Exponential Distribution With Mean Life Time

y

fx



x





x

fx fx



x

y

x

fx



x

x 1

2

3

4

5

6

7

8

9

x

If X ~ U(a,b) show:

(a) E[X] =

a+b 2

(b) Var[X] =

(b  a)² 12

fx fx



x

for x  0 ,  a positive constant

is said to follow an exponential distribution.

X ~ Exp().

y

The Graph of f x  x is shown The exponential distribution can be thought of as the continuous equivalent of the Poisson distribution. e.g. the number of cars arriving at NIST during one minute could be modeled as Poisson while the time between arrivals could be modeled as exponential. Note 1:

P(X < a) =

a ex  0 x

= [e



dx a

]0

= 1  ea

and hence P(X > a) = ea

i.e. the cumulative density function is P(X < x) = F(x) = 1  ex

Note 2:

x

P(X > a + b | X > a)

P(X > [a + b] X > a) = P(X > a) P(X > [a + b] ) P(X > a) (a+b) e = a e a e × eb = ea = eb = P(X >b) =

, x0

Exercise: 1. Show that f x

x

for x  0 is a p.d.f 1 2. If X ~ Exp() , show that: (i) E[X] =  

(ii) Var[X] =

1 2

Examples: 1. The lifetime of one type of light bulb follows an exponential distribution with mean life time of 1000 hours. (a) Find the probability a bulb is still working after 1300 hours (b) Find P(X > 1500 | X > 1300) (c) Find the standard deviation of the lifetime of this type of bulb

E[X] = 1000 =

1 

 

= 0.001 and



f(x) = 0.001e0.001x x  0

(a) P(X > 1300) = e0.001×1300 = 0.273 (3 sf) (b) P(X > 1500 | X > 1300)= P(X > 200) = e0.001×200 = 0.819 (3 sf) (c) Var[X] =

1 

2



SD =

1 



SD = 1000 hours

2. On Sukhumwhit road, accidents occur at random at the rate of 3 per day. Find the probability that, after a particular accident has occurred, at least one day goes by without another. If Y is the number of accidents in one day then Y ~ Po(3) If X is the time in days between successive accidents then

E[X] =

1 1 =  3

 =3

and hence f(x) = 3e3x x  0

P(X > 1) = e3 = 0.0498 Exercise 6G Page 362 Crawshaw and Chambers

Formula Book: