VI. Continuous Proaility Distriutions A. An Important Definition (reminder) Continuous Random Variale - a numerical description of the outcome of an experiment whose outcome can assume any numerical value in an interval or collection of intervals. B. Some Important Continuous Proaility Distriutions 1. (Continuous) Uniform Proaility - expresses the likelihoods of outcomes for a continuous random variale x for which all possile outcomes are equally likely. This distriution function is given y 1 for a x -a f( x ) = otherwise where a = minimum possile value of random variale x = maximum possile value of random variale x Characteristics of a (Continuous) Uniform proaility distriution are: - the random variale can assume any value within a range - all possile values within this range are equally likely - no value outside of this range can occur The mean (expected value) is a+ E(x) = and the variance is ( ) -a σ = 1
ut the proaility is given y d-c for a c,d -a P( c x d ) = otherw ise Why isn t f(x) = P(x)? Think aout the graph of the proaility distriution function for a (continuous) uniform random variale with a range of 6-15, i.e., 1 for 6 x 15 9 f( x ) = otherwise f(x).5.45.4.35.3.5..15.1.5. 1 3 4 5 6 7 8 9 1 11 1 13 14 15 16 x What is the area under the proaility distriution function at f(x)? Consider x = 8. We have that f(8) =.111, ut P(8) is the product of the height of f(x) (the proaility distriution) and the width at x = 8. Thus P(x) =.111(.) =. This leads to two conclusions aout working with proaility distriution functions for continuous random variales: - P(x) = for any single value of x - we can only talk meaningfully aout the proaility over a range of values
So how can we calculate proailities for continuous random variales over ranges? To find P(a x ) we integrate the proaility distriution function f(x) from a to, i.e., a ( ) f x d x The mean (expected value) is and the variance is σ = ( ) E(x) = µ = xf x dx all x all x ( ) (x-µ) f x dx Example - Suppose that some random variale x can take on any value etween 6 and 15 with equal proaility, and can take on no value outside of this range. What is the proaility that x is etween 7 and 1? 1-7 3 P ( 7 x 1 ) = = =.333 15-6 9 What is the proaility that x is etween 4 and 8? P( 4 x 8 ) =P( 4 x <6 ) +P( 6 x 8) 8-6 = + = + =. 15-6 15-6 9 9 Why didn t we have to integrate to find proailities for the (continuous) uniform proaility distriution? f(x).5.45.4.35.3.5..15.1.5. 1 3 4 5 6 7 8 9 1 11 1 13 14 15 16 x
. Exponential Proaility Distriution - expresses the likelihoods of outcomes for a continuous random variale x that represents the amount of time or space that passes etween consecutive occurrences of a Poisson random variale. This distriution function is given y 1 - x µ f(x) = e for x, µ > µ where µ = mean time or space etween consecutive occurrences e =.7188 Characteristics of an Exponential proaility distriution are: - the random variale can assume any positive value - the random variale represents the amount of the interval (time, space, etc.) that passes etween consecutive occurrences of a Poisson event The mean (expected value) is E(x) = µ and the variance is σ = µ ut the proaility is given y P(x ) = P( x ) = f(x)dx 1 = e dx = 1 - e µ - - µ µ Fortunately, the integration for this proaility distriution function is relatively easy and yields a closed-form result! This means we will not have to do the integration to find proailities for an exponential random variale?
Generically, the exponential proaility distriution curve looks like this: f(x) Proaility Curve - Exponential Distriution x Example: suppose that the time that passes etween customers arriving at an ATM is exponentially distriuted with a mean of 6 minutes. What is the proaility that no more.5 hours will pass etween arrivals of the next two customers? Call random variale x the amount of time (in minutes) that passes etween consecutive arrivals. We have that µ = 6. minutes etween consecutive customers and x =.5 hours =.5(6) minutes = 3. minutes so - 3. 6. P(x 3.) = 1 - e =.3935 OR IN HOURS µ = 6. minutes =.1 hours etween consecutive customers and x =.5 hours so -.5.1 P(x.5) = 1 - e =.3935
To illustrate the relationship etween the Exponential distriution and the Poisson distriution, consider the following example: Suppose that an ATM averages 1 customer arrivals per hour, and that these arrivals are Poisson distriuted. What is the proaility that no more three minutes will pass etween arrivals of the next two customers? Call random variale x the amount of time (in minutes) that passes etween consecutive arrivals. We have that so 6 µ = = 6. minutes etween customers 1-3. 6. P(x 3.) = 1 - e =.3935 Again, this same prolem could e done in terms of hours (instead of minutes). Call random variale y the amount of time (in hours) that passes etween consecutive arrivals. We have that so 1 µ = =.1 hours etween customers 1 -.5.1 P(x.5) = 1 - e =.3935 O.K. - we can find proailities of the form P(x ). What aout prolems of the form P(x ) or P(a x )? We can now use our original result for P(x ). Rememer that P(x ) = 1 - P(x ), so P(x )=1-P(x )=1-1-e =e - - µ µ Similarly, P(a x ) = P(x ) - P(x a), so P(a x ) = P(x ) - P(x a) - - - - = 1 - e - 1 - e = e - e a a µ µ µ µ
Note that Exponential Proailities Tales do exist for certain values of µ, ut these are not provided in your textook. Also note that and E(x) = µ Var(x) = σ = µ for the exponential distriution. x f(x)..166667 1..1418..1194 3..1188 4..8557 5..7433 6..61313 7..5191 8..43933 9..37188 1..31479 11..6647 1..556 13..1993 14..1616 15..13681 16..11581 17..983 18..898 19..74..5946 1..533..46 3..366 4..353 5..584 : : : : f(x) The proaility curve and distriution for the previous prolem (a Poisson distriuted random variale with a mean of µ = 7 customers per hour) look like this:..1. Proaility Curve - Exponential Distriution (µ=6). 1.. 3. 4. x 3. Normal Proaility Distriution - expresses the likelihoods of outcomes for a continuous random variale x with a particular symmetric and unimodal distriution. This distriution function is given y f(x) = 1 e πσ ( x-µ ) - σ where µ = mean σ = standard deviation π = 3.14159 e =.7188
ut the proaility is given y P(a x ) = f(x)dx a -( x-µ ) σ 1 = e dx a πσ This looks like a difficult integration prolem! Will I have to integrate this function every time I want to calculate proailities for some normal random variale? Characteristics of the normal proaility distriution are: - there are an infinite numer of normal distriutions, each defined y their unique comination of the mean µ and standard deviation σ - µ determines the central location and σ determines the spread or width - the distriution is symmetric aout µ - it is unimodal - µ = M d = M o - it is asymptotic with respect to the horiontal axis - the area under the curve is 1. - it is neither platykurtic nor leptokurtic - it follows the empirical rule Normal distriutions with the same mean ut different standard deviations: f(x) µ x
Normal distriutions with the same standard deviation ut different means: f(x) µ µ µ x The Standard Normal Proaility Distriution - the proaility distriution associated with any normal random variale that has µ = and σ = 1. There are tales that give the results of the integration P(a x ) = f(x)dx a -( x-µ ) 1 σ = e dx a πσ for the standard normal random variale (Tale 1, Appendix A). The Cumulative Standard Normal Distriution (Appendix B, Tale 1)..1..3.4.5.6.7.8.9-3..13.14.14.15.15.16.16.17.18.18 -.9.19.19..1.1..3.3.4.5 -.8.6.6.7.8.9.3.31.3.33.34 -.7.35.36.37.38.39.4.41.43.44.45 -.6.47.48.49.51.5.54.55.57.59.6 -.5.6.64.66.68.69.71.73.75.78.8 -.4.8.84.87.89.91.94.96.99.1.14 -.3.17.11.113.116.119.1.15.19.13.136 -..139.143.146.15.154.158.16.166.17.174 -.1.179.183.188.19.197..7.1.17. -..8.33.39.44.5.56.6.68.74.81-1.9.87.94.31.37.314.3.39.336.344.351-1.8.359.367.375.384.39.41.49.418.47.436-1.7.446.455.465.475.485.495.55.516.56.537-1.6.548.559.571.58.594.66.618.63.643.655-1.5.668.681.694.78.71.735.749.764.778.793-1.4.88.83.838.853.869.885.91.918.934.951-1.3.968.985.13.1.138.156.175.193.111.1131-1..1151.117.119.11.13.151.171.19.1314.1335-1.1.1357.1379.141.143.1446.1469.149.1515.1539.156-1..1587.1611.1635.166.1685.1711.1736.176.1788.1814 -.9.1841.1867.1894.19.1949.1977.5.33.61.9 -.8.119.148.177.6.36.66.96.37.358.389 -.7.4.451.483.514.546.578.611.643.676.79 -.6.743.776.81.843.877.91.946.981.315.35 -.5.385.311.3156.319.38.364.33.3336.337.349 -.4.3446.3483.35.3557.3594.363.3669.377.3745.3783 -.3.381.3859.3897.3936.3974.413.45.49.419.4168 -..47.447.486.435.4364.444.4443.4483.45.456 -.1.46.4641.4681.471.4761.481.484.488.49.496..5.54.58.51.516.5199.539.579.5319.5359
The Cumulative Standard Normal Distriution (Appendix B, Tale 1)..1..3.4.5.6.7.8.9..5.54.58.51.516.5199.539.579.5319.5359.1.5398.5438.5478.5517.5557.5596.5636.5675.5714.5753..5793.583.5871.591.5948.5987.66.664.613.6141.3.6179.617.655.693.6331.6368.646.6443.648.6517.4.6554.6591.668.6664.67.6736.677.688.6844.6879.5.6915.695.6985.719.754.788.713.7157.719.74.6.757.791.734.7357.7389.74.7454.7486.7517.7549.7.758.7611.764.7673.774.7734.7764.7794.783.785.8.7881.791.7939.7967.7995.83.851.878.816.8133.9.8159.8186.81.838.864.889.8315.834.8365.8389 1..8413.8438.8461.8485.858.8531.8554.8577.8599.861 1.1.8643.8665.8686.878.879.8749.877.879.881.883 1..8849.8869.8888.897.895.8944.896.898.8997.915 1.3.93.949.966.98.999.9115.9131.9147.916.9177 1.4.919.97.9.936.951.965.979.99.936.9319 1.5.933.9345.9357.937.938.9394.946.9418.949.9441 1.6.945.9463.9474.9484.9495.955.9515.955.9535.9545 1.7.9554.9564.9573.958.9591.9599.968.9616.965.9633 1.8.9641.9649.9656.9664.9671.9678.9686.9693.9699.976 1.9.9713.9719.976.973.9738.9744.975.9756.9761.9767..977.9778.9783.9788.9793.9798.983.988.981.9817.1.981.986.983.9834.9838.984.9846.985.9854.9857..9861.9864.9868.9871.9875.9878.9881.9884.9887.989.3.9893.9896.9898.991.994.996.999.9911.9913.9916.4.9918.99.99.995.997.999.9931.993.9934.9936.5.9938.994.9941.9943.9945.9946.9948.9949.9951.995.6.9953.9955.9956.9957.9959.996.9961.996.9963.9964.7.9965.9966.9967.9968.9969.997.9971.997.9973.9974.8.9974.9975.9976.9977.9977.9978.9979.9979.998.9981.9.9981.998.998.9983.9984.9984.9985.9985.9986.9986 3..9987.9987.9987.9988.9988.9989.9989.9989.999.999 Let s focus on a small part of the Cumulative Standard Normal Proaility Distriution Tale Example: for a standard normal random variale, what is the proaility that is etween - and.43?..1..3.4..5.54.58.51.516.1.5398.5438.5478.5517.5557..5793.583.5871.591.5948.3.6179.617.655.693.6331.4.6554.6591.668.6664.67.5.6915.695.6985.719.754.6.757.791.734.7357.7389 Example: for a standard normal random variale, what is the proaility that is etween and.? f().
Again, looking at a small part of the Cumulative Standard Normal Proaility Distriution Tale, we find the proaility that a standard normal random variale is etween - and.?..1..3.4 : : : : : : : : : : : : 1.3.93.949.966.98.999 1.4.919.97.9.936.951 1.5.933.9345.9357.937.938 1.6.945.9463.9474.9484.9495 1.7.9554.9564.9573.958.9591 1.8.9641.9649.9656.9664.9671 1.9.9713.9719.976.973.9738..977.9778.9783.9788.9793.1.981.986.983.9834.9838 Example: for a standard normal random variale, what is the proaility that is etween and.? Proaility =.977 f() Proaility =.5 Proaility =.977.5 =.477. P( ) = P(- ) - P(- ) =.977 -.5 =.477 What is the proaility that is at least.? Proaility = 1. -.977 =.8 f(). P( ) = P(- ) - P(- ) = 1. -.977 =.8
What is the proaility that is etween -1.5 and.? Proaility =? f() Proaility =? Proaility =.477-1.5. We need to find the proaility that is etween -1.5 and.! By symmetry of the standard normal distriution, we have equivalently Proaility =? f() Proaility =.5 Proaility =? 1.5 Note that we have simply inverted the previous prolem so that we can use the cummulative standard normal distriution tale! Again, looking at a small part of the Cumulative Standard Normal Proaility Distriution Tale, we find the proaility that a standard normal random variale is etween - and 1.5?..1..3.4 : : : : : : : : : : : : 1.3.93.949.966.98.999 1.4.919.97.9.936.951 1.5.933.9345.9357.937.938 1.6.945.9463.9474.9484.9495 1.7.9554.9564.9573.958.9591 1.8.9641.9649.9656.9664.9671 1.9.9713.9719.976.973.9738..977.9778.9783.9788.9793.1.981.986.983.9834.9838
so y symmetry of the standard normal distriution, we have equivalently Proaility =.933 f() Proaility =.5 Proaility =.933.5 1.5 P(-1.5.) = P(. 1.5) [ ] [ ] = P(- 1.5) - P(-.) =.933 -.5 =.433 What is the proaility that is etween -1.5 and.? Proaility =.433 +.477 f() Proaility = =.433 Proaility =.477-1.5. P(-1.5.) = P(-1.5.)+P(..) =.433 +.477 =.914 f() Notice we could find the proaility that is etween -1.5 and. another way! Proaility =.977 Proaility =.668-1.5. P(-1.5.) = P(-.) - P(- -1.5) =.977 -.668 =.914
f() There are often multiple ways to use the Cumulative Standard Normal Proaility Distriution Tale to find the proaility that a standard normal random variale is etween two given values! How do you decide which to use? - Do what you understand (make yourself comfortale) and - DRAW THE PICTURE!!! What is the proaility that is etween -1.5 and -.? f() Proaility =.668 Proaility =.8 Proaility =.5668 -.8 =.44 -. -1.5 P(-. -1.5) = P(- -1.5) - P(-.) =.668 -.8 =.44 What is the proaility that is etween -1.5 and -.? Proaility =.5 -.8 =.477 Proaility =.477.433 =.44 f() Proaility =.433 -. -1.5 P(-. -1.5) = P(-..) - P(-1.5.) [ ] [ ] = P(-.) - P(- -.) -.433 =.5 -.8 -.433 =.44
What is the proaility that is exactly 1.5? Proaility =.933 f() 1.5 P( = 1.5) = P(1.5 1.5) = P(- 1.5) - P(- 1.5) =.933 -.933 =. (why?) -Transformation - mathematical means y which any normal random variale with a mean µ and standard deviation σ can e converted into a standard normal random variale. - to make the mean equal to, we simply sutract µ from each oservation in the population - to then make the standard deviation equal to 1, we divide the results in the first step y σ The resulting transformation is given y x-µ = σ Example: for a normal random variale x with a mean of 5 and a standard deviation of 3, what is the proaility that x is etween 5. and 7.? Proaility f(x) } µ=5 7. x.
Using the -transformation, we can restate the prolem in the following manner: 5. - 5. x - µ 7. - 5. P(5. x 7.) = P 3. σ 3. = P(..67) then use the standard normal proaility tale to find the ultimate answer: P(..67) =.486 which graphically looks like this: f(x) Proaility =.486 } µ=5 7. x..67 Why is the normal proaility distriution considered so important? - many random variales are naturally normally distriuted - many distriutions, such as the Poisson and the inomial, can e approximated y the normal distriution - the distriution of many statistics, such as the sample mean and the sample proportion, are approximately normally distriuted if the sample is sufficiently large (Central Limit Theorem)
Yes Use (Continuous) Uniform Distriution 1 for a x a fx ( ) = otherwise d-c for a c,d ( ) a Pc x d = otherwise Are all Outcomes Equally Likely? Yes No Use Exponential Distriution 1 -x f ( x) = e µ µ x µ P(x x ) = 1 e Are you measuring the time etween occurrences of a relatively rare (Poisson) event? Yes No Do you have evidence that the variale is normally distriuted? No Use Normal Distriution ( x µ ) 1 σ f(x) = e πσ ( x µ ) 1 σ P(a x ) = e dx πσ a Use Relative Frequency Distriution or some other proaility distriution not discussed in class