Question Scheme Marks AOs
12 12.5
1(a) Q2 5 5 or (use of (n 1)) (5) 5 M1 1.1a
15 15
=9 or 9.166… awrt 9.17 A1 1.1b
(2)
2
(b) 5675 355
= 49.14...
M1 1.1a
x
30 30
= awrt 7.01 A1 1.1b
355
2
Accept 5675 30
s 30 7.1294...
x 29
(2)
t - 15
(c) x= or t = 2x + 15 M1 3.1b
2
Median = 2¥ "9"+ 15 = 33 (allow awrt 33.3 from “9.17” in (a)) A1ft 1.1b
Sd = 2¥ "7.01" = 14.02...(awrt 14.0) [allow awrt 14.3 if s used] A1ft 1.1b
(3)
(d) The median time is "33" and “33” < 35 so 50% (30) should finish in
35 minutes.
18 18 M1 2.4
ALT Probability of being < 35 mins is \ ¥ 60 = 36
30 30
applicants to choose from.
It is likely that they will fill all 25 positions [providing those offered
A1 2.2b
accept]
(2)
Notes: (9 marks)
(a) M1: For a suitable fraction ×5 (ignore end points)
A1: For 9 or awrt 9.17 if using n + 1
(b) M1: For a correct expression for x and s x or sx
A1: For awrt s x = 7.01or sx = awrt 7.13
t - 15
(c) M1: For realising x = and then rearranging to get a correct equation with t as the subject
2
May be implied by a correct answer for the median of t.
A1ft: ft their median
A1ft: ft their s x or sx . NB using s gives awrt14.3
(d) M1: For a suitable comparison following through their value for the median of t.
A1: A correct conclusion in context following through their value for the median of t.
, Question Scheme Marks AOs
2(a) P(5 X < 12) = P(X 11) – P(X 4) M1 1.1b
= 0.8939 – 0.0495 = awrt 0.844 A1 1.1b
(2)
(b) H0: p = 0.25 H1: p > 0.25 ( both correct in terms of p or p ) B1 2.5
Y ~ B(40, 0.25) M1 3.3
Method 1 Method 2
P(Y 16) = 1 – P(Y 15) P(Y 17) = 0.0116 M1 1.1b
= 1 – 0.9378 P(Y 18) = 0.0047
= 0.0262 CR: Y 18 A1 1.1b
0.0262 > 0.01 16 < 18 or 16 is not in the critical region or 16 is not
significant, accept H0. There is no significant evidence that the A1 cso 2.2b
proportion of people who bought organic eggs has increased
(5)
(c) There is evidence that the proportion of people who bought organic
B1ft 2.2b
eggs has increased [since 0.05 > 0.0262 or 16 is in critical region]
(1)
(8 marks)
Notes:
(a)M 1: For dealing with P(5 X < 12) they need to use the cumulative prob. Function on the calc.
A1: awrt 8.44 ( from calculator).
(b) B1: Both hypotheses correct using p or p and 0.25
M1: Realising that the model B(40, 0.25) is to be used. This may be stated or used.
M1: Using or writing 1 – P(Y 15) or 1 – P(Y < 16)
a correct CR or P(Y 17) = 0.0116 and P(Y 18) = 0.0047
A1: awrt 0.0262 or CR Y 18 or Y > 17
A1cso: A fully correct solution with a correct conclusion in context to include the idea of
proportion and increased plus referring to organic
(c) B1ft: For 0.0262 < 0.05 [ft their probability in part(b)] or a CR of 16 15 (allow 16 > 14)
and a correct contextual conclusion.
12 12.5
1(a) Q2 5 5 or (use of (n 1)) (5) 5 M1 1.1a
15 15
=9 or 9.166… awrt 9.17 A1 1.1b
(2)
2
(b) 5675 355
= 49.14...
M1 1.1a
x
30 30
= awrt 7.01 A1 1.1b
355
2
Accept 5675 30
s 30 7.1294...
x 29
(2)
t - 15
(c) x= or t = 2x + 15 M1 3.1b
2
Median = 2¥ "9"+ 15 = 33 (allow awrt 33.3 from “9.17” in (a)) A1ft 1.1b
Sd = 2¥ "7.01" = 14.02...(awrt 14.0) [allow awrt 14.3 if s used] A1ft 1.1b
(3)
(d) The median time is "33" and “33” < 35 so 50% (30) should finish in
35 minutes.
18 18 M1 2.4
ALT Probability of being < 35 mins is \ ¥ 60 = 36
30 30
applicants to choose from.
It is likely that they will fill all 25 positions [providing those offered
A1 2.2b
accept]
(2)
Notes: (9 marks)
(a) M1: For a suitable fraction ×5 (ignore end points)
A1: For 9 or awrt 9.17 if using n + 1
(b) M1: For a correct expression for x and s x or sx
A1: For awrt s x = 7.01or sx = awrt 7.13
t - 15
(c) M1: For realising x = and then rearranging to get a correct equation with t as the subject
2
May be implied by a correct answer for the median of t.
A1ft: ft their median
A1ft: ft their s x or sx . NB using s gives awrt14.3
(d) M1: For a suitable comparison following through their value for the median of t.
A1: A correct conclusion in context following through their value for the median of t.
, Question Scheme Marks AOs
2(a) P(5 X < 12) = P(X 11) – P(X 4) M1 1.1b
= 0.8939 – 0.0495 = awrt 0.844 A1 1.1b
(2)
(b) H0: p = 0.25 H1: p > 0.25 ( both correct in terms of p or p ) B1 2.5
Y ~ B(40, 0.25) M1 3.3
Method 1 Method 2
P(Y 16) = 1 – P(Y 15) P(Y 17) = 0.0116 M1 1.1b
= 1 – 0.9378 P(Y 18) = 0.0047
= 0.0262 CR: Y 18 A1 1.1b
0.0262 > 0.01 16 < 18 or 16 is not in the critical region or 16 is not
significant, accept H0. There is no significant evidence that the A1 cso 2.2b
proportion of people who bought organic eggs has increased
(5)
(c) There is evidence that the proportion of people who bought organic
B1ft 2.2b
eggs has increased [since 0.05 > 0.0262 or 16 is in critical region]
(1)
(8 marks)
Notes:
(a)M 1: For dealing with P(5 X < 12) they need to use the cumulative prob. Function on the calc.
A1: awrt 8.44 ( from calculator).
(b) B1: Both hypotheses correct using p or p and 0.25
M1: Realising that the model B(40, 0.25) is to be used. This may be stated or used.
M1: Using or writing 1 – P(Y 15) or 1 – P(Y < 16)
a correct CR or P(Y 17) = 0.0116 and P(Y 18) = 0.0047
A1: awrt 0.0262 or CR Y 18 or Y > 17
A1cso: A fully correct solution with a correct conclusion in context to include the idea of
proportion and increased plus referring to organic
(c) B1ft: For 0.0262 < 0.05 [ft their probability in part(b)] or a CR of 16 15 (allow 16 > 14)
and a correct contextual conclusion.
