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Mathematics 9709/51 October/November 2022: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_w22_qp_519709_w22_ms_519709/51/O/N/22

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/51/O/N/22
9709_w22_ms_51.pdf

9709/51 mark scheme, question by question

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 14 marks

Question

The probability distribution table for a random variable X is shown below. x −2 −1 0.5 1 2 P X = x 0.12 p q 0.16 0.3 Given that E X = 0.28, find the value of p and the value of q. [4]

Mark scheme

1 0.12 + p + q + 0.16 + 0.3 = 1 B1 Sum of probabilities = 1 p +q = 0.42 OE. −0.24 − p + 0.5q + 0.16 + 0.6 = 0.28 B1 Form equation using E(X) = 0.28 −p + 0.5q = −0.24 OE. Accept unsimplified. Attempt to solve their two equations in p and q M1 Either Substitution method to form a single equation in either p or q and finding values for both unknowns. Or Elimination method by writing both equations in the same form (usually ap + bq = c) and + or – to find an equation in one unknown and finding values for both unknowns. q = 0.12, p = 0.3 A1 CAO, both WWW. If M0 awarded SC B1 for both correct WWW. 4

Question 28 marks

Question

The residents of Persham were surveyed about the reliability of their internet service. 12% rated the service as ‘poor’, 36% rated it as ‘satisfactory’ and 52% rated it as ‘good’. A random sample of 8 residents of Persham is chosen. (a) Find the probability that more than 2 and fewer than 8 of them rate their internet service as poor or satisfactory. [3] A random sample of 125 residents of Persham is now chosen. (b) Use an approximation to find the probability that more than 72 of these residents rate their internet service as good. [5]

Mark scheme

2(a) [P(3, 4, …7) = 1 – P(0, 1, 2, 8)] M1 One term 8Cx p x (1 − p )8− x , for 0 < x < 8, 0 < p < 1 = 1 − ( 8C0 0.480 0.528 + 8C1 0.481 0.527 + 8C2 0.482 0.526 + 8C8 0.488 0.520 ) = 1 – (0.00534597 + 0.039478 + 0.127544 + 0.0028179) A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.825 B1 Mark the final answer at the most accurate value. 0.8248 < p ⩽ 0.825 WWW. Alternative method for Question 2(a) [P(3, 4, 5, 6, 7) =] M1 One term 8Cx p x (1 − p )8− x , for 0<x<8, 0<p<1 8 C3 0.483 0.525 + 8C4 0.484 0.524 + 8C5 0.485 0.523 + 8C6 0.486 0.522 + 8C7 0.487 0.521 A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.825 B1 Final answer 0.8248 < p ⩽ 0.825 WWW. 3 2(b) [Mean = 0.52 ×125 =]65, B1 65 and 31.2 seen, allow unsimplified. May be seen in [ var = 0.52 × 0.48 × 125 =] 31.2 standardisation formula. (5.585 < σ ⩽ 5.586 imply correct variance). 72.5 − 65 M1 Substituting their 65 and their 31.2 into ±standardisation [P(X > 72) = ]P( Z > ) [= P( Z > 1.343 )] 31.2 formula (any number for 72∙5), not their 31.2, √ their 5.586 . M1 Using continuity correction 72∙5 or 71∙5 in their standardisation formula . ±7.5 ±7.5 Note or seen gains M2 BOD 31.2 5.586 = 1 – 0.9104 M1 Appropriate area Φ, from final process, must be probability. 0.0896 A1 0.0896 ⩽ p ⩽ 0.0897 WWW. 5

Question 39 marks

Question

The Lions and the Tigers are two basketball clubs. The heights, in cm, of the 11 players in each of their first team squads are given in the table. Lions 178 186 181 187 179 190 189 190 180 169 196 Tigers 194 179 187 190 183 201 184 180 195 191 197 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with the Lions on the left. [4] (b) Find the median and the interquartile range of the heights of the Lions first team squad. [3] It is given that for the Tigers, the lower quartile is 183 cm, the median is 190 cm and the upper quartile is 195 cm. (c) Make two comparisons between the heights of the players in the Lions first team squad and the heights of the players in the Tigers first team squad. [2]

Mark scheme

3(a) Lions Tigers B1 Correct stem can be upside down, ignore extra values (not in 9 16 reverse). 9 8 17 9 9 7 6 1 0 18 0 3 4 7 B1 Correct Lions labelled on left, leaves in order from right to left 6 0 0 19 0 1 4 5 7 and lined up vertically, no commas or other punctuation. 20 1 B1 Correct Tigers labelled on same diagram, leaves in order and lined up vertically, no commas or other punctuation. If the correct data for Lions and Tigers is transposed, treat as a single error in Lions and condone in Tigers. Key 1|18|3 means 181 cm for Lions and 183 cm for Tigers B1 Correct single key for their diagram, need both teams identified and ‘cm’ stated at least once here or in leaf headings or title. SC If 2 separate diagrams drawn, SC B1 if both keys meet these criteria (Max B1, B0, B0, B1). 4 3(b) Median = 186 cm B1 [UQ = 190 cm, LQ = 179 cm] M1 189 ⩽ UQ ⩽ 190 – 178 ⩽ LQ ⩽ 180 IQR = 190 – 179 11[cm] A1 WWW 3 3(c) Tigers are (generally) taller B1 Comparison about central tendency in context. Heights of Tigers are slightly less consistent than heights of Lions B1 Comparison about spread in context. (Condone ‘similar spread’ in context.) 2

Question 49 marks

Question

In a large population, the systolic blood pressure (SBP) of adults is normally distributed with mean 125.4 and standard deviation 18.6. (a) Find the probability that the SBP of a randomly chosen adult is less than 132. [2] The SBP of 12-year-old children in the same population is normally distributed with mean 117. Of these children 88% have SBP more than 108. (b) Find the standard deviation of this distribution. [3] Three adults are chosen at random from this population. (c) Find the probability that each of these three adults has SBP within 1.5 standard deviations of the mean. [4]

Mark scheme

4(a) 132 − 125.4 M1 Use of ±standardisation formula with 132 and 125.4 substituted, P(X<132) = P( Z < ) = P( Z < 0.3548) 18.6 condone continuity correction 132±0.5 and use of 18.62, 18.6 0.639 A1 0.6385 < p ⩽ 0.639 If M0 scored, SC B1 for 0.6385 < p ⩽ 0.639 2 4(b) 108 − 117 B1 1.1749 < z ⩽ 1.175 or – 1.175 ⩽ z < −1.1749 = −1.175 σ M1 108 and 117 substituted in ±standardisation formula, no continuity correction, not σ2, √ σ, equated to a z-value. σ = 7.66 A1 7.659 ⩽ σ ⩽ 7.66 If M0 scored, SC B1 for 7.659 ⩽ σ ⩽ 7.66 3 4(c) P( −1.5 < Z < 1.5) M1 {Both 1.5 and –1.5 seen as z-values [Φ(1.5) – Φ(−1.5)] or appropriate use of 1.5 or −1.5} [= 2Φ (1.5 ) − 1 ] and {no other z-values in part}. = 2 × their 0.9332 −1 M1 Calculating the appropriate area from stated phis of z-values or their 0.9332 − (1 − their 0.9332 ) which must be ± the same number. or 2 × ( their 0.9332 − 0.5 ) Condone their 0.0668 as (1 – their 0.9332). 0.8664 A1 Accept answers wrt 0.866 If A0 scored SC B1 for answers wrt 0.866 0.86643 = 0.650[36] B1 FT FT their 4SF (or better) probability, accept final answers to 3SF. 4

Question 510 marks

Question

A game is played with an ordinary fair 6-sided die. A player throws the die once. If the result is 2, 3, 4 or 5, that result is the player’s score and the player does not throw the die again. If the result is 1 or 6, the player throws the die a second time and the player’s score is the sum of the two numbers from the two throws. (a) Draw a fully labelled tree diagram to represent this information. [2] Events A and B are defined as follows. A: the player’s score is 5, 6, 7, 8 or 9 B: the player has two throws (b) Show that P A = 13 . [3] (c) Determine whether or not events A and B are independent. [2] (d) Find P B | A′ . [3]

Mark scheme

5(a) B1 1st throw fully correct with probabilities and outcomes identified. 1 (Probabilities all and outcomes (1,2,3,4,5,6) on branches). 6 B1 2nd throw fully correct with probabilities and outcomes identified. 1 (Probabilities all and outcomes (1,2,3,4,5,6) on branches). 6 2 5(b) 1 1 1 7 B1 P(5) or P(7) identified and correct unsimplified, accept if 5 comes from 1+4 or 5: P(5) = × + = supported by correct scenarios shown or from tree diagram . 6 6 6 36 1 1 1 6 comes from 1+5 P(6) = × = 6 6 36 1 1 1 1 2 7 comes from 1+6 or 6+1 P(7) = × + × = 6 6 6 6 36 1 1 1 8 comes from 6+2 P(8) = × = 6 6 36 1 1 1 9 comes from 6+3 P(9) = × = 6 6 36 7 1 2 1 1 M1 Adding only the values from 5 correct scenarios. P(A) = + + + + 36 36 36 36 36 12 1 A1 Scenarios identified (may be on tree diagram in 5(a)), all = = probabilities seen, WWW AG. 36 3 3 5(c) 1 6 M1 Both identified and evaluated, consistent with their tree diagram P(B) = , P(A∩B)= or correct. 3 36 1 1 1 A1 P(A) × P(B) seen and evaluated, all notation present and correct. P(A)P(B) = × = Correct conclusion WWW. 3 3 9 6 1 ≠ , so not independent 36 9 2 5(d) 6 B1 6 P ( B A’) their 36 36 oe as numerator of a fraction. P(B | A’) = = P ( A′ ) 2 3 M1 6 their or correct 36 seen, consistent with their tree diagram. 1 their 1 − or correct 3 1 A1 , 0.25 4 3

Question 610 marks

Question

A Social Club has 15 members, of whom 8 are men and 7 are women. The committee of the club consists of 5 of its members. (a) Find the number of different ways in which the committee can be formed from the 15 members if it must include more men than women. [4] The 15 members are having their photograph taken. They stand in three rows, with 3 people in the front row, 5 people in the middle row and 7 people in the back row. (b) In how many different ways can the 15 members of the club be divided into a group of 3, a group of 5 and a group of 7? [3] In one photograph Abel, Betty, Cally, Doug, Eve, Freya and Gino are the 7 members in the back row. (c) In how many different ways can these 7 members be arranged so that Abel and Betty are next to each other and Freya and Gino are not next to each other? [3] 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

6(a) 5M0W 8C5 [× 7 C0] = 56 M1 8 Cx × 7 C5-x for x = 1, 2, 3, 4,or 5 4M1W 8C4 × 7 C1 = 490 3M2W 8C3 × 7 C2 = 1176 B1 Outcome for 4M1W or 3M2W correct and identified, accept unsimplified. M1 Add 3 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified. Addition may be implied by final answer. [Total =] 1722 A1 Value stated WWW. Alternative method for Question 6(a) 2M3W 8C2 × 7 C3 = 980 M1 8 Cx × 7 C5-x for x = 1, 2, 3, 4,or 5 1M4W 8C1 × 7 C4 = 280 0M5W 8C0 × 7 C5 = 21 B1 Outcome for 2M3W or 1M4W correct and identified, accept unsimplified. [Total = 15C5 – (980 + 280 + 21)] M1 Subtract 3 values of appropriate scenarios from their identified 3003 – (980 + 280 + 21) total or correct, no incorrect scenarios, no repeated scenarios, accept unsimplified. [Total =] 1722 A1 Value stated WWW. 4 15 6(b) C3 × 12 C5 [× 7C7] [= 455 × 792 ] M1 15 Cr × q, r = 3, 5, 7; q a positive integer >1 M1 15 Cs × 15-sCt [× 15-s-tCu] s = 3,5,7; t = 3,5,7 ≠ s; u = 3,5,7 ≠ s,t 360360 A1 Final answer. If A0 awarded SC B1 for final answer 360360. 3 6(c) Method 1: Total number of arrangements with AB together – Arrangements with AB and FG together 6! × 2 – 5! ×2 ×2 M1 a! × 2! × b, a = 5, 6; b = 1,2 seen. [ = 1440 – 480 ] M1 Either 6! × 2 – c ¸1 < c < 1440 or d – 5! ×2 ×2, 1440 < d 960 A1 Method 2: arrangements with AB together with F and G not together. 2 × 4! × 5 × 4 M1 2 × 4! × e, e positive integer >1 M1 f × 5 × 4, f positive integer >1 condone f × 20, f × 5C2, f positive integer >1 960 A1 3

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Where can I find the 9709/51 October/November 2022 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/51, October/November 2022 (9709_w22_ms_51), and the question paper (9709_w22_qp_51), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/51 for October/November 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.

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Yes. Upload your answers to 9709/51/O/N/22 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.