www.furthermaths.org.uk
The FMSP Wales
Is managed by
Rheolir FMSP
Cymru gan
in partnership with
mewn
partneriaeth â
funded by
ac fe’i
hariannir gan
REVISION SHEET – S1 (WJEC)
CONTINUOUS RANDOM VARIABLES
The main ideas are:
• Probability density
function f(x)
• Expectation (mean)
and variance
• Cumulative Distribution
Function F(x)
• Median and quartiles
The probability density function of a
continuous random variable can be
shown on a graph
Before the exam you should know:
•
Continuous random variables are used to create
mathematical models to describe and explain data
you might find in the real world.
•
A continuous random variable X takes values in a
given interval
•
Total probability is 1
•
Expectation = Mean = E(X) = µ = ∫ xf(x) dx
•
Variance = Var(X) = σ2 = E(X2) – [E(X)]2
∫f(x) dx= 1
= E(X2) – µ2
f(x)
= ∫ x2 f(x) dx – µ2
x
Notation
• A continuous random variable is
usually denoted by a capital
letter (X, Y etc).
• The area under the graph of
the pdf denotes probability.
Total area = ∫f(x) dx= 1
•
For a function g(X ) : E(g(X)) = ∫ g(x) f(x) dx
•
E(aX +b) = aE(X) + b
•
Var(aX + b) = a2 Var(X)
•
F(x) = P(X ≤ x) = ∫ f(t) dt
•
P(a ≤ X ≤ b) = F(b) – F(a)
•
F(m) = 0.5 when m = median
•
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support
Programme Wales can accept no responsibility for its content matching WJEC specifications exactly.
Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth
Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
1
www.furthermaths.org.uk
The FMSP Wales
Is managed by
Rheolir FMSP
Cymru gan
in partnership with
mewn
partneriaeth â
funded by
ac fe’i
hariannir gan
Example: X is a continuous random variable with probability density function f given
1 2
x for 1 ≤ x ≤ 4
by f(x) =
21
(a) Evaluate E(X)
(b) Find an expression for F(x), valid for 1 ≤ x ≤ 4, where F(x) denotes the
cumulative distribution function of X
(c) Calculate P(2 ≤ X ≤ 3)
(d) Find the median of X correct to 2d.p.
Answer:
4
4
1 x2
(a) E(X) = ∫ xf ( x ) dx = ∫ x
dx =
21
1
1
x
(b)
∫
F(x) =
−∞
x
f (t ) dt = ∫
1
4
4
1 3
256 1
85
 1 4
∫1 21 x dx =  84 x 1 = 84 − 84 = 28 = 3.04
x
1 2
1
1 
t dt =  t 3  = ( x 3 − 1)
21
 63 1 63
(c)
P(2 ≤ X ≤ 3) = F(3) – F(2) =
(d)
F (m) =
1
(m 3 − 1) = 0.5
63
1
19
= 0.3016
( 26 − 7) =
63
63
m3 − 1 = 31.5
m3 = 32.5
m = 3.19
Example: The continuous random variable X has cumulative distribution function F
given by
for x < 0
F ( x) = 0
F ( x ) = 4 x 3 − 3x 4
for 0 ≤ x ≤ 1
F ( x) = 1
for x ≥ 1
(a) Show that the lower quartile lies between 0.45 and 0.46
(b) Find an expression for f(x), valid for 0 ≤ x ≤ 1, where f denotes the probability
density function of X.
Answer:
(a) F (0.45) = 4 × 0.45 3 − 3 × 0.45 4 = 0.2415 < 0.25
F (0.46) = 4 × 0.46 3 − 3 × 0.46 4 = 0.2550
(b)
f(x) = F′(x) = 12x2 – 12x3
> 0.25
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support
Programme Wales can accept no responsibility for its content matching WJEC specifications exactly.
Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth
Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
2
www.furthermaths.org.uk
The FMSP Wales
Is managed by
Rheolir FMSP
Cymru gan
in partnership with
mewn
partneriaeth â
funded by
ac fe’i
hariannir gan
TAFLEN ADOLYGU – S1 (CBAC)
HAPNEWIDYNNAU DI-DOR
Y prif syniadau yw:
• Ffwythiant dwysedd
tebygolrwydd f (x)
• Disgwyliant (cymedr) ac
amrywiant
• Ffwythiant dosraniad
cronnus F(x)
• Canolrif a chwartelau
Cyn yr arholiad mae angen gwybod:
•
Defnyddir hapnewidynnau di-dor i greu modelau
mathemategol i ddisgrifo ac esbonio data gallwch
ddarganfod yn y byd iawn bob dydd.
•
Mae hapnewidyn di-dor X yn cymryd gwerthoedd
mewn cyfwng penedol.
•
Cyfanswm y tebygolrwyddau yw ∫ f (x) dx = 1
•
Disgwyliant = Cymedr = E(X) = µ = ∫ x f (x) dx
•
Amrywiant = Var(X) = σ2 = E(X2) – [E(X)]2
Gellir dangos y ffwythiant dwysedd
tebygolrwydd ar gyfer hapnewidyn didor gan graff.
= E(X2) – µ2
f (x)
= ∫ x2 f (x) dx – µ2
x
•
Ar gyfer ffywthiant g(X ) : E(g(X)) = ∫ g(x) f (x) dx
•
E(aX +b) = aE(X) + b
•
Var(aX + b) = a2 Var(X)
•
F(x) = P(X ≤ x) = ∫ f (t) dt
•
P(a ≤ X ≤ b) = F(b) – F(a)
•
F(m) = 0.5 pan fo m = canolrif
Nodiant
• Dynodir hapnewidyn di-dor gan brif
llythyren ( W,X,Y, Z a.y.b.).
• Mae’r arwynebdd o dan graff ff.d.t.
yn dynodi’r tebygolrwydd.
Cyfanswm arwynebedd = ∫ f (x)dx =1
•
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support
Programme Wales can accept no responsibility for its content matching WJEC specifications exactly.
Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth
Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
3
www.furthermaths.org.uk
The FMSP Wales
Is managed by
Rheolir FMSP
Cymru gan
in partnership with
mewn
partneriaeth â
funded by
ac fe’i
hariannir gan
Enghraifft: Mae X yn hapnewidyn di-dor sydd â ffwythiant dwysedd tebygolrwydd a
1 2
roddir gan
f (x) =
x for 1 ≤ x ≤ 4
21
(e) Enrhifwch E(X)
(f) Darganfyddwch fynegiad ar gyfer F(x), sy’n ddilys ar gyfer 1 ≤ x ≤ 4, lle F(x)
sy’n dynodi’r ffwythiant dosraniad cronnus ar gyfer X
(g) Cyfrifwch P(2 ≤ X ≤ 3)
(h) Darganfyddwch ganolrif X yn gywir i 2 le degol.
Ateb:
4
(e)
E(X) =
4
∫ xf ( x)dx =
∫x
1
1
1 x2
dx =
21
4
4
1 3
256 1
85
 1 4
∫1 21 x dx =  84 x 1 = 84 − 84 = 28 = 3.04
x
1 2
1 3
 1 3
(f) F(x) = ∫ f (t )dt = ∫ t dt =  t  =
( x − 1)
21
 63 1 63
−∞
1
x
(g)
x
P(2 ≤ X ≤ 3) = F(3) – F(2) =
(h) F ( m) =
Enghraifft:
1
( m 3 − 1) = 0.5
63
1
19
= 0.3016
( 26 − 7) =
63
63
m 3 − 1 = 31.5
m 3 = 32.5
m = 3.19
Mae gan yr hapnewidyn di-dor Xffwythiant dosraniad cronnus a roddir
gan
F ( x) = 0
ar gyfer x < 0
F ( x) = 4 x 3 − 3 x 4
ar gyfer 0 ≤ x ≤ 1
F ( x) = 1
ar gyfer x ≥ 1
(c) Dangoswch bod y cwartel isaf yn gorwedd rhwng 0.45 a 0.46
(d) Darganfyddwch fynegiad ar gyfer f (x), sy’n ddilys ar gyfer 0 ≤ x ≤ 1, lle mae
f yn dynodi’r ffwythiant dwysedd tebygolrwydd ar gyfer X.
Ateb:
(a) F (0.45) = 4 × 0.45 3 − 3 × 0.45 4 = 0.2415 < 0.25
F (0.46) = 4 × 0.463 − 3 × 0.464 = 0.2550 > 0.25
(b)
f (x) = F′(x) = 12x2 – 12x3
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support
Programme Wales can accept no responsibility for its content matching WJEC specifications exactly.
Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth
Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
4