1 stat 500 – statistics for managers stat 500 statistics for managers

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STAT 500 – Statistics for Managers

STAT 500Statistics for Managers

STAT 500Statistics for Managers

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Agenda for this Session

• Continuous Random Variables

• Uniform Distribution

• Exponential Distribution

• Applications

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• Applications

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Continuous Random Variables

•A continuous random variable takes a large set of possible values over an interval. •Examples:

The fuel efficiency of your car Your body temperature The value of your stock portfolio The daily sales value in a grocery store

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Continuous Vs Discrete Random Variables

•A discrete variable takes specific values while the continuous variable could take any value in a range or interval. •Look at the following questions:

•In what region of the country were you born? •How many siblings do you have? •What is the speed of travel of a jet plane?

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Probability Density Function

•When dealing with continuous random variables, we attempt to define a function, f(x) called probability density function, the graph of which approximates the relative frequency polygon for the population. A probability density function must satisfy two conditions:

•f(x) is non-negative. •The total area under the curve representing f(x) is 1

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•Note that f(x) is not probability, i.e., f(x) P(X = x)

a b

p (a < x < b)

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•The area under the curve (corresponding to the probability density function) between 2 points a and b gives the probability the random variable, x lies between a and b or •p(a < x < b) = Area under the curve

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Agenda for this Session# 3 Part 3




• Applications

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Uniform Distribution

The continuous uniform distribution for x with (l <= x <= h) has a density function: f(x) = 1 / (h – l) where h is the highest possible value of x l is the lowest possible value of x Mean = (l + h) / 2 Variance = (h – l) 2 / 12 P (a <= x <= b) = (b – a) / (h – l)

l h

{1/ (h – l)

l h

{1/ (h – l)

a b


P (a <= x <= b) = (b – a) / (h – l)

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Uniform Probability Distribution - Example

A manufacturer has observed that the time elapsed between the placement of an order with a just-in-time supplier and the delivery of the parts is uniformly distributed between 100 and 180 minutes.

a) Define and graph the density function b) What proportion of orders takes between 2

and 2.5 hours to be delivered.

100 180

{1/ 80

120 150


P (a <= x <= b) = (b – a) / (h – l)

P (120 <= x <= 150) = 30 / 80 = 0.375

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• Applications

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Exponential Distribution

A continuous probability distribution that is often useful in describing the time required to complete a task is the exponential probability distribution

Let Lambda, be the parameter of the distribution. The probability density function is: f (x) = 0, if x <= 0;

f(x) = e- x if x > 0

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f(x)

x0=5

=1

.5

1.0

1.5

2.0

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1

1

P (X <= x) = 1 – e-x

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Exponential Probability Distribution - ExampleThe mean time between arrivals of airplanes in

the JFK, New York is 2 minutes and assume the inter arrival time follows exponential distribution:

a) What is the probability that the time between arrivals is 2 minutes or less?

b) What is the probability that the time between arrivals is 1 minute or less?

c) What is the probability that the time between arrivals is 5 minutes or more?

Mean arrival time = 2 minutes

The parameter, , = ½ = 0.5

P ( X <= 2) = 1 – e- x = 0.6321

Or you can use Excel formula, =EXPONDIST(2,1/2,1)


The parameter, , = ½ = 0.5

P ( X <= 1) = 1 – e- x = 0.3935


P ( X <= 5) = 1 – e- x = 0.9179

Probability that inter arrival time exceeds 5 minutes

= 1 – 0.9179 = 0.0821

Exponential Probability Distribution - Example

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• Applications

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Uniform Probability Distribution - ApplicationAmerican Airlines publishes a scheduled flight

time of 4 hours 20 minutes for its flights from Washington D.C. to Los Angeles. Suppose we believe that the flight time is uniformly distributed between 4 hours and 5 hours.

a) What is the probability that the flight will be no more than 10 minutes late?

b) What is the probability that the flight will be no more than 30 minutes late?

c) What is the expected flight time?

240 300

{1/ 60


P (a <= x <= b) = (b – a) / (h – l)

a) P (240 <= x <= 270) = 30 / 60 = 0.5 b) P (240 <= x <= 290) = 50 / 60 = 0.833 c) Mean = (l + h) / 2 = 270 minutes

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Exponential Probability Distribution - Application

The lifetime of bulbs is a random variable with exponential distribution and has a mean lifetime of 50 days.

a) What is the probability that the bulb fails in the first 25 days of operation?

b) What is the probability that the bulb operates 100 or more days?

Mean life time = 50 days

The parameter, , = 1/50 = 0.02

P ( X <= 25) = 1 – e- x = 0.3935



The parameter, , = 1/50 = 0.02

P ( X <= 100) = 1 – e- x = 0.8647


P (X > 100) = 1 – 0.8647 = 0.1353

Exponential Probability Distribution - Example

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Continuous Random Variables



Applications

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Documents

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continuous variable

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