Chapter 3-PART I. QUANTITATIVE METHODS FOR BUSINESS DISCREETE PROBABILITY DISTRIBUTION. John Kooti. Chapter 3, Part I Discrete Probability Distributions. Random Variables Discrete Random Variables Expected Value and Variance Binomial Probability Distribution
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Chapter 3-PART I
DISCREETE PROBABILITY DISTRIBUTION
Let x = number of TV sets sold at the store in one day
where x can take on 5 values (0, 1, 2, 3, 4)
Let x = number of customers arriving in one day
where x can take on the values 0, 1, 2, . . .
We can count the customers arriving, but there is no finite upper limit on the number that might arrive.
f(x) > 0
Sf(x) = 1
Units Soldof Daysxf(x)
0 80 0 .40
1 50 1 .25
2 40 2 .20
3 10 3 .05
420 4 .10
0 1 2 3 4
Values of Random Variable x (TV sales)
E(x) = m = Sxf(x)
Var(x) = s2 = S(x - m)2f(x)
1.20 = E(x)
The expected number of TV sets sold in a day is 1.2
of a Discrete Random Variable
xx - m (x - m)2f(x)(x - m)2f(x)
_____ _________ ___________ _______ _______________
0-1.2 1.44.40 .576
1-0.2 0.04.25 .010
2 0.8 0.64.20 .128
3 1.8 3.24.05 .162
4 2.8 7.84.10 .784
1.660 = s 2
The variance of daily sales is 1.66 TV sets squared.
The standard deviation of sales is 1.29 TV sets.
Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.
Choosing 3 hourly employees a random, what is the probability that 1 of them will leave the company this year?
Let: p = .10, n = 3, x = 1
f(x) = the probability of x successes in n trials
n = the number of trials
p = the probability of success on any one trial
want the binomial probabilities to appear.
Choose Statistical from the FunctionCategory box.
Choose BINOMDIST from the FunctionName box.
Enter 1 in the Number_s box (value of x).
Enter 3 in the Trials box (value of n).
Enter .1 in the Probability_s box (value of p).
Enter false in the Cumulative box.
[Note: At this point the desired binomial probability of .243 is automatically computed and appears in the right center of the dialog box.]
Select OK (and .243 will appear in the worksheet cell requested in Step 1).
E(x) = m = np
Var(x) = s 2 = np(1 - p)
E(x) = m = 3(.1) = .3 employees out of 3
Var(x) = s 2 = 3(.1)(.9) = .27
f(x) = probability of x occurrences in an interval
m = mean number of occurrences in an interval
e = 2.71828
Patients arrive at the emergency room of Mercy
Hospital at the average rate of 6 per hour on weekend
evenings. What is the probability of 4 arrivals in 30
minutes on a weekend evening?
m = 6/hour = 3/half-hour, x = 4
The Poisson probability distribution can be used as
an approximation of the binomial probability
distribution when p, the probability of success, is small
and n, the number of trials, is large.
A random variable is uniformly distributedwhenever
the probability is proportional to the length of the
f(x) = 1/(b - a) for a<x<b
= 0 elsewhere
E(x) = (a + b)/2
Var(x) = (b - a)2/12
where: a = smallest value the variable can assume
b = largest value the variable can assume
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
f(x) = 1/10 for 5 <x< 15
= 0 elsewhere
x = salad plate filling weight
What is the probability that a customer will take
between 12 and 15 ounces of salad?
P(12 <x< 15) = 1/10(3) = .3
Salad Weight (oz.)
m = mean
s = standard deviation
p = 3.14159
e = 2.71828
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 an order. It has
been determined that leadtime demand is normally
distributed with a mean of 15 gallons and a standard
deviation of 6 gallons.
The manager would like to know the probability of a
stockout, P(x > 20).
z = (x - m)/s
= (20 - 15)/6
The Standard Normal table shows an area of .2967 for the region between the z = 0 line and the z = .83 line above. The shaded tail area is .5 - .2967 = .2033. The probability of a stockout is .2033.
Area = .2967
Area = .2033
Area = .5
Enter 20 in the x box.
Enter 15 in the mean box.
Enter 6 in the standard deviation box.
Enter true in the cumulative box.
At this point, .7967 will appear in the cell selected in Step 1, indicating that the probability of demand during lead time being less than or equal to 20 gallons is .7967. The probability that demand will exceed 20 gallons is 1 - .7967 = .2033.
If the manager of Pep Zone wants the probability of a
stockout to be no more than .05, what should the
reorder point be?
Let z.05 represent the z value cutting the tail area of .05.
Area = .05
Area = .5
Area = .45
We now look-up the .4500 area in the Standard Normal Probability table to find the corresponding z.05 value.
z.05 = 1.645 is a reasonable estimate.
The corresponding value of x is given by
x = m + z.05s
= 15 + 1.645(6)
A reorder point of 24.87 gallons will place the
probability of a stockout during leadtime at .05.
Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability under .05.
for x> 0, m > 0
wherem = mean
e = 2.71828
wherex0 = some specific value of x
The time between arrivals of cars at Al’s Carwash
follows an exponential probability distribution with a
mean time between arrivals of 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
P(x< 2) = area = .4866
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