a randomly selected candidate is accepted. [2] (iii) Find the probability that a randomly selected candidate is retested at least once, given that this candidate is accepted. [3] 3 Each weekday, Marta travels to school by bus. Sometimes she arrives late. L is the event that Marta arrives late. R is the event that it is raining. You are given that P( L) = 0. 15, P( R) = and P( L ; R) = (i) Use this information to show that the events L and R are not independent. [1] (ii) Find P( L + R). [2] (iii) Draw a Venn diagram showing the events L and R, and fill in the probability corresponding to each of the four regions of your diagram. [3] 4 There are 16 girls and 14 boys in a class. Four of them are to be selected to form a quiz team. The team is to be selected at random. (i) Find the probability that all 4 members of the team will be girls. [3] (ii) Find the probability that the team will contain at least one girl and at least one boy. [3] OCR 2014 G241/01 Jun14

3 5 The probability distribution of the random variable X is given by the formula 3 P( X = r) = k r for r = 1, 2, 3, 4, 5. 2 (i) Show that k = Using this value of k, display the probability distribution of X in a table. [3] (ii) Find E ( X ) and V ar( X ). [5] Section B (36 marks) 6 The weights, w grams, of a random sample of 60 carrots of variety A are summarised in the table below. Weight 30 G w G w G w G w G w 1 90 Frequency (i) Draw a histogram to illustrate these data. [5] (ii) Calculate estimates of the mean and standard deviation of w. [4] (iii) Use your answers to part (ii) to investigate whether there are any outliers. [3] The weights, x grams, of a random sample of 50 carrots of variety B are summarised as follows. 2 = n = 50 / x = / x (iv) Calculate the mean and standard deviation of x. [3] (v) Compare the central tendency and variation of the weights of varieties A and B. [2] 7 It is known that on average 85% of seeds of a particular variety of tomato will germinate. Ramesh selects 15 of these seeds at random and sows them. (i) (A) Find the probability that exactly 12 germinate. [3] (B) Find the probability that fewer than 12 germinate. [2] The following year Ramesh finds that he still has many seeds left. Because the seeds are now one year old, he suspects that the germination rate will be lower. He conducts a trial by randomly selecting n of these seeds and sowing them. He then carries out a hypothesis test at the 1% significance level to investigate whether he is correct. (ii) Write down suitable null and alternative hypotheses for the test. Give a reason for your choice of alternative hypothesis. [4] (iii) In a trial with n = 20, Ramesh finds that 13 seeds germinate. Carry out the test. [4] (iv) Suppose instead that Ramesh conducts the trial with n = 50, and finds that 33 seeds germinate. Given that the critical value for the test in this case is 35, complete the test. [3] (v) If n is small, there is no point in carrying out the test at the 1% significance level, as the null hypothesis cannot be rejected however many seeds germinate. Find the least value of n for which the null hypothesis can be rejected, quoting appropriate probabilities to justify your answer. [3] OCR 2014 G241/01 Jun14

6 2 Section A (36 marks) 1 (i) & (ii) OCR 2014

7 3 2 (i) 2 (ii) 2 (iii) OCR 2014 Turn over

8 4 3 (i) 3 (ii) 3 (iii) OCR 2014

9 5 4 (i) 4 (ii) OCR 2014 Turn over

10 6 5 (i) & (ii) OCR 2014

11 7 Section B (36 marks) 6 (i) OCR 2014 Turn over

12 8 6 (ii) 6 (iii) OCR 2014

13 9 6 (iv) 6 (v) OCR 2014 Turn over

14 10 7 (i) (A) 7 (i) (B) 7 (ii) OCR 2014

15 11 7 (iii) 7 (iv) OCR 2014 Turn over

17 G241 Mark Scheme June 2014 Question Answer Marks Guidance 1 (i) Cumulative frequencies Upper Bound B1 All correct Cumulative Freq G1 G1 G1 For plotted points (Provided plotted at correct UCB positions) For joining points (within ½ a square) For scales Plotted within ½ small square If cf not given then allow G1 for good attempt at cf. e.g. if they have 0,10,40,72,95,104,109,110 G1 For labels [5] All marks dep on good attempt at cumulative frequency, but not cumulative fx s or other spurious values. 1

18 G241 Mark Scheme June 2014 Question Answer Marks Guidance 1 (ii) Median = 45 B1 Allow answers between 44 Based on 60 th value and 46 without checking curve. Otherwise check curve. No marks if not using diagram. ft their curve (not LCB s) Allow 40 for m.p. plot without checking graph B0 for interpolation If max value wrong (eg 110) FT Q1 = 37 Q3 = 53 B1 For Q3 or Q1 Allow Q1 between 37 and 38 without checking Allow Q3 between 52 and 54 without checking Inter-quartile range = = 16 B1 For IQR providing both Q1 and Q3 are correct [3] 2 (i) Alternative version of tree Accept Accept diagram for Q2(i) Retest Retest 0.4 Accept their max value for all 3 marks Based on 30 th and 90 th values ft their curve (not LCB s) Allow Q1 = 32; Q3 = 48 without checking graph B0 for interpolation B2 for correct IQR from graph if quartiles not stated but indicated on graph Allow from mid-point plot Must be good attempt at cumulative frequency in part (i) to score any marks here Lines of best fit: B0 B0 B0 here. Also cumulative frequency bars: B0 B0 B0 here 0.5 Reject 0. 5 Reject 0.6 Reject 2

19 G241 Mark Scheme June 2014 Question Answer Marks Guidance 2 (i) Allow labels such as A, R, F(Fail) etc Accept Do a vertical scan and give: 0.2 G1 First column All probabilities correct Reject Retest Accept 0. 3 Retest Reject Accept Reject G1 Second column All probabilities correct G1 Final column Do not award if first two branches missing Branches two and three should come out of retest All probabilities correct If any labels missing or incorrect allow max 2/3 Do not allow misreads here as all FT (eg 0.3 and 0.5 reversed) [3] 2 (ii) P(Accepted) = ( ) + ( ) M1 For second or third product FT their tree provided correct numbers of terms and correct structure of 3, 3, 2 branches. = = A1 CAO Allow 37/125 oe [2] 2 (iii) P(At least one retest given accepted) = P(At least one retest and accepted) P(Accepted) M1 For numerator FT their tree provided correct numbers of terms and correct structure of 3, 3, 2 branches. for both M1 s = = M1 For denominator Both must be part of a fraction Allow 12/125 oe = A1 FT their and Allow 0.32 with working [3] Allow 12/37 oe 3

20 G241 Mark Scheme June 2014 Question Answer Marks Guidance 3 (i) Because P(L R) P(L) E1 If two or more methods given and only one correct, do not award the mark Either P(L R) (= 0.099) P(L) P(R), provided in (ii) or (=0.033) Allow Look out for complement methods, etc [1] 3 (ii) P(L R) = P(L R) P(R) = M1 For product Allow if done correctly in part(i) = A1 CAO Allow 99/1000 [2] 3 (iii) G1 For two labelled intersecting circles, L R provided no incorrect labelling. No need for box G1 For at least 2 correct probabilities FT their P(L R) from part (ii) provided 0.15 G1 [3] For remaining probabilities. FT their P(L R) providing probabilities between 0 and 1. Condone labels such as P(L) etc Allow other shapes in place of circles FT from in (ii) gives 0.117, 0.033, 0.187, In general 0.15 x, x, 0.22 x, x May also see , , ,

21 G241 Mark Scheme June 2014 Question Answer Marks Guidance 4 (i) M1 For P(All four are girls) = M = A1 CAO For product of other three correct fractions Without extra terms Allow with working but not 0.07 OR / M1 for either term in correct position in a fraction M1 for correct fraction A1 CAO Allow full marks for unsimplified fractional answers SC2 for [3] SC2 for /

22 G241 Mark Scheme June 2014 Question Answer Marks Guidance 4 (ii) P(All four are boys) = M1 For P(All four are boys) OR Without extra terms / M1 for this then as per scheme. P(At least one girl and at least one boy) = 1 ( ) M1 For 1 ( ) FT their sensible' probabilities = A1 CAO Allow answer from use of NB Watch for ( ) ( ) = = Gets just M1 for [3] Accept 0.90 work working, but not 0.9 OR P(3b,1g) + P(2b,2g) +P(1b,3g) = M1 For any one product, even if coefficient missing Or 14 C 3 16 C 1 / 30 C C 2 16 C 2 / 30 C C 1 16 C 3 / 30 C 4 = = M1 for any one term = M1 For sum of all three (all correct) M1 for sum of all three (all correct) = A1 CAO A1 CAO [3] 6

23 G241 Mark Scheme June 2014 Question Answer Marks Guidance 5 (i) k k k k k = 1 M1 For equation in k 5k = 1 k = 0.09 A1 NB Answer Given Allow substitution of k = 0.09 to show probabilities add to 1 with convincing working r P(X= r) [3] 5 (ii) E(X) = (1 0.1) (2 0.13) (3 0.18) (4 0.25) (5 0.34) M1 For Σrp (at least 3 terms correct Provided 5 reasonable probabilities seen. = 3.6 A1 CAO E(X 2 ) = (1 0.1) (4 0.13) (9 0.18) ( ) ( ) Var(X) = M1* dep = 1.78 A1 CAO B1 Complete correct table Must tabulate probabilities, though may be seen in part(ii) M1* For Σr 2 p (at least 3 terms correct) [5] for their (E[X])² FT their E(X) provided Var( X ) > 0 If probs wrong but sum = 1 allow max M1A0M1M1A1. If sum 1 allow max M1A0M1M0A0 (provided all probabilities 0 and <1) No marks if all probs =0.2. Use of E(X-µ) 2 gets M1 for attempt at (x-µ) 2 should see (- 2.6) 2, (-1.6) 2, (-0.6) 2, 0.4 2, 1.4 2, (if E(X) wrong FT their E(X)) (all 5 correct for M1), then M1 for Σp(x-µ) 2 (at least 3 terms correct with their probabilities) Division by 5 or other spurious value at end and/or rooting final answer gives max M1A1M1M1A0, or M1A0M1M1A0 if E(X) also divided by 5. Unsupported correct answers get 5 marks (Probably from calculator) 7

24 G241 Mark Scheme June 2014 Question Answer Marks Guidance 6 (i) M1 can be also be gained from Weight Frequency Group Width Frequency density freq per , 10, 18, 14, 7 For fd s - at least 3 correct (at least 3 correct) or similar. 30 w < Accept any suitable unit for If fd not explicitly given, M1 50 w < M1 fd such as eg freq per 10g. A1 can be gained from all 60 w < heights correct (within half a 70 w < square) on histogram (and M1A0 80 w < if at least 3 correct) A1 G1 linear scales on both axes and labels Vertical scale starting from zero (not broken - but can get final mark for heights if broken) Linear scale and label on vertical axis IN RELATION to first M1 mark ie fd or frequency density or if relevant freq/10, etc (NOT eg fd/10). However allow scale given as fd 10, or similar. Accept f/w or f/cw (freq/width or freq/class width) Ignore horizontal label Can also be gained from an accurate key G0 if correct label but not fd s. G1 width of bars Must be drawn at 30, 50 etc NOT 29.5 or 30.5 etc NO GAPS ALLOWED Must have linear scale. No inequality labels on their own such as 30 W<50, 50 W<60 etc but allow if 30, 50, 60 etc occur at the correct boundary position. See additional notes. Allow this mark even if not using fd s 8

25 G241 Mark Scheme June 2014 Question Answer Marks Guidance Height of bars must be linear vertical scale. FT of heights dep on at least 3 heights correct and all must agree with their fds G1 [5] height of bars If fds not given and at least 3 heights correct then max M1A0G1G1G0 Allow restart with correct heights if given fd wrong (for last three marks only) 9

26 G241 Mark Scheme June 2014 Question Answer Marks Guidance 6 (ii) Mean = M1 For midpoints For midpoints (at least 3 correct) (40 11) (55 10) (65 18) (75 14) (85 7) 3805 Products are 440, 550, No marks for mean or sd unless , 1050, 595 using midpoints = 63.4 (or 63.42) A1 CAO (exact answer ) x 2 f = S xx (40 11) (55 10) (65 18) (75 14) (85 7) M1 For attempt at S xx Should include sum of at least 3 correct multiples fx 2 Ʃ x 2 /n Answer must NOT be left as improper fraction as this is an estimate Accept correct answers for mean and sd from calculator even if eg wrong Sxx given Allow M1 for anything which rounds to s A1 At least 1dp required Use of mean 63.4 leading to answer of with S xx = gets full credit. Allow SC1 for RMSD 14.1 ( ) from calculator. Only penalise once in part (ii) for over specification, even if mean and standard deviation both over specified leads to Do not FT their incorrect mean (exact answer ) If using (x x ) 2 method, B2 if 14.2 or better (14.3 if use of 63.4), otherwise B0 [4] 10

27 G241 Mark Scheme June 2014 Question Answer Marks Guidance 6 (iii) x 2s = 63.4 (2 14.2) = 35 M1 For either 6 (iv) No marks in (iii) unless using x + 2s or x 2s Only follow through numerical values, not variables such as s, so if a candidate does not find s but then writes here limit is standard deviation, do NOT award M1 x + 2s = (2 14.2) = 91.8 A1 For both (FT) Do not penalise for overspecification So there are probably some outliers at the lower end, but none at the upper end E1 Must have correct limits to get this mark Mean = S xx = = 72.5g (or exact answer 72.49g) = 2676 [3] B1 Must include an element of doubt and must mention both ends CAO Ignore units M1 For S M1 for their xx mean 2 BUT NOTE M0 if their S xx < 0 s = = = 7.39g A1 [3] CAO ignore units Allow 7.4 but NOT 7.3 (unless RMSD with working) For s 2 of 54.6 (or better) allow M1A0 with or without working. For RMSD of 7.3 (or better) allow M1A0 provided working seen For RMSD 2 of 53.5 (or better) allow M1A0 provided working seen 11

28 G241 Mark Scheme June 2014 Question Answer Marks Guidance 6 (v) Variety A have lower average than Variety B oe E1 FT their means Do not condone lower central tendency or lower mean 7 (i) (A) X ~ B(15, 0.85) Allow on the whole or similar in place of average. Variety A have higher variation than Variety B oe E1 FT their sd Allow more spread or similar but not higher range or higher variance Condone less consistent. P(exactly 12 germinate) = = A1 CAO OR OR [2] M1 M1 For For from tables: M2 For = A1 CAO [3] 7 (i) (B) P(X <12) = P(X 11) = M1 For P(X 11) or P( 11) (With no extras) A1 CAO (as final answer) May see alternative method: = their wrong answer to part (i) scores M1A0 [2] p q

29 G241 Mark Scheme June 2014 Question Answer Marks Guidance 7 (ii) Let p = probability of a seed germinating (for the population) B1 For definition of p See below for additional notes H 0 : p = 0.85 B1 For H 0 H 1 : p < 0.85 B1 For H 1 H 1 has this form because the test is to investigate whether the proportion of seeds which germinate is lower. E1 Dep on < 0.85 used in H 1 Do not allow just Germination rate will be lower or similar. [4] 7 (iii) Let X ~ B(20, 0.85) M1* For probability (provided P(X 13) = not as part of finding P(X=13)) Ignore notation > 1% M1* For comparison dep So not enough evidence to reject H 0. A1 * For not significant oe Not significant. Conclude that there is not enough evidence to indicate that the E1 * For conclusion in context proportion of seeds which have germinated has decreased. dep Must mention decrease, not just change For use of 0.15 as P(not germinating), contact team leader E0 for simply stating H 1 in words No further marks if point probs used - P(X = 13) = DO NOT FT wrong H 1, but see extra notes Allow accept H 0 or reject H 1 Must include sufficient evidence or something similar such as to suggest that ie an element of doubt either in the A or E mark. ALTERNATIVE METHOD follow method above unless some mention of CR seen Critical region method LOWER TAIL P(X 13) = > 1% P(X 12) = < 1% No marks if CR not justified Condone {0,1,2,, 12}, X 12, oe but not P(X 12) etc M1 For either probability Could get M1A0A1E1 if poor notation for CR Do not allow just 13 not in CR So critical region is {0,1,2,3,4,5,6,7,8,9,10,11,12} A1 cao dep on at least one correct comparison with 1% 13 not in CR so not significant A1 * - Must say not significant or accept H 0 or similar There is insufficient evidence to indicate that the proportion of seeds which have germinated has decreased. E1 * dep [4] 13

30 G241 Mark Scheme June 2014 Question Answer Marks Guidance 7 (iv) 33 < 35 M1 For comparison Allow 33 lies in the CR So there is sufficient evidence to reject H 0 A1* Must include sufficient evidence or something similar such as to suggest that ie an element of doubt either in the A or E mark. Conclude that there is enough evidence to indicate that the proportion of seeds which have germinated has decreased. E1* dep For conclusion in context Must mention decrease, not just change Do not FT wrong H 1 : In part (iv) ignore any interchanged H 0 and H 1 seen in part (ii) If use a calculator to find P(X 33) = and compare with 1% then B2 for P(X 33) = < 0.01 so reject H 0 then final E1 as per scheme. [3] 7 (v) For n = 3, P(X 0) = < 0.01 M1 For P(X 0) = Allow For n = 2, P(X 0) = > 0.01 M1 For P(X 0) = So the least value of n for which the critical region is not empty and thus H 0 could be rejected is 3. A1 CAO Condone just n = 3 for final A mark dep on both M marks If wrong H 1 allow max M2A0 if correct probabilities seen. ALTERNATIVE METHOD using logs 0.15 n < 0.01 n > log 0.01 / log 0.15 n > Least n = 3 M1 M1 A1 [3] 14

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