Question 8:
8 | (a) | ∞
∞  k 
∫ kx −ndx= 
1 (1−n)xn−1 
1
k
= = 1 so k = n – 1
n−1 | M1
B1
A1
[3] | 1.1
1.1
1.1 | Integral attempted, correct limits
Correct indefinite integral
Correctly obtain k = n – 1, www | Don’t need full details of
lim(a → ∞)
(b) | (i) | 1
∫3x −4dx=− +c
x3
x = 1, F(x) = 0 so c = 1. Hence 1 – x–3.
0 x<1,

F(x)= 1
1− x≥1
 x3 | M1
A1
B1
[3] | 1.1
1.1
1.1 | Needs + c or definite integral
between 1 and x, oe
Fully correct active part of CDF
“0 for x < 1” stated and no wrong
ranges (doesn’t need M1 or A1)
Allow ≤ for <, and/or > for ≥ | Wrong k: can get M1A0B1
Ignore ranges here
Or “0 otherwise” if “x ≥ 1”
stated in active part
(ii) | P[(X >7)∩(X >5)] P(X >7)
=
P(X >5) P(X >5)
1−F(7)
=
1−F(5)
= 125 or 0.364(431…)
343 | M1*
A1
*depM1
A1ft
[4] | 3.1a
3.1a
3.3
1.1 | Use conditional probability method
P[(X > 7) ∩ (X > 5)] = P(X > 7)
Convert probabilities into F(X), not
using P(X > 7) × P(X > 5)
Any exact fraction or awrt 0.364, ft
on 1 – a/x3, a ≠ 0, 1 | [1−F(7)][1−F(5)]
:
1−F(5)
M1A0M0A0
Allow from F(x) = 1 – a/x3,
otherwise www
(c) | ∞
 kx3−n 
∞
E(X2)=∫ kx2−ndx=  (n ≠ 3)
1 (3−n)
1
If n = 3, E(X 2) = lim[2ln(x)], not defined
x→∞
Infinite integral does not converge if 3 – n ≥ 0
∞
 kx2−n 
If n ≥ 4 then E(X)=  converges
(2−n)
1
Therefore Var(X) is not defined if and only if
n = 2 or 3. | M1*
B1
*depM1
B1
A1
[5] | 2.1
1.1
2.2a
2.3
2.2a | Correct limits needed somewhere
n−1
Correct indefinite integral or
n−3
No marks just for this unless last 3
marks all zero, then if this (or for
n = 2) is shown, award SC B1
Make deduction based on
convergence, ft
Consider convergence of E(X)
Shown not defined for n = 2 or 3 and
only for those | n−1
SC: E(X2)= , M1B1
n−3
n−1
E(X)= ⇒ n ≠ 2 or 3:
n−2
(not valid, must consider ln
if n = 2 or 3): B0
No limits used: M0B1M0B0
SC: Var(X) < 0 when n < 3:
M1B1M1 (B0) A0
But no need to state “if and
only if”
PMT
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