Find the probability that the site is completed within 215 workdays




Yüklə 16.54 Kb.
tarix10.04.2016
ölçüsü16.54 Kb.

  1. (6 pts.) Site preparation for a new housing development with tracts for several homes is normally distributed with a mean of 200 workdays and a standard deviation of 10 workdays.




  1. Find the probability that the site is completed within 215 workdays. .9332

P(X<= 215) = P(Z<=1.5)=.9332


  1. Find the number of workdays that represents a probability of 0.99 that the site is completed. 223.3 P(Z<=2.33)=.99 so value=200+2.33*10=223.3

  2. Find the probability that the site is completed between 205 and 215 workdays. .2417

P(X<=205)=P(Z<=.5)=.6915, P(X<=215)=.9332 (above) so soln=.9332-.6915=.2417


Note that because it says site preparation is normally distributed then site preparation is continuous. So there is no need for continuity correct factor or for rounding days to integers.


  1. (8 pts.) The table below presents the probability distribution for X = the number of email messages you will receive in the next 10 minutes:

X P(X)

  1. .35

  2. .35

  3. .25

  4. .05

  1. What is the probability that no emails will be received? .35

P(X=0)=.35


  1. What is the probability that at least one email will be received? .65

P(X>=1) = P(X=1)+P(X=2)+P(X=3)=.35+.25+.05=.65


  1. What is E(X), the expected number of emails that will be received? 1

  2. What is V(X), the variance in the number of emails that will be received? .80




  1. (13 pts.) The Census Bureau has found that 26% of U.S. families are headed by a single parent. If 80 families are chosen at random to be interviewed:

  1. What is the expected number of families headed by a single parent? 20.8=E(X)

  2. What is the standard deviation of the number of families that are headed by a single parent? 3.92 = stdev of X

  3. What is the expected proportion of families headed by a single parent? .26=pi

  4. What is the standard deviation of the proportion of families headed by a single parent? .049 = stdev of pi

  5. Find the approximate probability that at least 25 of the families are headed by a single parent, using the normal distribution and the continuity correction. .1728

P(X>=25) approx = P(X>=24.5) using normal


= P(Z>= (24.5-20.8)/3.92) = P(Z>= .95)=1-.83=.17

  1. Suppose that 33 families in your sample are each headed by a single parent; is this result surprising? Why? Yes. X=33 is 3.11 standard deviations from E(X)=20.8




  1. (6 pts.) Suppose that you are the owner of a service station in Seattle where gasoline customers arrive randomly on weekday afternoons at an average of 3.2 every 4 minutes.

  1. What is the expected number of arrivals in a 4-minute interval on a weekday afternoon? 3.2 = mu

  2. What is the standard deviation of the number of arrivals in a 4-minute interval on a weekday afternoon? 1.79 (std dev is the square root of mu)

  3. What is the probability of having more than 1 customer in a 4-minute interval on a weekday afternoon? .83 = 1-P(X=0) –P(X=1) =1-.04 - .13= 1-.17 (using the Poisson function)

Note that on part c, some of you instead used the normal approximation to the Poisson, using a normal with mean of 3.2 and std dev of 1.79. This can work because mu is not too small. But, you need the continuity correction factor for it to work well.

P(X>1)=1-P(X<=1) Where X is Poisson

Approx = 1-P(X<= 1.5) Where X is normal



=1-P(Z<= (1.5-3.2)/1.79) = 1- P(Z<= -.95) = 1-.17=.83 ( very good approximation!)


  1. (6 pts.) The Environmental Protection Agency is trying to decide which of two industrial sites to investigate. The regional director estimates that the probability of a federal law violation is 0.3 for the first site and 0.25 for the second site. The director also believes that the occurrences of violations at the two sites are mutually exclusive.

P(I)=.3, P(II)=.25, P(I and II)=0 for I and II mutually exclusive


  1. Find the probability of federal law violation at the first site or the second site or both sites. .55=.3+.25-0

  2. Given there is a federal law violation at the first site, find the probability that there is also a federal law violation at the second site. 0




  1. (8 pts.) The Manager of Operations at an assembly plant is analyzing employee absenteeism. Ten percent of all plant employees work in the finishing department, 20% of all plant employees are absent excessively, and 7% of all plant employees work in the finishing department and are absent excessively.

P(fin)=.10, P(abs)=.20, P(fin and abs)=.07


  1. Find the probability that an employee selected at random works in the finishing department. .10 =P(fin)

  2. Find the probability that an employee selected at random works in the finishing department and is absent excessively. .07 =P(fin and abs)

  3. Find the probability that an employee selected at random either works in the finishing department or is absent excessively. .23=P(fin or abs)=P(fin)+P(abs)-P(fin and abs)= .10 + .20 - .07 = .23

  4. Given that the employee selected at random works in the finishing department, find the probability that the employee is absent excessively. .70=.07/.10 =P(abs/fin)

  5. Given that the employee selected at random is absent excessively, find the probability that the employee works in the finishing department. .35=.07/.20 = P(fin/abs)

  6. Are the events “working in the finishing department” and “absent excessively” mutually exclusive? No. P(fin and abs)=.07

  7. Are the events “working in the finishing department” and “absent excessively” independent? No. P(fin)=.10 and P(fin/abs)=.35

  8. If the percentage of employees who worked in the finishing department and were absent excessively was found to be 2% instead of the original 7%, are the events “works in the finishing department” and “absent excessively” independent?

P(fin and abs)=.02 = P(fin)*P(abs)=.10*.20 so Yes…. independent





  1. (9 pts.) A mail-order company has estimated sales for a popular item to be normally distributed with a mean of 180,000 units and a standard deviation of 15,000 units.

a. What is the probability of selling all the units on hand if the inventory equals 200,000 units? .0912 =P(X>=200000)=P(Z>=1.33)=1-.9082

They sell all units when X is greater than what is on-hand, the inventory of 200000.

b. What inventory should the company have on hand if they want the probability of running out of stock to be 5%? 204600

P(Z>=1.64)=.05 so, value=180000+1.64*15000=204600


They run out when X exceeds inventory. Find inventory so that the probability the X exceeds inventory is .05.

c. What is the optimal stocking level if overstocking costs $2 per item and under-stocking costs $4 an item? 186600



The intuitive solution is to say that it is twice as expensive to run out of stock as it is to have stock left over. If those events had the same cost then you’d stock the average, 180000. But since it is cheaper to have stock left over than to run out, you should stock more than average. How much more than average? Use the costs given to decide: stock so that the chance of having stock left over is 4/6=.67…. so that you are twice as likely to have stock left over as you are to run out. That’s 186600.
Or, the expected-value calculation leads to a more complicated approach:

overstock when X< inventory and stock out when X>=inventory

set inventory so that expected cost of overstock is same as expected cost of stock out

2*P(X=inventory)=4*(1-P(X

2*P(X

6*P(X

P(X

Inventory = 180000 + .44*15000 = 186600




  1. (9 pts.)The manufacturer of a commercial television monitor guarantees the picture tube for one year (8760 hours = 24 hours a day * 365 days a year). The monitors are used in airport terminals for flight schedules, and they are continuously in use with the power on. The mean life for these tubes is 20,000 hours, and their time to failure has an exponential distribution. It costs the manufacturer $200 to manufacture, sell, and deliver a monitor that is sold for $300. It costs $150 to replace a failed tube, including materials and labor.

  1. What is the probability that a monitor must be replaced under warranty? .35

P(t<8760)= .35 using the exponential function.

b. How likely is it that a monitor lasts less than or equal to the mean life of 20,000 hours?



.6321 = P(t<200000) using the exponential function.

c. What is the manufacturer’s expected profit? $46.80

The profit is either 300-200= 100, if it lasts beyond the warranty, or 300-200-150= -50 if it is replaced under warranty. We found in part a the probability that one is replaced.
profit t prob.

-50 t<8760 .35



100 t>=8760 .65
Expected(profit)= .35*(-50) + .65*100 = $46.80

BA 500- Statistics-Exam 1- Solns to quantitative questions- Pilcher- page of


Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©azrefs.org 2016
rəhbərliyinə müraciət

    Ana səhifə