M.BILAL
TARIQ
SUFIAN
IQBAL
M.NAVEED
Exponential Probability
Distribution
The exponential probability distribution is useful in describing the
time it takes to complete a task.
The exponential random variables can be used to describe:
Time between vehicle arrivals at a toll booth
Time required to complete a questionnaire
Distance between major defects in a highway
The time between goals scored in a World Cup
soccer match
f x e x
( ) /
 1


for x > 0,  > 0
Density Function
where:  = mean
e = 2.71828
Cumulative Probabilities
P x x e x
( ) /
   
0 1 o 
where:
x0 = some specific value of x
Example: Al’s Full-Service Pump
The time between arrivals of cars at Al’s full-service gas pump follows
an exponential probability distribution with a mean time between
arrivals 3 minutes. Al would like to know the probability that the time
between two successive arrivals will be 2 minutes or less.
P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866
Solution:
Example:
If jobs arrive every 15 seconds on average, λ = 4 per minute,
what is the probability of waiting less than or equal to 30
seconds, i.e .5 min? P(T ≤ .5).
Characteristics :
The mean and standard deviation of exponential distribution are equal.
The distribution is extremely skewed and there does not exist any mode.
.
Graph of the exponential distribution with μ = 1
QUESTION

Exponential probability distribution

  • 1.
  • 2.
  • 3.
    The exponential probabilitydistribution is useful in describing the time it takes to complete a task. The exponential random variables can be used to describe: Time between vehicle arrivals at a toll booth Time required to complete a questionnaire Distance between major defects in a highway The time between goals scored in a World Cup soccer match
  • 4.
    f x ex ( ) /  1   for x > 0,  > 0 Density Function where:  = mean e = 2.71828
  • 5.
    Cumulative Probabilities P xx e x ( ) /     0 1 o  where: x0 = some specific value of x
  • 6.
    Example: Al’s Full-ServicePump The time between arrivals of cars at Al’s full-service gas pump follows an exponential probability distribution with a mean time between arrivals 3 minutes. Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less. P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866 Solution:
  • 7.
    Example: If jobs arriveevery 15 seconds on average, λ = 4 per minute, what is the probability of waiting less than or equal to 30 seconds, i.e .5 min? P(T ≤ .5).
  • 8.
    Characteristics : The meanand standard deviation of exponential distribution are equal. The distribution is extremely skewed and there does not exist any mode.
  • 9.
    . Graph of theexponential distribution with μ = 1
  • 11.