STAT 211
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Handout 4 (Chapter 4): Continuous Random Variables
A r.v. X is said to be continuous if its set of possible values is an entire interval
of numbers. Then a probability distribution of X is f(x) (pdf: probability density
b
function) such that for any two numbers a and b, P(a  X  b)   f ( x)dx.
a
Conditions to be a legitimate probability distribution:
(i)
f(x)  0, for all x

(ii)
 f ( x)dx  1
(area under the entire graph of f(x)).

If X is a continuous r.v., then for any number c, P(X=c)=0.
x
Cumulative Distribution Function for X (cdf) : F(x)= P( X  x) 
 f ( y)dy.

F(-)=0, F()=1, P(a  X  b)  P(a  X  b)  F (b)  F (a) , P(X>a)=P(Xa)=1-F(a)
Example 1: A college professor never finishes his lecture before the bell rings to end the
period and always finishes his lectures within 2 min after the bell rings. Let X=the time
that elapses between the bell and the end of the lecture and suppose the pdf of X is
k  x 2 , 0  x  2
.
f ( x)  
otherwise
0,
(a) Find the value of k which makes f(x) a legitimate pdf. (Answer:3/8)
(b) The cumulative distribution function, F(X)?
F(0) =
F(0.5) =
F(1) =
F(3) =
Example 2: The amount of bread (in hundreds of pounds) that a certain bakery is able to
sell in a day is a random variable with probability function,
0 x5
 Ax,

f ( x)   A(10  x),
5  x  10
0,
otherwise

(a) Find the value of A which makes f(x) a legitimate pdf.
10
5
10
 x2 5 
x2  
0 Axdx  5 A(10  x)dx  A 2  10 x  2    25 A . Then A=1/25
5 
 0
(b) What is the probability that the number of pounds of bread that will be sold tomorrow
is
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(i)
more than 500 pounds
10
1
1 
x2 
10 x    0.5
P(X>5)=  (10  x)dx 
25
25 
2 5
5
(ii)
less than 500 pounds
10
5
1
1 x2
P(X<5)=  xdx 
25
25 2
0
5
 0.5
0
(iii)
between 250 and 750 pounds
7.5
5
1
1
P(2.5<x<7.5)=  xdx +  (10  x)dx =0.75
25
25
5
2.5
(c) What is the F(x)?
x0
0,

x
2
0  1 xdx  x ,
0 x5
 0 25
50
F(x)= 
x
2
 1  1 (10  x)dx  1 10 x  x  50  5  x  10
 2  25
25 
2
2 
5

1,
x  10
Obtaining f(x) from F(x) : If X is a continuous r.v. with pdf f(x) and cdf F(x), then at
every x at which the derivative exists, F`(x)=f(x).
Percentile of a continuous distribution: Let p be a number between 0 and 1. the
(100p)th percentile of the distribution of a continuous r.v. X, denoted by r(p), is defined
r ( p)
by p  F (r ( p))  P( X  r ( p)) 
 f ( y)dy.

P(Xmedian) = P(X>median) = 0.50
Expected value for the continuous random variable, X:   E ( X ) 

 x  f ( x)dx.

Variance for the continuous random variable, X:

  Var ( X )  E[( X   ) ]   ( x   ) 2  f ( x)dx or
2
2

the shortcut is   Var ( X )  E ( X )  
2
2

2
where
E( X ) 
2
x

If h(X) is the function of random variable X,

E (h( X )) 
 h( X )  f ( x)dx

2
 f ( x)dx
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
Var (h( X )) 
 (h( X )  E (h( X )))
2
 f ( x)dx

If h(X) is a linear function of X, the rules of the mean and the variance can directly be
used instead of going through the mathematics.
Example 1(continue): Go back to the example with the college professor in this chapter
and show that
(i)
90th percentile of X is 1.931
Set F(r)=0.90 and solve for r to find it. r is the 90th percentile of X.
(ii)
the median is 1.5874
Set F(median)=0.50 and solve for median to find it.
(iii)
the expected value of X is 1.5
2
E ( X )   x  f ( x)dx   x(kx2 )dx =1.5
0
(iv)
the variance of X is Var(X)= 0.15
2
E ( X 2 )   x 2  f ( x)dx   x 2 (kx2 )dx =2.4
0
where Var(X)= E(X2)-[E(X)]2
(v)
how to obtain f(x) from F(x).
Example 2 (continue):
(i)
90th percentile of X is 7.7639
(ii)
the median is 5
(iii) the expected value of X is 5
(iv)
the variance of X is 4.1667
Uniform Distribution: X ~U[a,b]
1
f ( x) 
, axb
ba
ab
(b  a) 2
E(X)=
and Var(X)=
2
12
Example 3:
We would like to plan classes so that the lengths of these are uniformly distributed
between 50 min and 52 min on Monday-Wednesday-Friday (MWF). Let X be the length
of the randomly selected MWF class.
(i) What is the pdf for x?
(ii) Find the probability that a randomly selected class will last longer than 51.5 minutes.
(iii) What is the expected length of the class?
(iv) What is the variance in the length of the class?
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Normal Distribution: X ~ N ( , 2 )
A continuous r.v. X is said to have a normal distribution with parameters  and 2
where - <  <  and  > 0. The pdf of X is
f ( x;  ,  ) 
1
 e ( x   )
2
/ 2 2
,
2 
It is symmetric and bell-shaped.
   x  ,
The standard normal random variable Z 
X 
     ,
2 0
has =0 and 2 =1.
Z ~ N(0,1)

The cdf of Z is  (z ) =P(Zz). Appendix table A.3 can be used to compute  (z ) .
The (100p)th percentile of X with N ( ,2) =  + [The (100p)th percentile of Z with N (0,1)]
Example 4: In mathematics test, if we assume that your test scores (X) were
approximately normally distributed with mean of 76.8 and standard deviation of 13.94.
a. If a score below 60 represents a grade of F (failure), approximately what percent of
students failed the test?
60  76.8 

P(X<60)= P Z 
 =P(Z<-1.21)=0.1131=11.31%
13.94 

b. If the cutoff for a grade of A is the lowest score of the top 15%, what is that cutoff
point?
x * 76.8 

P(Xx*)=0.15 then P(Zz*)= P Z 
 =0.15.
13.94 

x * 76.8
 1.04 then x*=91.2976
By looking at the table, z*=
13.94
c. How many points must be added to the student scores so that only 10% fail (less than
60 be the failing grade)?
x * 76.8 

P(X<x*)=0.10 then P(Z<z*)= P Z 
 =0.10.
13.94 

x * 76.8
 1.28 then x*=58.9568. 1.0432 should be added.
z*=
13.94
d. If the cutoff for a grade of C is the lowest score of the top 45%, what is that cutoff
point?
x * 76.8 

P(Xx*)=0.45 then P(Zz*)= P Z 
 =0.45.
13.94 

x * 76.8
 0.13 then x*=78.6122.
z*=
13.94
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e. What is the 90th percentile of test scores (x)?
x * 76.8 

P(Xx*)=0.90 then P(Zz*)= P Z 
 =0.90.
13.94 

x * 76.8
 1.28 then x*=94.6432.
z*=
13.94
Example 5:A chemical plant superintendent orders a process shutdown and setting
readjustment whenever the pH of the final product falls below 6.9 or above 7.1. The
sample pH is normally distributed with unknown  and standard deviation 0.05.
Determine the probability
(a) of readjusting when the process is operating as intended and =7
P(X<6.9 or X>7.1)=P(X<6.9)+P(X>7.1)=P(Z<-2)+P(Z>2)=0.0228+0.0228=0.0456
(b) of failing to readjust when the process is too alkaline and the mean pH is =7.15
P(6.9 X  7.1)=P(X 7.1)-P(X< 6.9)=P(Z -1)-P(Z<-5)=0.1587-0=0.1587

IF X is a random variable whose logarithm is normally distributed then X has a
lognormal distribution.
Chebyshev's Inequality : P(| X   |)  k   ) 
1
k2
is the probability that the value of
X lies at least k standard deviations from its mean is at most
1
where k is a positive
k2
number greater than 1.
Example 6: If k=2 then the probability that the value of X lies at least 2 standard
deviations from its mean is at most 1/4=0.25=25% or 75% of the values are within two
standard deviation of the mean.
Empirical Rule: If the population distribution of a random variable is (approximately)
normal, then
 Roughly 68% of the values are within one standard deviation of the mean
 Roughly 95% of the values are within two standard deviations of the mean
 Roughly 99.7% of the values are within three standard deviations of the mean
SAT scores have a distribution that is roughly unimodal and symmetric, and are designed
to have an overall mean of 500 and a standard deviation of 100. In any one year, the
mean and standard deviation may differ from these target values by a small amount. But
they are a good overall approximation. Suppose you earned 600 on an SAT test, where
do you stand among all students who took the SAT?
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According to the empirical rule, if those scores are normally distributed, roughly 68% of
the scores are within one standard deviation of the mean, (400,600). A score 600 is one
standard deviation above the mean. About 32% of those who took the test were more
than one standard deviation from the mean, but only 16% of those were higher than 600.
Exercise: Let’s confirm the findings on the empirical rule using the z-scores.
If those scores are not normally distributed, at least 75% of the scores are within 2
standard deviations of the mean (between 300, 700) according to the chebyshev’s rule
Normal Approximation to the Binomial Distribution:
Let X be a binomial r.v. based on n trials with success probability p. Then if the binomial
probability histogram is not too skewed, X has approximately a normal distribution with
 x  0.5  np 


  P Z  x  0.5  np 
 = np and   np(1  p) then P( X  x)  
 np(1  p) 

np(1  p) 



(check if np10 and n(1-p)10 to use the formula).
Example 7 (Exercise 4.50, 6th edition, Exercise 4.48, 5th edition): Suppose that 10% of all
steel shafts produced by a certain process are nonconforming but can be reworked (rather
than having to be scrapped). Consider a random sample of 200 shafts and let X denote the
number among these that are nonconforming and can be reworked. What is the
approximate probability that X is
(a) At most 30?
P(X30)=P(Z2.48)=0.9934
(b) Less than 30?
P(X<30)= P(X29)=P(Z2.24)=0.9875 (because X is discrete)
(c) Between 15 and 25 (inclusive)?
P(15X25)=P(X25)-P(X14)= P(Z1.30)-P(Z-1.30)=0.9032-0.0968=0.8064
The Gamma Distribution: X ~ Gamma (  ,  )
1
 x  1  e  x /  ,
x  0,
  ( )
It is a skewed distribution. Note that
f ( x) 


( )   x  1 e  x dx,
,   0
 0
0
 (  1)  (  1),
 (  1)!,
(1 / 2)   ,

 1
 is any positive integer
E( X )     ,
Var ( X )     2
If =1 then it is called standard gamma distribution.
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
When the random variable is a standard gamma r.v. then the cdf is called the
incomplete gamma function (Appendix Table A.4). F(x;,)=F(x/;)

If  = 1 then it is Gamma ( =1 , =1/ )~Exponential(=1/). It is used as a model
for the distribution of times between the occurrence of successive events. The gamma
distribution is used for the segment of time or and space occurring until some
specified number of events has transpired where  being the average process rate and
 being the specified number of events that must transpired as X is reached.
The
pdf
for
exponentially
distributed
random
variable
is
 x
f ( x)    e ,
x  0,   0
P(X>x*)= e  x*
E(X)= 1/ and Var(X)=1/ 2

If  = r / 2 and  = 2 then it is Gamma ( =r/2 , =2 ) ~ Chi-squared(r). (Appendix
table A.7)

If the random variable, X is distributed Exponential() then Y=X1/ has Weibull(,).
Example 8 (Exercise 4.56, 6th edition, Exercise 4.57, 5th edition): Suppose the time spent
by a randomly selected student who uses a terminal connected to a local time sharing
facility has a gamma distribution with mean 20 and variance 80 min2.
(a) What are the values of  and ?
E(X)==20 and Var(X)=2=80 then =4 and =5
(b) What is the probability that a student uses the terminal for at most 24 min?
P(X24)=P(X/6)=0.715 using the table of incomplete gamma function
(c) What is the probability that a student spends between 20 and 40 min using the
terminal?
P(20X40)=P(5X/10)= P(X/10)- P(X/<5)=0.971-0.56=0.411 using the
table of incomplete gamma function
Example 9: Find the probability that the time taken for the next two cars to arrive at a
tollbooth will be 1 minute or less when =2 per minute.
X~Gamma(=2,=1/2)
P(X 1)=P(X/2)=0.594 using the table of incomplete gamma function
1
Or using the pdf of gamma  4 xe 2 x dx =0.594
0
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Example 10: Flaws in a reel of high-fidelity radar recording tape occur on the average of
once every 10 feet. Determine the probability that the next recording will begin on a
flawless stretch of tape over 5 feet long.
X~Exponential(=0.1)

P(X>5)=  0.1e 0.1x dx =0.6065
5
Example 11: A series system consists of 100 independent units, each with exponential
distribution with =0.005. Find the system reliability over a span of t=10.
Rs(10)= P( X  10)
100


=   0.005e 0.005x dx 
10

100
=(0.9512)100=0.0067
Example 12 (Exercise 4.61, 6th edition, Exercise 4.60, 5th edition): Extensive experience
with fans of a certain type used in diesel engines has suggested that the exponential
distribution provides a good model for time until failure. Suppose the mean time until
failure is 25000 hours. What is the probability that
1
f ( x) 
e  x / 25000 , x>0
25000
(a) A randomly selected fan will last at least 20000 hours?
P(X20000)= e 20000/ 25000
A randomly selected fan will last at most 30000 hours?
P(X30000)= 1- e 30000/ 25000
A randomly selected fan will last between 20000 and 30000 hours?
P(20000X30000)= e 20000/ 25000 - e 30000/ 25000
(b) The lifetime of a fan exceeds the mean value by more than 2 standard deviations?
P(X>+2)=P(X>25000+2(25000))= e 75000/ 25000
The lifetime of a fan exceeds the mean value by more than 3 standard deviations?
P(X>+3)=P(X>25000+3(25000))= e 100000/ 25000
Exponential distribution shares memoryless property of the geometric distribution then
P(Xt+t0|Xt0)=P(Xt). It simply is the distribution of additional lifetime is exactly the
same as the original distribution of lifetime.
Probability plots: Order the n-sample observations from smallest to largest. Then the ith
smallest observation in the list is taken to be the [100(i-0.5)/n]th sample percentile.
STAT 211
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Example 13 (Exercise 4.82, 6th edition, Exercise 4.80, 5th edition): Ten observations on
bearing lifetime (in hours) are collected. Construct a normal probability plot and
comment on the plausibility of the normal distribution as a model for bearing lifetime.
Data
152.7
172
172.5
173.3
193
i
1
2
3
4
5
(i-0.5)/n
.05
.15
.25
.35
.45
z
-1.645
-1.036
-0.675
-0.385
-0.126
Data
204.7
216.5
243.9
262.6
422.6
i
6
7
8
9
10
(i-0.5)/n z
.55
0.126
.65
0.385
.75
0.675
.85
1.036
.95
1.645
data
450
400
350
300
250
200
150
100
50
0
-2
-1
0
1
2
z
Normal Probability Plot for lifetime
ML Estimates - 95% CI
99
ML Estimates
95
Mean
221.38
StDev
74.6053
90
Goodness of Fit
Percent
80
AD*
70
60
50
40
30
20
10
5
1
0
200
Data
400
1.902
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Example 14 (Exercise 4.91, 6th edition, Exercise 4.87, 5th edition): The failure time
observations (1000’s of hours) resulted from accelerated life testing of 16 integrated
circuit chips of a certain type. Use the corresponding percentiles of the exponential
distribution with =1 to construct a probability plot. Comment on the sample having
been generated from any exponential distribution.
F(x)= 1  e x  1  e  x , x  0
That means if the smallest 5 th percentile is observed then F(x)=0.05 and we are trying to
find what x is. x can be found as -ln(1-F(x)).
Data
11.6
26.5
82.8
179.7
204.6
212.6
229.9
242
244.8
304.3
307.8
359.5
366.7
379.1
502.5
558.9
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
(i-0.5)/n
5/160
15/160
25/160
35/160
45/160
55/160
65/160
75/160
85/160
95/160
105/160
115/160
125/160
135/160
145/160
155/160
x
0.031749
0.09844
0.169899
0.24686
0.330242
0.421213
0.521297
0.632523
0.757686
0.900787
1.067841
1.268511
1.519826
1.856298
2.367124
3.465736
4
3.5
3
2.5
2
1.5
1
0.5
0
0
100
200
300
400
500
600