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Uniform random variable
• A random variable is said to be uniformly distributed
over the interval (a, b) if its probability density function is
1
given by
, axb
ba
f(x) 
0, otherwise
• Theorem. If U is uniformly distributed over (0, 1) and a
and b are real numbers with a < b, then X = (b – a)U + a is
uniformly distributed over (a, b).
• The latter theorem will allow us to use Maple’s random
number generator to generate any uniformly distributed
random variable that we want.
Uniform random variable, continued
• Theorem. If X is uniformly distributed over (a, b), then
(b  a) 2
ab
E[X] 
, Var(X) 
.
2
12
• Problem. Buses arrive at a specified stop at 7:15am and
7:30am. If a passenger arrives at the stop at a time that is
uniformly distributed between 7 and 7:30, find the
probability that she waits less than 5 minutes for the bus.
Solution. We use a uniform r. v. on (0, 30). The desired
probability is given by:
P{10  X  15}  P{25  X  30}

15

1
10 30
dx 

30
1
25 30
dx  13 .
Normal random variables
• X is a normal random variable (or normally distributed)
with parameters (, 2) if the density of X is
1
( x   ) 2 / 2 2
f(x) 
e
,  x  .
2
• We often write X ~ N( ,  2 ).
• Theorem. If X is normally distributed (, 2), then
1. Z = (X – )/ is normally distributed (0, 1), that is,
Z is a standard normal random variable.
2. E[X] = .
3. Var(X) = 2.
Cumulative distribution function of a standard normal r. v.
• The c. d. f. of a standard normal random variable is
traditionally denoted by (x). That is,
x
1
 y2 / 2
 (x) 
e
dy.
2 
 y2 / 2
• Theorem. It follows from the symmetry of e
that

(x)  1  (x).
• (x) is tabulated in the textbook in Tables 1 and 2.
• Example. Let X be a normal random variable with
parameters (3, 9). We “standardize” to get the following
probability:
23 X3 53
P{2  X  5}  P{


}
3
3
3
 (2 / 3)  (1 / 3)  0.3779
68-95-99.7 Rule
• If X~ N( ,  2 ), then
P(  -   X     )  0.68
P(  - 2  X    2 )  0.95
P(  - 3  X    3 )  0.997.
Scottish Soldier’s Chest Size is Normally Distributed
• Let X be the chest measurement in inches of the chest size
of a Scottish soldier. Belgian scholar Quetelet has
determined that X ~ N(39.8, 2.052) .
• What is the probability that a randomly selected Scottish
soldier has a chest size of 40 or more inches?
X  39.8 40  39.8
P(X  40)  P(

)
2.05
2.05
 P(Z  0.10)  1  (0.1)  0.46
• Note the use of Z to denote a standard normal r. v.
Scores on an Achievement Test
• The scores on an achievement test given to 100,000
students are normally distributed, N(500, 1002). What
should the score of a student be to place him among the top
10% of all students?
• We want to find the score t such that
P(X  t)  0.10 or P(X  t)  0.90.
• This becomes
X  500 t  500
t  500
P(

)  0.90, or (
)  0.90.
100
100
100
• From Table 2,
t  500
 1.28 so that t  628.
100
Normal approximation to the binomial distribution
• Theorem (DeMoivre-Laplace) If Sn denotes the number of
successes in n independent trials, each resulting in a success
with probability p, then for any a < b,
Sn  np
P{a 
 b}  (b)  (a), as n  .
np(1  p)
• Problem. Let X be the number of times a coin, flipped 40 times,
lands heads. Find P{X=20} using the normal approximation
with the continuity correction.
Solution.
P{X  20}  P{19.5  X  20.5}
 P{.16 
X  20
 .16}
10
 (.16)  (.16)  0.1272
Exponential random variable
• A continuous random variable whose density is
f(x) 
 e  x , x  0
0 , x0
for some  > 0 is an exponential random variable with
parameter .
• Theorem. For an exponential r. v. with parameter ,
F(a) 
1  e  a , a  0
0 , a0
Also, E[X] = 1/  and Var(X) = 1/ 2.
Exponential random variable continued
• The exponential random variable is often the amount of time until
some specific event occurs. For example, it could be the amount
of time (starting now) until an earthquake occurs.
• Problem. Suppose that the length of an ATM session in minutes
is an exponential random variable with  = 0.1. If someone
arrives immediately ahead of you at an ATM, find the probability
that you will wait between 10 and 20 minutes.
Solution.
P{10  X  20}  F(20)  F(10)  e 1  e 2  0.233
• A nonnegative random variable X is memoryless if
P{X  s  t | X  t}  P{X  s} for all s, t  0.
• Theorem. Exponentially distributed random variables are
memoryless.
Proof of memoryless property for exponential random variable
• We want to show that
P(X  s  t, X  t)
 P(X  s).
P(X  t)
• This is equivalent to
P(X  s  t)  P(X  s)P(X  t).
• The result follows using the following:
P(X  s  t)  1  [1  e  (s t) ]  e  (s t)
P(X  s)  1  [1  e s ]  e s
P(X  t)  1  [1  et ]  et
The Cauchy distribution
• A random variable has a Cauchy distribution with parameter 
if its density is given by
1
1
f( x) 
,    x  .
2
 1 (x   )
• Theorem. If the angle  in the figure below is uniformly
distributed from –/2 to /2, then X has the Cauchy distribution
with  = 0.
flashlight

1
x-axis
X
Flashlight Interpretation
• Suppose that a narrow beam flashlight is spun around its
center, which is located a unit distance from the x-axis (see
the figure on the previous slide). When the flashlight has
stopped spinning, consider the point X at which the beam
intersects the x-axis. (If the beam is not pointing toward
the x-axis, repeat the experiment.)
• F(x) = P(X  x)  P(tan   x)  P(   arctan x)
1 1
  arctan x,
2 
where the latter equality follows since α is uniformly
distributed over (–/2, /2). Upon differentiation of F(x),
we see that the density function of X is the density
function of the Cauchy distribution with  = 0.