Modeling Randomess Practice Test Answers
1.
(a)
1
(= 0.111)
9
(i)
P(Alan scores 9) =
(A1)
(ii)
1  1 
P(Alan scores 9 and Belle scores 9) =     
 9   81 
2
(= 0.0123)
(A1)
2
(b)
(i)
2
2
 1 
 2 
 2 
 6 
 +   +…+   +…+  
 36 
 36 
 36 
 36 
2
2
P(Same score) = 
2
 1 

 36 
73
=
(= 0.113)
648
+ 
(ii)
P(A>B) =
=
(c)
(i)
(M1)
(A1)
1 
73 
1 –

2  648 
(M1)
575
(= 0.444)
1296
(A1)
P(One number  x) =
x
(with some explanation)
6
 x
6
(R1)
4
P(X  x) = P(All four numbers  x) =  
(M1)(AG)
4
(ii)
4
 x
 x – 1
P(X = x) = P(X  x) – P(X  x – l) =   – 

6
 6 
4
x
1
2
3
4
5
6
P(X = x)
1
1296
15
1296
65
1296
175
1296
369
1296
671
1296
(A1)(A1)(A1)
Note: Award (A3) if table is not completed but calculation of
E(X) in part (iii) is correct.
1
(iii)
1
15
671
+2×
+…+6×
1296
1296
1296
6797
=
(= 5.24)
1296
E(X) = 1 ×
(M1)
(A1)
7
[13]
2.
 P(X  x)  1
all x
1 2 1
 
+x=1
5 5 10
3
Therefore,
x=
10
1 1
2 3 
P(scoring six after two rolls) =     2    
 10 10 
 5 10 
1
=
4
Therefore,
(A1)
(M1)
(A1) (C3)
[3]
3.
Note: Throughout the whole question, students may be using their graphic display calculators
and should not be penalized if they do not show as much work as the marking scheme.
(a)
(b)
Let X denote the number of flaws in one metre of the wire.
(2.3) 2
Then E(X) = 2.3 flaws and P(X = 2) = e–2.3
(M1)(M1)
2!
= 0.265.
(A1)
Note: Award (C3) for a correct answer from a graphic display
calculator.
Let Y denote the number of flaws in two metres of wire.
Then Y has a Poisson distribution with mean E(Y) = 2 × 2.3 = 4.6
flaws for 2 metres.
Hence, P(Y  1) = 1 – P(Y = 0) = 1 – e–4.6
= 0.990 (3 sf)
–4.6
Note: Accept 1 – e
(M1)
(M1)
(A1)
3
3
.
[6]
2
4.
Let m be the median.
m

Then
0
1
x (4 – x2)dx = 0.5.
4
 4 x – x dx
(M1)
m
=>
3
=2
(A1)
0
1 4 m
x ]0 = 2
4
1
=> 2m2 – m4 = 2
4
=> [2x2 –
(M1)
=> m4 – 8m2 + 8 = 0
(A1)
(G2)
m = 1.08
OR
8  64 – 32
8  32
=
=4
2
2
=> m = 4 – 8  4 – 2 2 


m2 =
8 (4  2 2 )
(M1)
(A1) (C6)
Note: Award (C5) if other solutions to the equation
m4 – 8m2 + 8 = 0 appear in the answer box.
[6]
5.
(a)
The given condition implies that
3
6
e–  = e–  +  e– 
(M1)
 3 – 6 – 6 = 0
   2.8473
(b)
(A1)
(G1)
P(2  X  4) = P(X = 2) + P(X = 3) + P(X = 4)
2 –
P(X = 2) =
e
3
(M1)
3 –
= 0.235, P(X = 3) =
2
4 –
e
P(X = 4) =
= 0.159
24
Hence P(2  X  4) = 0.617
OR
P(2  X  4) = P(X  4) – P(X  1)
= 0.8402 – 0.2231
= 0.617
e
6
= 0.223,
(G1)
(A1)
(M1)
(G1)
(G1)
3
[6]
3
6.
(a)
mean for 30 days: 30  0.2 = 6.
P X  4  
6 4 6
e  0.134
4!
(b)
P(X > 3) = 1  P(X  3) = 1  e6(1 + 6 + 18 + 36) = 0.849
(c)
EITHER
mean for five days: 5  0.2 = 1
P(X = 0) = e1 (= 0.368)
(A1)
(M1)A1
N3
(M1)A1
N2
(A1)
A1
N2
OR
mean for one day: 0.2
(A1)
P(X = 0) = (e0.2 )5 = e1 (= 0.368)
(d)
Required probability = e0.2  e0.2  (1  e0.2)
= 0.122
(e)
Expected cost is 1850  6 = 11100 Euros
(f)
On any one day P(X = 0) = e 0.2

 5
Therefore,   e 0.2
1
 1  e   0.407
4
0.2
A1
N2
M1A1
A1
N3
A1
M1A1
N2
[13]
7.
Let (z) = 0.017
then (–z) = 1 – 0.017 = 0.983
z = –2.12
But z =
x
Therefore


1  1.02
1  1.02


(M1)
(A1)
where x = 1 kg
= –2.12
  = 0.00943 kg = 9.4 g (to the nearest 0.1 g)
(M1)
(A1) (C4)
[4]
4
8.
Note: In all 3 parts, award (A2) for correct answers with no working.
Award (M2)(A2) for correct answers with written evidence of the
correct use of a GDC (see GDC examples).
(a)
(i)
Let X be the random variable “the weight of a bag of salt”.
Then X ~ N(110, σ2), where σ is the new standard deviation.
Given P(X < 108) = 0.07, let Z =
108  110 

Then P  Z 
 = 0.07



2
Therefore,
= –1.476

X  110

(M1)
(M1)(A1)
Therefore, σ = 1.355.
(A1)
GDC Example: Graphing of normal cdf with σ as the variable,
and finding the intersection with p = 0.07.
(ii)
(b)
Let the new mean be µ, then X ~ N(µ, 1.3552).
108   

Then P  Z 
(M1)
 = 0.04
1.355 

108  
Therefore,
= 1.751
(M1)(A1)
1.355
Therefore, µ = 110.37
(A1)
GDC Example: Graphing of normal cdf with µ as the variable
and finding intersection with p = 0.04.
If the mean is 110.37 g then X ~ N(110.37, 1.3552), P(A < X < B) = 0.8
Then P(X < A) = 0.1, and P(X < B) = 0.9.
A  110.37
Therefore,
= –1.282
(M1)
1.355
A = 108.63
(A1)
B  110.37
Therefore,
= 1.282
(M1)
1.355
B = 112.11
(A1)
GDC Example: Graphing of normal cdf with X as the variable
and finding intersection with p = 0.1 and 0.9.
4
4
4
[12]
5
n
9.
(a)
 e λ 
P(Z = n) =
k 0
e     
n!
=
=
λk
 nk
 e  
n  k !
k!
n
n!
 k !n  k ! λ
k
M1A1
nk
e     
   n
n!
A1
This shows that Z is Poisson distributed with mean ( + ).
(b)
M1A1
k 0
R1
The result is (trivially) true for n = 1. A1
k
Assuming it to be true for n = k, ie
U
r 1
k 1
Consider
k
U  U
r 1
r
r 1
r
r
~ Pokm
 U k 1
M1
M1A1
which, using (a) is Po(km + m) ie Po([k + 1]m)
A1
Hence proved by induction since true for n = k  true for
n = k + 1 and we have shown true for n = 1.
R1
[12]
6