Class worksheet #1
a. Probability rules
b. Binomial distribution
c. Normal distribution
1.Casino Windsor conducts survey to determine the opinion of its customers. Among other questions, respondents are asked to give their opinion about “Your overall impression of Casino Windsor”. The responses are: Excellent, Good, Average, Poor. In addition, the gender of the respondent is noted. After analyzing the results the following table of joint probabilities was produced:
RatingWomenMen
Excellent.27.22
Good.14.10
Average.06.12
Poor.03.06
(i)What proportion of customers rate Casino Windsor as excellent?
(ii)Determine the probability that a customer is a male and rate Windsor as
excellent.
(iii)Find the probability that a customer who rates Windsor as excellent is a man.
(iv)Are gender and rating independent? Explain.
2.A paint store carries three brands of paint. Two customers buy one gallon of paint each.
(a)What is the probability they buy the same brand?
(b)A customer arrives at the store to buy one gallon to match the paint she previously purchased. She (obviously) knows the color she wants but she can’t remember the brand name she used last time. Therefore, she decides to rely on her luck, and selects two of the three brands of paint at random and buys them. What is the probability she will match the paint previously purchased?
3.In a campus restaurant, it was found that 35% of all customers order hot meals, and 50% of all customers are students. Further, 25% of all customers who are students order hot meals.
(a)What is the probability that a randomly chosen customer is both a student and orders a hot meal?
(b)If a randomly selected customer orders a hot meal, what is the probability that the customer is a student?
(c)What is the probability that a randomly chosen customer is not a student or does not order a hot meal?
(d)Are the events “customer orders a hot meal”, and “customer is a student” independent? Explain by calculating the proper probabilities.
4.A CBS news survey reported that 67% of adults said the US treasury should continue making pennies. If adults form their opinion independently regarding this matter, what is the probability that for 6 adults selected at random:
(a)Exactly 5 adults would want the treasury to continue making pennies?
(b)More than two adults would want the treasury to continue making pennies?
(c)Four or less adults would want the treasury to continue making pennies?
(d)Between 2 and 5 adults would want the treasury to continue making pennies?
(e)More than 4 adults would not want the treasury to continue making pennies, given that not all the adults surveyed wanted the treasury to continue making pennies.
5.A commuter airline has studied the passenger counts on a flight between Boston and Atlanta, and found that the number of passenger is approximately normally distributed with a mean of 72 and a variance of 25. The capacity on the plane is 85 passengers:
(a)What percentage of the time would the flight “sell-out”?
(b)What percentage of the time would the flight take-off less than 50% full?
(c)If management is planning changes in the number of seats such that 80% of the flights fly at least at 75% capacity, how many seats should be built in each plane?
Bonus:
(d)The catering manager who is responsible for the snack and beverage provisions on the flight plans to stock 90 snack packs. Assuming that each passenger gets one snack pack and all passengers who buy a ticket show for the flight. What is the probability that there will be 8 or more snacks left over?
(e)Answer part (d) assuming that 20% of the passengers ask for a second snack.
6.Customers arrive at a bank and sometimes need to wait in line before get served. The branch manager has some concerns regarding service level provided to her customers, and decides to study this matter. A sample of customer waiting times was collected, and the mean waiting time was found to be 3.5 minutes. The standard deviation of the waiting time was 1.2 minute.
(a)If approximately 5% of the customers waited more than 7.5 minutes, the manager
would add a teller. Assume the waiting times form a bell shape histogram. Should
a teller be added?
(b)If at most 10% of the customers waited more than 7 minutes, a teller will not be
added. Do not assume the waiting times were mound shaped. Should a teller be
added?
Solution
1.(i)P(Excellent) = P(Excellent and woman) + P(Excellent and man) = .27+ 22 = .49
(ii)P(Man and excellent) = .22
(iii)P(Man|Excellent) = P(Man and Excellent)/P(Excellent) = .22/.49 = .449
(iv)There is more than one way to determine whether gender and rating are
independent.For example we can find
P(Man) and compare it to P(Man|Excellent) already calculated. Specifically,
P(Man) = .22+.10+.12+.06 = .40. Since this result is different than
P(Man|Excellent) = .449, we conclude that gender and rating are not independent.
2.(a)Define B1, B2, and B3 as the events that describe the brand purchased by a
customer. Then we need to calculate P(B1 and B1or B2 and B2 or B3 and B3) =
P(B1 and B1) + P(B2 and B2) + P(B3 and B3) = (B1)P(B1)+P(B2)P(B2)+P(B3)P(B3).
Now, P(B1) = P(B2) = P(B3) = 1/3, so the required probability is 3(1/3)2= 1/3.
(b)There are 3 possible pairs she can select with equal probabilities (1/3). The probability of a match for, say, the pair B1 and B2 is P(B1 purchased last time and B1and B2 are purchased this time) + P(B2 purchased last time and B1 and B2 are purchased this time) = (1/3)(1/3)+(1/3)(1/3) = 2/9. This is true for any one of the three possible purchases made last time. Therefore, the probability of a match is 3(2/9) = 2/3.
3.Define: H – a customer orders a hot meal, so P(H) = .35
S – a customer is a student, so P(S) = .50. Also known: P(H|S) = .25
(a)P(H and S) = P(H|S)P(S) = .25(.50) = .125
(b)P(S|H) = P(H and S)/P(H) = .125/.35
(c)P(Sc or Hc) = P(Sc) + P(Hc) – P(Sc and Hc).
To find P(Sc and Hc) we can proceed as follows: The marginal probability of Hc is P(Hc) = P(Hc and S) + P(Hc and Sc), thus, P(Hc and Sc) = P(Hc) – P(Hc and S). Now, by the rule of multiplication, P(Hc and S) = P(Hc|S)P(S). Since P(H|S) = .25, then P(Hc|S) = .75. P(Hc and S) can now be determined: P(Hc and S) = (.75)(.50) = .375. Finally, P(Sc and Hc) = .65 - .375, and the required probability [P(Sc or Hc)] can now be found. In this case we could also use a probability tree quite efficiently. Observe the graphical demonstration below.
(d)To find if the two events are independent we can use the rule of multiplication.
P(H)P(S) = .35(.5) = .175; P(H and S) = calculated before = .125. Since
.175 .125 the two events are not independent. We can also look at the conditional probability and compare it to the marginal probability. For example: P(H|S) = .25 (given). P(H) = .35 (given). Again, since .25 .35 the two events are not independent.
- X is binomial with n = 6 and p = .67.
(a)P(X=5) can be calculated using the formula or by Excel.
Using the formula:
Using Excel: = binomdist(5,6,.67,false)
(b)P(X>2) = 1 – P(X2) Using Excel: = 1 – binomdist(2,6,.67,true)
(c)P(X 4) = binomdist(4,6,.67,True)
(d)P(2X5) = Binomdist(5,6,.67,True) – Binomdist(1,6,.67,True)
(e)If more than 4 adults do not want the government….then less than 2 adults want the government…. So, P(X ) = … use Excel.
5.X is normal with = 72 and = 5.
(a)P(X85) = P(Z(85 – 72)/5) = P(Z2.6) = 1 – P(Z2.6).
Using the table: P(Z2.6) = .5 + .4953 = .9953. So, 1 – P(Z2.6) = 1 - .9953
(b)The flight is less than 50% full if there are 42 passengers or less.
P(X<42) = P(Z<(42 – 72)/5) = P(Z<-6)
(c)Define K as the number of seats. Management requires that P(X>.75K) = .80 that is P(X<.75K) = .20. After standardization we have P(Z<(.75K-72)/5) = .20. For convenience let us write P(Z<Z0) = .20, and find Z0 from the table (see the graph and the explanation below.
Z0 Z1
6.(a)Since the distribution of the waiting times is bell shaped, we can apply the empirical rule.
- Approximately 68% of the waiting times fall in the interval [3.5-1.2, 3.5+1.2], that is in the interval [2.3, 4.7]
- Approximately 95% of the waiting times fall in the interval
[3.5-2(1.2), 3.5+2(1.2)], that is [1.1, 5.9]. - Approximately 99% of the waiting times fall in the interval
[3.5-3(1.2), 3.5+3(1.2)], that is [0, 7.1].
This means that (even in the worst case scenario) only 1% is waiting more than 7.1 minutes. So, less than 1% is waiting more than 7.5 minutes. A teller should not be added.
(b)Since we do not assume a bell shaped distribution for the waiting times, we need to use Chebychev’s rule. To have at least 90% of the waiting times fall inside an interval of K standard deviations around the mean we need to have K satisfy:
1 – 1/K2 = .90. Solving for K we have 1/K2 = .1, K2 = 10 so K = 3.16. The interval of waiting times that corresponds to this K is [3.5-3.16(1.2), 3.5+3.16(1.2)] or
[0, 7.292]. Since at least 90% of all the waiting times fall in this interval, at most 10% does not.
A waiting time of 7 minutes is located 2.92 standard deviations away from the mean of 3.5 (3.5 +K(1.2)=7, K=2.92). By Chebychev’s rule, a proportion of at least 1-1/K2 = 1-1/2.922= .883 of the waiting times fall in the interval [3.5-2.91(1.2), 3.5+2.9 (1.2)] or [.01, 7]. So, at most .117 of the waiting times are longer than 7 minutes. Since the criterion was that not more than 10% of the waiting times are longer than 7 minutes, the manager does not have a proof that the service level is met, and a teller should be added.