Lect05 BT1211.pdf

Statistics (BT1111)
Lecture 5: Continuous
Probability Distributions

Lect05 Continuous Probability Distributions 2
Agenda
◼ Lecture 4: discrete probability distr. (Keller Ch.7)
◼ random variables and probability distributions
◼ expected value and variance (discrete random variable)
◼ binomial distribution and Poisson distribution
◼ Lecture 5: continuous probability distr. (Keller Ch.8)
◼ cont. prob. distr. and probability density functions
◼ uniform distribution (continuous)
◼ normal distribution and use of standard normal table
◼ exponential distribution and its relation to Poisson distr.
N.B. Sect.8.4: self-reading (student t, chi-quared and F
distribution); will be discussed in later sessions

Continuous random variables
– Sect.8.1
◼ Discrete random variables:
◼ finite number of possible values (e.g. binomial variable; x =
0, 1, 2, …, n)
◼ countably infinite number of possible values (e.g. Poisson
variable; x = 0, 1, ...)
◼ Continous random variables:
◼ uncountably infinite number of possible values in an interval
(e.g. x can take all values in the interval [a, b])
◼ example (Lecture 4): amount of time (seconds) workers on
an assembly line take to complete a particular task → x can
take all non-negative real numbers

Continuous random variables
– cont. – Sect.8.1
◼ Discrete random variables:
◼ all possible values of random variable X can be listed
◼ it is meaningful to consider probability P(X = x)
◼ Continous random variables:
◼ all possible values cannot be listed: there is always another
possible value between any two of its values
◼ probability that a continuous random variable X will assume
any particular value is zero: P(X = x) = 0 !
◼ the only meaningful events for a continuous random variable
are intervals
◼ comparable situation: a line segment has a positive length,
while no single point on the line segment does

Probability density functions
– Sect.8.1
◼ The probability that X falls between a
and b is found by calculating the area
under the graph of f(x) between a and b:
Area = 1
f(x)

=
=
=


b
x
a
x
dx
x
f
b
X
a
P )
(
)
(
a b
f(x)
◼ To calculate probabilities we define a
probability density function f(x)
◼ The density function satisfies the
following conditions:
◼ f(x) is non-negative
◼ the total area under the curve
representing f(x) equals 1
Important: f(x)
 P(X = x)

Uniform distribution –
continuous case – Sect.8.1
◼ obtained as follows:
◼ (Continuous) uniform distribution:
◼ example of continuous random variable (r.v.)
◼ domain of continuous uniform r.v. X: a  x  b
◼ probability density function (p.d.f.):
◼ hence, outcomes are distributed evenly
◼ expected value and variance:
x b
1
f(x) a
b a
 
 
 
 
=
−
2
(b a)
a b
E(X) V(X)
2 12
−
+
= =
( )
x b x b 2
x a x a
E(X) xf(x)dx V(X) x f(x)dx
= =
= =
= = −
 

Uniform distribution
– example – Sect.8.1
◼ Solution:
◼ Example (= example 7.1 5th ed.):
◼ the time elapsed between the placement of an order and the
delivery time (X, measured in minutes) is uniformly
distributed between 100 and 180 minutes
◼ define the density function and draw the graph
◼ what proportion of orders takes between 2 and 2.5 hours to
deliver:
100 180
1/80
x
120 150
P(120 X  150) = (150-120)(1/80) = .375
?
)
150
120
( =

 X
P
f(x) = 1/80 100  x  180  probability density function

Normal distribution
– Sect.8.2
◼ Most important continuous distribution in statistics:
◼ many random variables can be properly modeled as normally
distributed
◼ many distributions can be approximated by a normal
distribution
◼ the normal distribution is the cornerstone distribution in
statistical inference; the (asymptotic) distribution of sample
means is normal (see next lecture)

Normal distribution
– definition – Sect.8.2
◼ Notation:
◼ Normal distribution:
◼ a random variable X is said to have a normal
distribution with parameters μ and σ if its p.d.f. is:
2
)
(
)
( 
=

= X
V
X
E
...
71828
.
2
and
...
14159
.
3
with
2
1
exp
2
1
)
(
2
=
=





−
















−
−


=
e
x
x
x
f
◼ Also:
( )

,
~ N
X

Normal distribution
– shape – Sect.8.2
◼ The p.d.f. of the normal distribution is bell shaped
and symmetrical around :
σ
μ

Normal distribution
– shape (cont.) – Sect.8.2
How does the standard deviation affect the shape of f(x)?
= 2
 =3
 =4
 = 10  = 11  = 12
How does the expected value affect the location of f(x)?

Standard normal distribution
– Sect.8.2
◼ Two facts help calculate normal probabilities:
◼ the normal distribution is symmetrical
◼ any normal distribution can be transformed into a specific
normal distribution, called standard normal distribution, with
mean 0 and variance 1
◼ Example:
◼ the time it takes to complete a standard entrance exam is
assumed to be normally distributed, with a mean of 60
minutes and a standard deviation of 8 minutes
◼ what is the probability that a student will complete the exam
within 70 minutes?

◼ Solution:
◼ if X denotes the time taken to complete the exam, we look
for the probability P(X < 70) where μ = 60 and σ = 8
◼ this probability can be calculated by creating a new normal
variable, the standard normal variable (Z):
E(Z) = 0
V(Z) = 1
Z measures the number of standard deviation
units (σ) that X deviates from its mean (μ)
Therefore, once probabilities for Z are calculated,
probabilities of any normal variable can be found
Every normal variable X with some μ
and σ, can be transformed into this Z
X
Z
− 
=


◼ To complete the calculation we need to compute the
standard normal probability P(Z < 1.25)
◼ Solution (cont.):
P(X  70) = P(X –   70 – 60)
= P(Z  1.25)
X 70 60
P
8
−  −
 
= 
 

 

Standard normal table
– Table 3 App. B – Sect.8.2
◼ standard normal probabilities
have been calculated and are
provided in a table:
The tabulated
probabilities
correspond to the
area between Z = −
and some Z = z0
z0
0
P(Z < z0)
Table 3 - Appendix B (pages B-8 and B-9)
z 0.03 0.04 0.05 0.06 0.07
1.0 0.8485 0.8508 0.8531 0.8554 0.8577
1.1 0.8708 0.8729 0.8749 0.8770 0.8790
1.2 0.8907 0.8925 0.8944 0.8962 0.8980
1.3 0.9082 0.9099 0.9115 0.9131 0.9147
1.4 0.9236 0.9251 0.9265 0.9279 0.9292

Standard normal table
– Table 3 App. B – Sect.8.2
z0
0
z 0.03 0.04 0.05 0.06 0.07
1.0 0.8485 0.8508 0.8531 0.8554 0.8577
1.1 0.8708 0.8729 0.8749 0.8770 0.8790
1.2 0.8907 0.8925 0.8944 0.8962 0.8980
1.3 0.9082 0.9099 0.9115 0.9131 0.9147
1.4 0.9236 0.9251 0.9265 0.9279 0.9292
In this example
z0 = 1.25
P(X  70) = P(X –   70 – 60)
= P(Z  1.25)
X 70 60
P
8
−  −
 
= 
 

 
= 0.8944
0.8944

−1.25 z
0 1.25
Calculating normal
probabilities – Sect.8.2
◼ Probabilities of the type P(Z > z0) can easily be
derived from Table 3 by applying the complement rule
(see Lecture 3), hence:
◼ P(Z>1.25) = 1 – P(Z<1.25) = 1 – 0.8944 = 0.1056
◼ From the symmetry of the normal distribution it
follows that P(Z < –1.25) = P(Z > 1.25) = 0.1056...
◼ …as also follows from Table 3
◼ Hence, P(Z < –z) + P(Z < z) = 1
for any z (verify yourself in Table 3!)
z 0.03 0.04 0.05 0.06 0.07
-1.4 0.0764 0.0749 0.0735 0.0721 0.0708
-1.3 0.0918 0.0901 0.0885 0.0869 0.0853
-1.2 0.1093 0.1075 0.1056 0.1038 0.1020
-1.1 0.1292 0.1271 0.1251 0.1230 0.1210
-1.0 0.1515 0.1492 0.1469 0.1446 0.1423
0.1056
0.1056
0.1056
P(Z < −z) = P(Z > +z)

Calculating normal prob.
– examples – Sect.8.2
◼ (a) find the probability P(Z > 1.47)
P(Z > 1.47) = 1 – P(Z < 1.47) = 1 – 0.9292 = 0.0708
Tip: always first
sketch the
distribution and
the relevant area !
1.47 z
0
0.9292
0.0708

Calculating normal prob.
– examples – Sect.8.2
◼ (b) find the probability P(−2.25 < Z < 1.85)
Hence, P(−2.25 < Z < 1.85) = 0.9678 − 0.0122 = 0.9556
P(Z < −2.25) = 0.0122 P(Z < 1.85) = 0.9678
−2.25 z
0 1.85
P(−2.25 < Z < 1.85) = P(Z < 1.85) – P(Z < –2.25)
0.9556

Normal distribution
– applications – Sect.8.2
◼ the rate of return (X) on an investment is normally distributed
with mean of 30% and standard deviation of 10%
◼ what is the probability that the return will exceed 55%?
0 z = 2.5
=1 – P(Z < 2.5) = 1 – 0.9938 = 0.0062
µ = 30% x = 55%
P(X > 55) = P(Z > ) = P(Z > 2.5)
55 - 30
10
0.9938
0.0062
0.0062 0.9938

P(X < 22) = P(Z < )
22 30
10
−
Normal distribution
◼ Example (cont.):
◼ what is the probability that the return will be less than 22%?
30% x
22%
0 z
−0.8
= P(Z < −0.8) = 0.2119
0.2119
0.2119

Normal distribution
◼ if Z is a standard normal variable, determine the value z for
which P(Z < z) = .6331
◼ → use Table 3 backward:
since P(Z < 0.34) = 0.6331
z 0.02 0.03 0.04 0.05 0.06
0.1 0.5478 0.5517 0.5557 0.5596 0.5636
0.2 0.5871 0.5910 0.5948 0.5987 0.6026
0.3 0.6255 0.6293 0.6331 0.6368 0.6406
0.4 0.6628 0.6664 0.6700 0.6736 0.6772
0.5 0.6985 0.7019 0.7054 0.7088 0.7123
0.6331
0.6331
0 z
z = 0.34

z 0.04 0.05 0.06 0.07 0.08
1.7 0.9591 0.9599 0.9608 0.9616 0.9625
1.8 0.9671 0.9678 0.9686 0.9693 0.9699
1.9 0.9738 0.9744 0.9750 0.9756 0.9761
2.0 0.9793 0.9798 0.9803 0.9808 0.9812
2.1 0.9838 0.9842 0.9846 0.9850 0.9854
Normal distribution
◼ Example (= example 7.5 5th ed.): determine z0.025 (critical
value)
◼ Solution: the value zA is defined as the z-value for which the
area to the right(!) of zA under the standard normal curve is A:
P(Z > zA) = A (right-tail probability)
=1.96
-1.96= −z0.025
0.025
Table 3: P(Z < 1.96) = 0.975
→ z0.025 = 1.96
1.9
0.06
0.9750
P(Z < z0.025) = 0.975 (= 1 – A)
z0.025
0
P(Z > z0.025) = 0.025 (= A)

Normal distribution
– properties – Sect.8.2
and
◼ Special case: X (= X1) and Y (= X2) are identically
distributed:
and X and Y are independent random variables
then
◼ If
and a = b = 1/2, then it follows that
(X1 + X2)/2 is sample
mean of X1 and X2
Verify
yourself !
( ) ( )
X X Y Y
X ~ N , and Y ~ N ,
   
( )
2 2 2 2
X Y X Y
aX bY ~ N a b , a b
+  =  +   =  + 
( )
2 2 2 2
X Y X Y
aX bY ~ N a b , a b
−  =  −   =  + 
( )
1 2 X x
X , X ~ N ,
 
1 2 x
X
X X
~ N ,
2 2
+ 
 
 =   =
 
 

Exponential distribution
– Sect.8.3
◼ The exponential distribution is frequently applied
when interested in elapsed time intervals
◼ Examples:
◼ the length of time between telephone calls
◼ the length of time between arrivals at a service station
◼ the life-time of electronic components
◼ When the number of occurrences of an event has a
Poisson distribution with average µ per unit of time,
the time between the occurrences has an exponential
distribution with average 1/µ time unit

◼ Notation:
◼ A random variable X is said to be exponentially
distribution with parameter  if its p.d.f. is:
◼ The expected value and the variance of X are:
( ) x
f x e x 0
where 0
−
=  
 
( ) ( ) 2
1 1
E X V X
= =
 
( )
X ~ Exp 

( ) ( )
( )
t a
t a
t t
t 0
t 0
a
P X a 1 P 0 X a
1 e dt 1 e
e
=
=
− −
=
=
−
 = −  
 
= −  = +  
=

◼ Also, the probability that an exponentially distributed
r.v. X will take a value greater than a specified
nonnegative number a can be easily computed using:

– shape – Sect.8.3
for  = 0.5, 1, 2
0
0.5
1
1.5
2
2.5
f(x) = 2e -2x
f(x) = 1e -1x
f(x) = 0.5e -0.5x
0 1 2 3 4 5
0
0.5
1
1.5
2
2.5
0 a b
P(a < X < b) =
= P(X > a) − P(X > b)
= e−a - e−b

– application – Sect.8.3
◼ Example (= example 7.6 5th ed.; see also example 6.11 5th ed.
mentioned in Lecture 4):
◼ cars arrive randomly and independently at a tollbooth at an
average of 360 cars per hour
◼ let Y = number of arrivals per minute, then we know already
from Lecture 4 that Y ~ Poisson( = 6) [average arrival rate
is 360/60 = 6 cars per minute]
◼ (a) use the exponential distribution to find the probability
that the next car will not arrive within half a minute
◼ (b) use the Poisson distribution to find the probability that
no car will arrive within the next half minute

◼ Solution (a):
◼ let X = time (in minutes) that elapses before the next car
arrives, then the desired probability is P(X > 0.5), i.e. it
takes more than 0.5 minute before next car arrives
◼ if cars arrive at an average rate of 6 cars per minute, then
the average inter-arrival time is 1/6 minute [= E(X) = 1/],
hence X ~ Exp( = 6)
◼ it follows that P(X > 0.5) = e−60.5 = e−3 = 0.0498
◼ note that the parameter of the Poisson variable Y ( = 6)
always equals the parameter of the corresponding
exponential variable X ( = 6), when same units of time
(here: minutes) are used

◼ Solution (b):
◼ now we use Y = number of arrivals per minute
◼ we know that Y ~ Poisson( = 6 per minute), or Y ~
Poisson( = 3 per half minute)
◼ desired probability P(Y = 0 (per half minute)):
Comment: If the first car will not arrive within the next half minute
then no car will arrive within the next half minute. Therefore, not
surprisingly, the probability found here is the exactly same
probability found in the previous question
( )
3 0
3
e 3
P Y 0 e 0.0498
0!
−
−
= = = =

◼ the lifetime of a transistor is exponentially distributed, with a
mean of 1,000 hours
◼ what is the probability that the transistor will last between
1,000 and 1,500 hours
◼ Solution:
◼ let X denote the lifetime of a transistor (in hours)
◼ E(X) =1000 = 1/, so  = 1/1000 = 0.001
◼ P(1000 < X < 1500) = e−0.0011000 − e−0.0011500 = 0.1448

End of Lecture 5
◼ Workshop exercises:
◼ Keller (9th int. stud. ed.) 8.6; 8.50; 8.63; 8.74; 8.78
◼ Self-study exercises:
◼ Keller (9th int. stud. ed.) 8.13; 8.31; 8.61; 8.79; 8.73
◼ Lecture 6:
◼ Sampling and Sampling Distributions (Keller Ch.5 and Ch.9
excl. 9.3)

Lect05 BT1211.pdf

Recommended

Recommended

More Related Content

Similar to Lect05 BT1211.pdf

Similar to Lect05 BT1211.pdf (20)

Recently uploaded

Recently uploaded (19)

Lect05 BT1211.pdf