National Taiwan University - Calculus 2 for Class 11-14
Worksheet 3: Applications in Probability
Name:
賴賴怡怡帆帆
ID:
binerlblzrerl112v
Department: acot
Introduction.
• Given a continuous random variable X whose domain is the whole real line, a probability density
function
Z 1(p.d.f.) of X is a non-negative function f (x) defined on R such that
(i)
f (x) dx = 1,
1
(ii) For any interval [a, b], the probability of X 2 [a, b] is given by P(X 2 [a, b]) =
Z b
f (x) dx.
a
• Given a probability density function f (x) of a continuous random variable X, we define its expected
value E(X), its variance var(X) and its standard deviation (X) by
Z 1
Z 1
E(X) =
xf (x) dx ;
var(X) =
(x E(X))2 f (x) dx ;
1
(X) =
1
In particular, these definitions are defined by improper integrals.
p
var(X).
Question 1. Many random variables X are modelled by a normal distribution whose probability density func-
1
tion equals
(x µ)2
1
e 2 2 , where µ, are constants and > 0.
式 x ( x ) 20
2⇡
Z 1
Z 1
(x µ)2
p
2
1
p
(a) It is known that
e x dx = ⇡. By a suitable substitution, deduce that
e 2 2 dx = 1.
2⇡
1
1
(b) Compute the expected value and standard deviation of X.
(x µ)2
x2
x
1
(c) Let f (x) = ce 8 + 4 where c is a constant. Complete the square and rewrite f (x) in the form of p
e 2 2 .
2⇡
Z
fX (x) = p
…
1
Find c so that f (x) is a probability density density function (i.e.
f (x) dx = 1) and find its expected
1
value and standard deviation by just reading the simplified form of f (x).
昌 exp - ( ⾳] ] dx lety
ca )
[
程鹿 exp [ yr) 冷
=
=
]
→
虎和 eidy 完玩
=
By definition
E
σ
(x )
=
{‰
x
.5
=
=
量 dy 叔 dx
=
dy
1
dx
dx λ 鹿exp [ - [晨 ] dx 0 lety x dy 鹿
/
xx )-
*
晶
x = U + vroy
U 鹿 eRody+ roy eEody
o
=µ 鹿念 eydy
u
.
品
=
( C)
+
:
yez . dd (
=
By definition Var ( ) 念[ x- Ex ] fx ] dx
M roy
ety *
{ 是 x µ= expt ( “Jydx
ye - idy
oy 7 e 哈 vdy = 晶吃⼼是
=
x
,
鹿
=
鼠
)
=
/
=
晶
λ=
+
-
1
=
九
}
=
2
?
=
=
=
yi " dy + let
Yz dy 亞
玩
=
dz
:
=
奇函數)
兒了 ze 起 dz
。
=
=
學 iiizdz
⼼
2 信 ] 器嘻 =
是 E
2
O [ x ]i =σ
Distribution Function.
For a random variable X with p.d.f. f (x), we define its distribution function F (x) by
Z x
F (x) = P(X  x) =
f (t) dt.
1
Using the Fundamental Theorem of Calculus (F.T.C.), we have that F 0 (x) = f (x). This provides us a way to
recover the probability density function from its distribution function.
For example, if we define a new random variableY = aX + b where a > 0, then its distribution function is
✓
◆ Z y b
a
y b
F (y) = P(Y  y) = P(aX + b  y) = P X 
=
f (x)dx
a
1
and hence the p.d.f of Y is given by
d
dy
F 0 (y) =
Z ya b
f (x)dx
F.T.C. 1
=
1
a
f
✓
y
b
a
◆
.
Question 2. Suppose that a random variable X has probability density function
(x µ)2
1
fX (x) = p
e 2 2 , where µ, are constants and
2⇡
Define Y = aX + b where a, b are constants and a > 0.
> 0.
(a) Write down the distribution function of Y , F (y) = P(Y  y), as an integration of fX (x).
(b) Apply the Fundamental Theorem of Calculus and find the probability density function of Y . What is the
expected value and standard deviation of Y ?
(c) Find constants c and d such that the random variable Z = cX + d has expected value 0 and standard
deviation 1. (In this case, Z is called to have the standard normal distribution.)
( a)
Fy ( y )
Y
≤
=
P (Y ε y )
a 70
ax + b ≤ y
y
P (Y ≤ y )
=
=
ax ≤ y
-
b
号
→ x ≤Y
P( x ≤ )
{器 5
fy ( y ] = Fy " y )
=
⼆
x( u
= 品 exp 燕ldx
) dx
嘿器了六
鹿exp ( l
anl
}
texp 1 apy
R
_
Yis
a
normal distribution
b
withexpected Value
=
b + aU
and standard deviation = a σ
) Z = cx +
d 影恐 ) ,{ 跟台
a
+
cm陷些灣
"
2
Question 3. A person with IQ 157 is considered as very smart in a population. In this question, we will clarify this
claim in terms of probability.
(a) Indeed IQ is an example of a continuous random variable. Find the following information regarding IQ
online. What is its probability density function? What are the expected value and standard deviation?
(b) Find constants a and b such that a IQ + b is the standard normal distribution.
(c) Estimate P(IQ
157) as follows : firstly find a constant c such that P(IQ
157) = P(X
c) where X
has the standard normal distribution. Then look up the later probability from a table of standard normal
distribution.
functionofIeis
[ a) Theprobabilitg density
(x M o
exp 1 Ʃ
)
-
F 2 Q 1x 3
=
nO.⼆.
Where UzQ
and
σzQ
= ( 5↑
zvrzu
theexpectedvalue
is
100
=
{ the
bisolve d Mrntb = o
1 lal
standard cdeviation
we
a=
⽅
.
|
σra = 1
may as { um
b=
a
-
loo atbao
lal
.
.
.
2
⾄ 0 ,
then
1 o=
looa 的
=
. u5 = 1
号
tundard normal
《]
[ UIQ b 2 a . 15
IQZ 157 ⑤
P [ Q 2 [5] ] = P ( x 23 8 ] = 1
.
=
1
-
b=
y
+
5
” 管⼏
_
p ( λ< 3 8 )
.
-
0
.
99 q } = 0 00 07
.
3
( ige
Question 4.
Suppose X is a continuous random variable with standard normal distribution and set Y = X 2 .
(a) Write down the distribution function of Y as an integral. (You don’t need to evaluate it.)
(b) Find the probability density function of Y . This probability density function is called the Chi squared
⾃由度為 1 的卡⽅分佈
distribution.
(c) Compute the expected value and variance of Y .
@)
Fr ( y )
=
P (y ≤y ]
=
Pt } ≤ x ≤ 可 )
whem y 20
Yzy
Fyly)
=
}
{河 ⼦
xi≤
x (x )
y Ʃ 河≤x ≤可
dx
-
=
/ 荷 e 吃 dx wem
瓷塔 dx = 了 ^dx
whenyco
P [ Y ≤ y ] = P ( d]
b)
=
0
fyly ) Fily )
=
.
whenyzofyly ] 式與
=
whenyco fely )
,
tucy
¢1
By
Eu
)
d
等
境塔前
E
=
0
)
"
⾺彭 ]
(
let t 吃 y
dy zdt
=
⽅平均 平均平⽅
-
=} 最 yfycy )dy =xy 鹿 y 拉
鹿S: y
活
蔬走 y ( 爾 王⼼
{
=
}
=
=
= 鹿 Yeiy
光
^
=
王]
definition
]
tz
[ recall Rz) { % etdt
vary
9
dy
( Yy fy yidy
=
_
VwlY ]
=
y
=
=
dy
鹿徵
{ [y E ( y ) ] gy ( y )
-
1]
到
{+
Elyp
}
=
4
_
)
店 yzfylyldy = 。 鹿 y ly
iy 鹿了瓦 eidz 虎已信 ]
=
(
鹿 : l 2tietzdt
=
3
dy 流
=
Var ( y
)
= 三三 2
啦