Continuous Probability Distributions For discrete RVs, f (x) is the probability density function (PDF) is not the probability of x but areas under it are probabilities is the HEIGHT at x For continuous RVs, f (x) is the probability distribution function (PDF) is the probability of x is the HEIGHT at x

Continuous Probability Distributions The probability of the random variable assuming a value within some given interval from x 1 to x 2 is defined to be the area under the graph of the probability density function that is between x 1 and x 2. x Uniform x1 x1x1 x1 x2 x2x2 x2 x Normal x1 x1x1 x1 x2 x2x2 x2 x2 x2 Exponential x x1 x1 P ( x 1 < x < x 2 ) = area

E( x ) = ( b + a )/2 = (15 + 5)/2 = 10 Uniform Probability Distribution Example: Slater's Buffet Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. a b = 1/10 x f ( x ) = 1/( b – a ) = 1/(15 – 5) Var( x ) = ( b – a ) 2 /12 = (15 – 5) 2 /12 = 8.33  = = 2.886

Uniform Probability Distribution for Salad Plate Filling Weight f(x)f(x) x 1/10 Salad Weight (oz.) Uniform Probability Distribution

f(x)f(x) x 1/10 Salad Weight (oz.) P(12 < x < 15) = ( h )( w ) = (1/10)(3) =.3 What is the probability that a customer will take between 12 and 15 ounces of salad? Uniform Probability Distribution 12

f(x)f(x) x 1/10 Salad Weight (oz.) f(x)f(x) P( x = 12) = ( h )( w ) = (1/10)(0) = 0 What is the probability that a customer will equal 12 ounces of salad? Uniform Probability Distribution 12

x   Normal Probability Distribution The normal probability distribution is widely used in statistical inference, and has many business applications.  ≈ … e ≈ … x is a normal distributed with mean  and standard deviation  skew = ?

Normal Probability Distribution The mean can be any numerical value: negative, zero, or positive.  = 4  = 6  = 8  = 2

Normal Probability Distribution The standard deviation determines the width and height  = 4  = 6  = 3  = 2 data_bwt.xls

z is a random variable having a normal distribution with a mean of 0 and a standard deviation of 1. Standard Normal Probability Distribution  = 1  = 0 z

Use the standard normal distribution to verify the Empirical Rule: Standard Normal Probability Distribution 99.74% of values of a normal random variable are within 3 standard deviations of its mean % of values of a normal random variable are within 2 standard deviations of its mean % of values of a normal random variable are within 1 standard deviations of its mean.

Standard Normal Probability Distribution  = 1 z ? Compute the probability of being within 3 standard deviations from the mean First compute P(z < -3) = ?

Standard Normal Probability Distribution Z P ( z < -3) =.0013 row = -3.0 column =.00 P(z < -3.00) = ?

Standard Normal Probability Distribution  = 1 z Compute the probability of being within 3 standard deviations from the mean P(z < -3) =.0013

Standard Normal Probability Distribution  = 1 z ? Compute the probability of being within 3 standard deviations from the mean Next compute P(z > 3) = ?

Standard Normal Probability Distribution Z P ( z < 3) =.9987 row = 3.0 column =.00 P(z < 3.00) = ?

Standard Normal Probability Distribution  = 1 z Compute the probability of being within 3 standard deviations from the mean P(z < 3) =.9987P(z > 3) = 1 –.9987

3.00 Standard Normal Probability Distribution  = 1 z Compute the probability of being within 3 standard deviations from the mean % of values of a normal random variable are within 3 standard deviations of its mean.

Standard Normal Probability Distribution  = 1 z ? Compute the probability of being within 2 standard deviations from the mean First compute P(z < -2) = ?

Standard Normal Probability Distribution Z P ( z < -2) =.0228 row = -2.0 column =.00 P(z < -2.00) = ?

Standard Normal Probability Distribution  = 1 z 0 Compute the probability of being within 2 standard deviations from the mean P(z < -2) =

Standard Normal Probability Distribution  = 1 z ? Compute the probability of being within 2 standard deviations from the mean Next compute P(z > 2) = ?

Standard Normal Probability Distribution Z P ( z < 2) =.9772 row = 2.0 column =.00 P(z < 2.00) = ?

Standard Normal Probability Distribution  = 1 z Compute the probability of being within 2 standard deviations from the mean P(z < 2) =.9772P(z > 2) = 1 –

Standard Normal Probability Distribution  = 1 z 0 Compute the probability of being within 2 standard deviations from the mean % of values of a normal random variable are within 2 standard deviations of its mean

Standard Normal Probability Distribution  = 1 z 0 ? Compute the probability of being within 1 standard deviations from the mean First compute P(z < -1) = ?

Standard Normal Probability Distribution Z P ( z < -1) =.1587 row = -1.0 column =.00 P(z < -1.00) = ?

Standard Normal Probability Distribution  = 1 z 0 Compute the probability of being within 1 standard deviations from the mean P(z < -1) =

Standard Normal Probability Distribution  = 1 z 0 Compute the probability of being within 1 standard deviations from the mean Next compute P(z > 1) = ? 1.00 ?

Standard Normal Probability Distribution Z P ( z < 1) =.8413 row = 1.0 column =.00 P(z < 1.00) = ?

Standard Normal Probability Distribution  = 1 z Compute the probability of being within 1 standard deviations from the mean P(z < 1) =.8413P(z > 1) = 1 –

Standard Normal Probability Distribution  = 1 z 0 Compute the probability of being within 1 standard deviations from the mean % of values of a normal random variable are within 1 standard deviations of its mean

Probabilities for the normal random variable are given by areas under the curve. Verify the following: The area to the left of the mean is.5 Standard Normal Probability Distribution  = 1 z 0 ? P(z < 0) = ?

Standard Normal Probability Distribution Z P ( z < 0) =.5000 row = 0.0 column =.00 P(z < 0.00) = ?

Probabilities for the normal random variable are given by areas under the curve. Verify the following: The area to the left of the mean is.5 Standard Normal Probability Distribution  = 1 z P(z < 0) =

Probabilities for the normal random variable are given by areas under the curve. Verify the following: The area to the right of the mean is.5 Standard Normal Probability Distribution  = 1 z P(z > 0) = 1 –.5000

.5000 Probabilities for the normal random variable are given by areas under the curve. Verify the following: The total area under the curve is 1 Standard Normal Probability Distribution  = 1 z

Standard Normal Probability Distribution  = 1 z ? What is the probability that z is less than or equal to P(z < -2.76) = ?

Standard Normal Probability Distribution Z P ( z < -2.76) =.0029 row = -2.7 column =.06 P(z < -2.76) = ?

Standard Normal Probability Distribution  = 1 z P(z < -2.76) =.0029 What is the probability that z is less than -2.76? What is the probability that z is less than or equal to -2.76? P(z < -2.76) =.0029

Standard Normal Probability Distribution  = 1 z P(z > -2.76) = 1 –.0029 What is the probability that z is greater than or equal to -2.76? What is the that z is greater than -2.76?.9971 =.9971 P(z > -2.76) =.9971

Standard Normal Probability Distribution  = 1 z ? P(z < 2.87) = ? What is the probability that z is less than or equal to 2.87

Standard Normal Probability Distribution Z P ( z < 2.87) =.9979 row = 2.8 column =.07 P(z < 2.87) = ?

Standard Normal Probability Distribution  = 1 z P(z < 2.87) =.9979 What is the probability that z is less than or equal to 2.87 What is the probability that z is less than 2.87? P(z < 2.87) =.9971

Standard Normal Probability Distribution P(z > 2.87) = 1 –.9979 What is the probability that z is greater than or equal to 2.87? What is the that z is greater than 2.87? =.0021 P(z > 2.87) =.0021  = 1 z

Standard Normal Probability Distribution  = 1 z ? What is the value of z if the probability of being smaller than it is.0250? P(z < ?) =.0250

Standard Normal Probability Distribution Z row = -1.9 column =.06 P(z < -1.96) =.0250 z = What is the value of z if the probability of being smaller than it is.0250?

Standard Normal Probability Distribution What is the value of z if the probability of being greater than it is.0192? P(z > ?) =.0192  = 1 z ? ? less

Standard Normal Probability Distribution Z row = 2.0 column =.07 P(z < 2.07) =.9808 z = 2.07 What is the value of z if the probability of being greater than it is.0192?.9808? less

z Standard Normal Probability Distribution  = 1 z -z If the area in the middle is.95 What is the value of -z and z if the probability of being between them is.9500? then the area NOT in the middle is.05 and so each tail has an area of.025

Standard Normal Probability Distribution Z row = -1.9 column =.06 P(z < -1.96) =.0250 z = What is the value of -z and z if the probability of being between them is.9500?

Standard Normal Probability Distribution  = 1 z By symmetry, the upper z value is 1.96 What is the value of -z and z if the probability of being between them is.9500?

To handle this we simply convert x to z using Normal Probability Distribution We can think of z as a measure of the number of standard deviations x is from . z is a random variable that is normally distributed with a mean of 0 and a standard deviation of 1 Let x be a random variable that is normally distribution with a mean of  and a standard deviation of . Since there are infinite many choices for  and , it would be impossible to have more than one normal distribution table in the textbook.

Normal Probability Distribution Example: Pep Zone Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for a replenishment order.

It has been determined that demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. Normal Probability Distribution Example: Pep Zone The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 gallons? P ( x > 20) = ?

x p = ? Step 1: Draw and label the distribution Normal Probability Distribution Note: this probability must be less than 0.5 Example: Pep Zone  = 6

15 z = ( x -  )/  = ( )/6 =.83 Step 2: Convert x to the standard normal distribution. Normal Probability Distribution x 20 Note: this probability must be less than 0.5 =.83 z E( z ) = 0 Example: Pep Zone p = ?

Normal Probability Distribution z P ( z <.83) =.7967 Step 3: Find the area under the standard normal curve to the left of z =.83. row =.8 column =.03 Example: Pep Zone

0.83 z Normal Probability Distribution Step 4: Compute the area under the standard normal curve to the right of z =.83. P ( x > 20) = P ( z >.83) = 1 –.7967 =.2033  = 6 x Example: Pep Zone

Normal Probability Distribution If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than.05, what should the reorder point be? Example: Pep Zone P(x > x 1 ) =.0500 x 1 = ?

x Normal Probability Distribution Example: Pep Zone.05 x1 x1  = 6 ? z  = 1 0

Step 1: Find the z -value that cuts off an area of.05 in the right tail of the standard normal distribution. Normal Probability Distribution Example: Pep Zone z left z.05 = 1.64 z.05 = 1.65 or z.05 = 1.645

 = 1 z z Normal Probability Distribution Example: Pep Zone.05   x = 15  x = 6 x 15 A reorder point of gallons will place the probability of a stockout at 5%

Exponential Probability Distribution The exponential probability distribution is useful in describing the time it takes to complete a task. where:  = mean x > 0  > 0) e ≈ Cumulative Probability: x 0 = some specific value of x

Exponential Probability Distribution 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 of 3 minutes. Al wants to know the probability that the time between two arrivals is 2 minutes or less. P ( x < 2) = ? x0x0  1 – e –2/  =