Standard +0.8 This question requires transformation of random variables using the Jacobian method, finding the CDF through integration, calculating both median (from CDF) and mean (from transformed PDF), then computing a probability. While the PDF is simple (uniform), the cubic transformation creates a non-trivial density requiring careful handling of the transformation formula and multiple integration steps. This is moderately challenging for Further Maths students.
The continuous random variable \(X\) has probability density function f given by
$$f(x) = \begin{cases} \frac{1}{2} & 1 \leq x \leq 3, \\ 0 & \text{otherwise.} \end{cases}$$
The random variable \(Y\) is defined by \(Y = X^3\). Find the distribution function of \(Y\). [5]
Sketch the graph of the probability density function of \(Y\). [3]
Find the probability that \(Y\) lies between its median value and its mean value. [4]
Consistent conclusion (AEF, on two t values): [Accept H0];
mean lengths are the same B1
Find confidence interval for the difference: 13⋅25 – 12⋅65 ± t s√(8 –1 + 6 –1 ) M1
Answer
Marks
Evaluate C.I. with t12,0.975 = 2⋅179, to 2 d.p.: 0⋅6 ± 1⋅84 or [– 1⋅24, 2⋅44] A1
5
7
Answer
Marks
2
[14]
Question 10:
10 | 1
Find F(x) for 1 < x < 3: F(x) = (x – 1) B1
2
Find G(y) from Y = X 3 for 1 < x < 3: G(y) = P(Y < y) = P(X 3 < y)
y1 y1
= P(X < 3) = F( 3)
1 y1
(result may be stated) = ( 3 – 1); (1 < y < 27) M1 A1; B1
2
State G(y) for other values of y: 0 (y < 1) and 1 (y > 27) B1
y− 3 2 1
Find g(y) for 1 < y < 27 ( on G(y)): g(y) = or B1
6 6y2
3
Sketch g(y) for 1 < y < 27 B1
with g(y) = 0 on either side of this interval B1
y1
3
Find mean of Y: E(Y) = ∫ y g(y) dy = ∫ ( ) dy
6
27
y4
3
(no need to find median = 8) = = 10 M1 A1
8
1
Find probability Y lies between median and mean:
1
G(10) – G(8) or G(10) –
2
1( )
101 −81
= 3 3
2
1( ) 1
(2 s.f. sufficient) or 101 3 −1 − = 0⋅077 [2]
2 2
M1 A1 | 5
3
4 | [12]
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Cambridge International A Level – October/November 2014 | 9231 | 22
11a (i)
(ii) | EITHER: State or find (by ⊥ axes) MI of X about AB:
1
2
I X = 2 mr M1 A1
1 3mr2
2 2
State or find MI of Y (or Z) about AB: IY = mr + mr = M1 A1
2 2
State or find (by ⊥ axes) MI of W about AB:
1 1 2
IW = 3mR 2 = 3mr 2 (1 + √3) 2
2 2 3
1
= (7 + 4√3) mr 2 M1 A1
2
1 3 7
Find MI of object about AB: I = ( + 2× + + 2√3) mr 2
2 2 2
= (7 + 2√3) mr 2 A.G. M1 A1
OR: State or find MI of X, Y or Z about centre O:
2
2r 7mr2
2
IX = mr + m = (M1 A1)
3 3
2
State or find MI of W about O: I W = 3mR 2 = 3mr 2 (1 + √3) 2
3
= (7 + 4√3) mr 2 (M1 A1)
Find MI of object about O: IO = 3IX + IW = (14 + 4√3) mr 2
(M1 A1)
Find (by ⊥ axes) MI of object about AB:
1
I = IO = (7 + 2√3) mr 2 A.G.
2
(M1 A1)
Find new MI of object plus particle about AB: I′ = I + 9mR 2
= I + 3 (7 + 4√3) mr 2
= 14 (2 + √3) mr 2 M1 A1
1
Find eqn for angular speed ω using energy: I′ ω 2 = 9mg × R sin 60° M1 A1
2
9mgR 3
2
Substitute and simplify to find ω: ω =
I'
9g g
(AEF) ω = or 0⋅802 M1 A1
14r r | 8
6 | [14]
Page 12 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9231 | 22
11b | 1062
1422.34−
8
2
Estimate population variance using A’s sample: sA =
7
446
(allow use of biased: σ A,8 2 = 2⋅23 or 1⋅493 2 ) = or 2⋅549 or 1⋅596 2 M1 A1
175
106 s 2
Find confidence interval: ± t X M1
8 8
State or use correct tabular value of t: t7,0.975 = 2⋅36 [5] A1
Evaluate C.I. correct to 1 d.p.: 13⋅25 ± 1⋅335 or [11⋅9, 14⋅6] A1
State suitable assumptions (A.E.F.): Distribution of B is Normal with
same population variance B1
a
State hypotheses (B0 for …) e.g.: H0: µA = µB , H1: µA > µB B1
( )
971.53− 75.92
2 6
Estimate (or imply) B’s population variance: sB =
5
(allow use of biased: σ B,6 2 = 1⋅899 or 1⋅378 2 )
= 2⋅279 or 1⋅510 2
2
and find pooled estimate of common variance s :
( )
7s 2 +5s 2
2 A B
s =
12
(17.84+11.395)
=
12
= 2⋅436 or 1⋅561 2 M1
(13.25−12.65)
Calculate value of t (to 2 d.p., either sign): t =
( )
s 8−1+6−1
0.6
= = 0⋅712 M1 A1
0.8430
State or use correct tabular t-value (to 2 d.p.): t12,0.95 = 1⋅782 B1
(or compare 0⋅6 with 1⋅782 s√(8 –1 + 6 –1 ) = 1⋅50)
Consistent conclusion (AEF, on two t values): [Accept H0];
mean lengths are the same B1
Find confidence interval for the difference: 13⋅25 – 12⋅65 ± t s√(8 –1 + 6 –1 ) M1
Evaluate C.I. with t12,0.975 = 2⋅179, to 2 d.p.: 0⋅6 ± 1⋅84 or [– 1⋅24, 2⋅44] A1 | 5
7
2 | [14]
The continuous random variable $X$ has probability density function f given by
$$f(x) = \begin{cases} \frac{1}{2} & 1 \leq x \leq 3, \\ 0 & \text{otherwise.} \end{cases}$$
The random variable $Y$ is defined by $Y = X^3$. Find the distribution function of $Y$. [5]
Sketch the graph of the probability density function of $Y$. [3]
Find the probability that $Y$ lies between its median value and its mean value. [4]
\hfill \mbox{\textit{CAIE FP2 2014 Q10 [12]}}