Question 4:
4 | 9 | 11.03 | 15 | 15.75 | 0.0357 | 14.2857
5 | 6 | 4.72
Totals | 4.6461 | 204.6461
4. (a)
(b) | Let X  weight of a sack of potatoes, X  N(25.6,0.242)
Attempt at D andD  N(0, ..)
So D  X  X  N(0, 2(0.24)2) or D  N(0,0.1152)
1 2 (0.24)2 (0.24)2; 0.1152
 P D  0.5   2P(D  0.5) 2P(D  0.5) can be implied
 0.5 
= 2P Z 
 
 0.1152
= 2PZ 1.4731... or = 2(10.9292)
= 0.1416 awrt 0.141 or awrt 0.142
Let Y  weight of an empty pallet, Y  N(20.0,0.322)
So T  X  X  ...  X Y
1 2 30
30(25.6)  20 or 788
T  N(30(25.6)  20, 30(0.24)2  0.322)
30(0.24)2  0.322
T  N(788,1.8304) N and 1.8304 or awrt 1.83
 785 788
P(T  785)  P Z 
 
 1.8304 
= P(Z  2.2174...)
= 0.9868 awrt 0.987 | M1
A1; A1
dM1
dM1
A1
[6]
B1
M1
A1
M1
A1
[5]
(Total 11)
Notes
(a)
(b) | 1st M1: For clear definition of D and normal distribution with mean of 0 (Can be implied by 3rd M1)
1st A1: for correct use of Var(X  X ) formula
1 2
2nd A1: for 0.1152
2nd dM1: For realising need 2P(D  0.5) (Dependent on 1st M1 i.e. must be using suitable D)
3rd dM1: Dep on 1st M1 for standardising with 0.5, 0 and their s.d.(0.24)Must lead to P(Z  ve)(o.e.)
P(Z > 1.47) implies 1st M1 1st A1 2nd A1 and 3rd M1
Correct answer only will score 6 out of 6
B1: For a mean of 30(25.6)  20. Can be implied by 788.
1st M1: For30(0.24)2  0.322. Can be implied by 1.8304 or awrt 1.83
Allow M1 for swapping error i.e. 300.322 0.242 if the expression is seen
1st A1: For normal and correct variance of 1.8304 or awrt 1.83.
Normality may be implied by standardisation
2nd M1: For standardising with 785 with their mean and st. dev..(0.24) Must lead to P(Z  ve) oe.
2nd A1: awrt 0.987
Correct answer only will score 5 out of 5
Note: Calculator answers are (a) 0.14071... , (b) 0.98670...
5. | H : Grades and gender are independent (or not associated) “grades” and “gender”
0
H : Grades and gender are dependent (or associated) mentioned at least once.
1
Observed Male Female An attempt to convert percentages
Distinction 37 44 to observed frequencies.
Merit 127 96
All observed frequencies
Unsatisfactory 36 20
are correct.
Some attempt at
Expected Male Female Totals (Row Total)(Column Total)
Distinction 45 36 81 (Grand Total)
Merit 123.889 99.111 223 Can be implied by a correct E
i
Unsatisfactory 31.111 24.889 56
All expected frequencies are
Totals 200 160 360 correct to nearest integer.
At least 2 correct terms for
(O  E)2 O2 (O  E)2 O2
Observed Expected or or correct
E E E E
37 45 1.422 30.422 expressions with their E .
i
44 36 1.778 53.778 Accept 2 sf accuracy
for the M1 mark.
127 123.889 0.078 130.189
(O  E)2 O2
96 99.111 0.098 92.987 All correct or terms
E E
36 31.111 0.768 41.657
to either 2 dp or better.
20 24.889 0.960 16.071
Allow truncation.
Totals 5.104 365.104 (by awrt 5.1 if 3rd M1 seen)
(O  E)2 O2
X2  or  360; awrt 5.1 awrt 5.1
E E
(31)(2 1)  2 ( ) 2 (Can be implied by 5.991)
c2(0.05) 5.991  CR: X2  5.991 For 5.991 only
2
Since X2 5.1 does not lie in the CR, then there is insufficient evidence to reject H
0
Business Studies grades and gender are independent or
There is no association between Business Studies grades and gender. Or
Head of department’s (belief ) is correct | B1
(1)
M1
A1
M1
A1
M1
A1
A1
(7)
B1
B1
M1
A1ft
(4)
[12]
(Total 12)
Notes
5.10 only | Final M1: For a correct statement linking their test statistic and their critical value ( > 3.8)
Note: Contradictory statements score M0. E.g. “significant, do not reject H ”.
0
Final A1ft: For a correct ft statement in context –
must mention “grades” and “gender” or “sex” or “head of department”
Condone “relationship” or “connection” here but not “correlation”.
e.g. “There is no evidence of a relationship between grades and gender”
Just seeing 5.10... only can imply 1st 3 Ms but loses 1st 3 As so can score 4 out of 7 (Qu says “show..”)
Note: Full accuracy gives X2 5.104356... and p-value 0.0779
Observed | Male | Female
Distinction | 37 | 44
Merit | 127 | 96
Unsatisfactory | 36 | 20
Expected | Male | Female | Totals
Distinction | 45 | 36 | 81
Merit | 123.889 | 99.111 | 223
Unsatisfactory | 31.111 | 24.889 | 56
Totals | 200 | 160 | 360
Observed | Expected | (O  E)2
E | O2
E