Mathematics 9709/62 February/March 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability and Statistics, February/March 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
On each day that Tamar goes to work, he wears either a blue suit with probability 0.6 or a grey suit with probability 0.4. If he wears a blue suit then the probability that he wears red socks is 0.2. If he wears a grey suit then the probability that he wears red socks is 0.32.
(i) Find the probability that Tamar wears red socks on any particular day that he is at work. [2]
(ii) Given that Tamar is not wearing red socks at work, find the probability that he is wearing a grey suit. [3]
Mark scheme
1(i) 0.6 × 0.2 + 0.4 × 0.32 M1 Addition of 2 two-factor terms 0.6 × a + 0.4 × b
31 A1 CAO
= 0.248,
125
2
1(ii) Method 1
0.4 × 0.68 B1 Correct [unsimplified] numerator seen in fraction
P(GS|Not Red socks) =
1 − (i)
M1 1 – their (i) as denominator in fraction
17 A1
= 0.362,
47
Method 2
0.4 × 0.68 B1 Correct [unsimplified] numerator seen in fraction
P(GS|Not Red socks) =
0.6 × 0.8 + 0.4 × 0.68
M1 Correct or (their (i))’ as denominator in fraction
17 A1
= 0.362,
47
3
For 40 values of the variable x, it is given that Σ x − c2 = 3099.2, where c is a constant. The standard deviation of these values of x is 3.2.
(i) Find the value of Σ x − c. [3]
(ii) Given that c = 50, find the mean of these values of x. [1]
Mark scheme
2(i) ∑( x − c)
2
∑ ( x − c)
2 M1 Use correct formula with values substituted
σ =
2
−
n n
3099.2 ∑ ( x − c )
2
2
3.2 = −
40 40
∑ ( x − c)
2 M1 ∑ ( x − c)
2
= 67.24 : Rearrange to make their the subject, unsimplified.
40 40
∑ ( x − c ) = 40 × 67.24
= 328 A1 Exact value, cao
3
2(ii) ∑ x − 40c = their (i) B1FT FT their (i)
their (i)
Mean = + 50
40
= 58.2
1
The times taken, in minutes, for trains to travel between Alphaton and Beeton are normally distributed with mean 140 and standard deviation 12.
(i) Find the probability that a randomly chosen train will take less than 132 minutes to travel between
Alphaton and Beeton. [3]
(ii) The probability that a randomly chosen train takes more than k minutes to travel between
Alphaton and Beeton is 0.675. Find the value of k. [3]
Mark scheme
3(i) 132 − 140 M1 Using ± standardisation formula, no continuity correction, not σ2 or
P(X< 132) = P Z < = P( Z < −0.6667) √σ
12
= 1 – 0.7477 M1 Appropriate area Φ from standardisation formula P(z<….) in final
solution
= 0.252 awrt A1 Condone linear interpolation = 0.25243
3
3(ii) P(time>k) = 0.675, z = –0.454 B1 ±0.454 seen
k − 140 M1 An equation using the standardisation formula with a
= −0.454 z-value (not 1 – z), condone σ2 or √σ
12
k = 135, 134.6, 134.55 A1 B0M1A1 max from –0.45
3
The random variable X takes the values −1, 1, 2, 3 only. The probability that X takes the value x is kx2 , where k is a constant.
(i) Draw up the probability distribution table for X , in terms of k, and find the value of k. [3]
(ii) Find E X and Var X . [3]
Mark scheme
4(i) B1 Probability distribution table with correct values of x, no additional
x −1 1 2 3 values unless with probability 0 stated, at least one correct
probability including k
p k k 4k 9k
15k = 1, M1 Equating Σp = 1, may be implied by answer
1 A1 If 0 scored, SCB2 for probability distribution table with correct
k= numerical probabilities.
15
3
4(ii) Method 1
35 7 B1FT FT if 0< their k<1
E(X) = 8k + 27k = 35k = =
15 3
Var(X) = ( k + k + 16k + 81k ) − ( 35k ) M1 Correct formula for variance, in terms of k at least – must have ‘–
2
mean2’(ft).
52 A1
= 1.16 ,
45
Method 2
8 27 35 7 B1FT FT if 0< their k<1
E(X) = + = =
15 15 15 3
1 1 16 81 7
2 M1 Subst their values in correct var formula – must have
Var(X) = + + + − ‘– mean2’(ft) (condone probs not summing to exactly 1)
15 15 15 15 3
= 1.16 (= 52/45) A1 Using their values from (i)
3
The weights, in kg, of the 11 members of the Dolphins swimming team and the 11 members of the
Sharks swimming team are shown below.
Dolphins 62 75 69 82 63 80 65 65 73 82 72
Sharks 68 84 59 70 71 64 77 80 66 74 72
(i) Draw a back-to-back stem-and-leaf diagram to represent this information, with Dolphins on the left-hand side of the diagram and Sharks on the right-hand side. [4]
(ii) Find the median and interquartile range for the Dolphins. [3]
Mark scheme
5(i) B1 Correct stem can be upside down, ignore extra values,
Dolphins Sharks
B1 Correct Dolphin must be on LHS,
5 9
B1 Correct Sharks on either LHS or RHS of back-to-back. Alignment ±
9 5 5 3 2 6 4 6 8 half a space, no late entries squeezed in, no crossing out if shape is
changed. Condone a separate RHS stem-and-leaf diagram
5 3 2 7 0 1 2 4 7
B1FT Correct single key for their single diagram, need both teams
2 2 0 8 0 4 identified and ‘kg’ stated at least once here or in leaf headings or
title.
Key: 3|6|4 means 63 kg
for Dolphins and 64 kg
for Sharks
4
5(ii) Median = 72 B1 72<UQ<82 – 62<LQ<72
LQ = 65, UQ = 80,
IQR = 80 – 65 M1 nfww
= 15 A1 SCB1 if M0 scored for LQ = 65 and UQ = 80
3
The results of a survey by a large supermarket show that 35% of its customers shop online.
(i) Six customers are chosen at random. Find the probability that more than three of them shop online. [3]
(ii) For a random sample of n customers, the probability that at least one of them shops online is greater than 0.95. Find the least possible value of n. [3]
(iii) For a random sample of 100 customers, use a suitable approximating distribution to find the probability that more than 39 shop online. [5]
Mark scheme
6(i) P(4, 5, 6) = 6C4 0.3540.652 + 6C5 0.3550.651 + 0.356 M1 Binomial term of form 6Cxpx(1 – p)6– x 0 < p < 1 any p, x≠6,0
A1 Correct unsimplified answer
= 0.117 A1
3
6(ii) 1 − 0.65n > 0.95 M1 Equation or inequality involving ‘0.65n or 0.35n’ and ‘0.95 or 0.05’
0.65n < 0.05
log 0.05 M1 Attempt to solve their exponential equation using logs or Trial and
n> = 6.95 Error.
log 0.65
n=7 A1 CAO
3
6(iii) Mean = 0.35 ×100 = 35 B1 Correct unsimplified np and npq,
Variance = 0.35 × 0.65 × 100 = 22.75
39.5 − 35 M1 Substituting their µ and σ (condone σ2) into the ±Standardisation
P z > = P ( z > 0.943) Formula with a numerical value for ‘39.5’.
22.75
M1 Using continuity correction 39.5 or 40.5
= 1 − 0.8272 M1 Appropriate area Φ from standardisation formula P(z>….) in final
solution, (>0.5 if z is -ve, <0.5 if z is +ve)
= 0.173 A1 Final answer
5
Find the number of different arrangements that can be made of all 9 letters in the word CAMERAMAN in each of the following cases.
(i) There are no restrictions. [2]
(ii) The As occupy the 1st, 5th and 9th positions. [1]
(iii) There is exactly one letter between the Ms. [4]
Three letters are selected from the 9 letters of the word CAMERAMAN.
(iv) Find the number of different selections if the three letters include exactly one M and exactly one A. [1]
(v) Find the number of different selections if the three letters include at least one M. [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
7(i) 9! M1 9! alone on numerator, 2! and/or 3! on denominator
2!3!
= 30240 A1 Exact value, final answer
2
7(ii) A^^^A^^^A B1 Final answer
6!
Arrangements = = 360
2!
1
7(iii) M^M^^^^^^ M1 7! in numerator, (considering letters not M)
7!
= ×7
3!
M1 Division by 3! only (removing repeated As)
M1 Multiply by 7 (positions of M-M)
= 5880 A1 Exact value, final answer
Method 2 (choosing letter between Ms)
6! 6! M1 6! in sum of 2 expressions a6! + b6!
1× ×7 + 4× ×7
2! 3!
M1 Multiply by 7 in both expressions (positions of M-M)
= 2520 + 3360 M1 c d
+ seen (removing repeated As)
2! 3!
= 5880 A1 Exact value
7(iii) Method 3
(MAM) ^ ^ ^ ^ ^ ^ = 7!/2! = 2520 M1 7! in numerator (considering 6 letters + block)
(MA’M) ^ ^ ^ ^ ^ ^ = 7!/3! × 4 = 840 × 4 = 3360 M1 Division by 2! and 3! seen in different terms
Total = 2520 + 3360 M1 Summing 5 correct scenarios only
= 5880 A1 Exact value
4
7(iv) M A ^ = 4C1 = 4 B1 Final answer
1
7(v) M ^ ^ : 4C2 =6 M1 Either option M M ^ or M ^ ^ correct, accept unsimplified
M M ^ : 4C1 =4
MMA: =1 M1 Add 4 or 5 correct scenarios only
MAA: =1
(M A _ :4C1 = 4)
Total = 16 A1 Value must be clearly stated
Method 2
M M ^ = 5C1 =5 M1 Either option M M ^ or M ^ ^ correct, accept unsimplified
M ^ ^ = 5C2 = 10 M1 Adding 2 or 3 correct scenarios only
MAA= =1 Total = 16 A1 Value must be clearly stated
3
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Mathematics 9709/62 for February/March 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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