IE 111 Fall 2010 Solutions to Exam 4
Question 1 (15 points) p oints) We Deliver, Inc. offers next day delivery of packages weighing Sec. 10 Question between 2 and 20 pounds pounds in a certain city. They have found that the weights of the packages they deliver are (continuous) uniformly distributed between 2 and 20 pounds. Given that a package weighs less than 10 pounds, what is the probability that it weighs less than 5 pounds? SOLUTION X:Weight of package in lb, ~Uniform(2,20) We know that for the Uniform distribution FX (x )
x a b a
52 P (X 5 AND X 10) P (X 5) 20 2 3 P (X 5 | X 10) P (X 10) P (X 10) 10 2 8 20 2 amount of time that a randomly chosen student spends spends on Sec. 11 Question 1 (15 points) The amount homework per week is (continuous) (continuous) uniformly distributed from 30 to 120 minutes. minutes. What is the probability that a randomly r andomly chosen student spends at least 60 minutes per week on homework knowing that he/she will spend at most 80 minutes per week on homework? SOLUTION X:Time spent on homework, Uniform(30,120)
P (X 60 | X 80)
P (X 60 AND X 80) P (X 80) P (X 60) P (X 80) P (X 8 0)
80 30 60 30 20 1 2 0 3 0 120 30 90 0.4 80 30 30 50 90 120 30 In both of the above problems, you are asked to to find conditional probabilities. The PDF of the distributions should not change and you should make your calculations using the definition of the conditional probability. Question 2 (15 points) The annual rainfall (in inches) in a certain region is BOTH SECTIONS Question normally distributed distr ibuted with μ = 40, σ = 4. What is the probability that in 2 of the next 4 years the rainfall will exceed 50 inches? inches? Assume that the rainfalls in different different years are independent. SOLUTION X:Annual rainfall, ~N(40,4 2)
Solutions to Exam 4
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50 40 P (X 50) P Z P (Z 2.5) P (Z 2.5) 0.00621 4 Therefore, the probability of “success” in a year is 0.00621 and we have a binomial distribution with n=4 and p=0.00621 4 2 2 P (2 of the next 4 yrs. rainfall exceeds 50in.) 0.00621 1 0.00621 2 0.000229 Sec. 10 Question 3 (15 points) A bank wishing to increase its customer base advertises that it has the fastest service and that virtually all of its customers are served in less than 10 minutes. A management scientist has studied the service times and concluded that service times are exponentially distributed with a mean of 5 minutes. Determine what the bank means when it claims “virtually all” its customers are served under 10 minutes. SOLUTION
X:Service time at the bank, ~Exponential(1/5) as E[X]=5=1/ λ x ( ) ( ) 1 P X x F x e We know that X 1 5
10
P (X 10) 1 e
0.8647
Even though the bank is claiming they are serving virtually all the customers under 10 minutes, they only serve 86.47% of them. Sec.11 Question 3 (15 points)Toll booths on the New York Thruway are often congested because of the large number of cars waiting to pay. A consultant working for the state concluded that of service times are measured from the time a car stops in line until it leaves, service times are exponentially distributed with a mean of 2.7 minutes. What proportion of cars can get through the toll booth in less than 3 minutes?
SOLUTION X:Service time at the toll booth, ~Exponential(1/2.7) as E[X]=2.7=1/ λ . We will again use the CDF given in the above problem.
P (X 3) 1 e
1 3 2.7
0.6708
Thus, 67.08% of the cars go through the booth in less than 3 minutes Sec. 10 Question 4 (20 points) According to US Census Bureau, in 2008, 13.2% of population was below the poverty level. Assuming that this percentage is still valid today, what is the probability that between 100 and 150, inclusive, of the next 1,000 randomly selected people will be below the poverty level?
SOLUTION P(A person below poverty level) = p = 0.132, n = 1000
Solutions to Exam 4
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We can check to see if Normal distribution approximation is ok to use in this problem. We need not only np > 5 but n(1-p) > 5 as well. np = 132 > 5 and n(1-p) = 868 > 5 so we use the Normal distribution. Then, E[X]= np = 100(0.132) = 132 and V(X) = np(1-p) = 1000(0.132)(0.868)=114.576 σ = 10.704, therefore Xc~N(132,114.576)
P (100 X 150) P (99.5 Xc 150.5) P (Xc 150.5) P (X c 99.5) 150.5 132 99.5 132 P Z P Z 10.704 10.704 Using your book’s z-tables the above probability is 0.958185 – 0.001183 = 0.957002
Using EXCEL 0.958035 – 0.001198 = 0.956837 Sec. 11 Question 4 (20 points) According to a study published by a group of sociologists at the University of Massachusetts, approximately 49% of the Valium users in the state of Massachusetts are white collar workers. What is the probability between 482 and 510, inclusive, of the next 1,000 randomly selected Valium users from this state will be white collar workers?
SOLUTION P(Valium user is a white collar worker) = p = 0.49, n = 1000 np = 490 > 5 and n(1-p) = 510 > 5 => Normal approximation will work E[X]= np = 100(0.49) = 490 and V(X) = np(1-p) = 1000(0.49)(0.51)=249.9 σ = 15.808, therefore Xc~N(490,249.9)
P (482 X 510) P (481.5 Xc 510.5) P (Xc 510.5) P (X c 481.5) 510.5 490 481.5 490 P Z P Z 15.808 15.808 Using your book’s z-tables the above probability is 0.903199 – 0.294599 = 0.6086
Using EXCEL 0.902652 – 0.295391 = 0.607261 Sec. 10 Question 5 Pasta sauces are sold in jars containing a stated weight of 500g of sauce. The jars are filled by a machine. The actual weight of sauce in each jar is normally distributed with mean 505 g.
a) (10 points) If 13.8% of the jars have less than 495g of sauce, find the standard deviation (use 2 decimal points)
495 505 495 505 1.09 9.17 P (X 495) P Z 0.138 b) (10 points) Find the probability that a jar containing less than the stated weight Since 500 is smaller than the mean, i.e. 505, the probability you are going to find should be less than 0.5! Solutions to Exam 4
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500 505 P (X 500) P Z P (Z 0.55) 0.29116 9.17 c) (5 points) In a box of 30 jars, find the expected number of jars containing less than the stated weight Expected # of jars containing less than < 500g = 30(0.29116) = 8.73 d) (10 points) The mean weight of sauce is changed so that 1% of the jars contain less than the stated weight. The standard deviation stays the same. Find the new mean weight of sauce.
500 P (X 500) P Z 0.01 Use either -2.33 or -2.32 9.17 500 2.33 521.37 g 9.17 Sec. 11 Question 5 A vending machine dispenses coffee into cups. A sign on the machine indicates that each cup contains 50 ml of coffee. The machine actually dispenses coffee that is normally distributed with mean 55 ml per cup and
a) (10 points) If 3.8% of the jars contain less than 48ml coffee, find the standard deviation (use 2 decimal points)
48 55 48 55 1.77 3.95 P (X 48) P Z 0.038 b) (10 points) Find the probability that a cup containing less than the stated amount Since 50 is smaller than the mean, i.e. 55, the probability you are going to find should b e less than 0.5!
50 55 P (Z 1.27) 0.102042 P (X 50) P Z 3.95 c) (5 points) If the machine has 100 cups, find the expected number of cups containing less than the stated amount Expected # cups containing < 50 ml = 100*0.102042 = 10.20 d) (10 points) Following complaints, the owners of the machine make adjustments. Only 2.5% of cups now contain less than 50 ml. The standard deviation of the amount dispensed is reduced to 3ml. Assuming the amount of coffee dispensed is still normally distributed, find the new mean amount of coffee per cup.
50 50 0.025 1.96 55.88ml P (X 50) P Z 3 3
Solutions to Exam 4
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BONUS (10 points) Find the PDF of the following CDF
0 x 0 2 0 x 4 0.5kx FX (x ) 0.1x 8k 0.4 4 x 10 14kx 0.5kx 2 82k 0.6 10 x 14 16k 0.6 x 14
FX (x ) fX (u )du , so the above functions have been already integrated! x
You know that when x upper bound, FX (x ) 1 16k 0.6 1 k Range 0 x 4 d d fX (x ) (0.05 0.025)x 2 0.0125x 2 0.025x dx dx 2 F X (4) 0.0125 4 0.2
Range 4 x 10 You don't need to calculate this sum here, as the derivative of a constant is zero
fX (x )
d (0.1x dx
8(0.025) 0.4 0.2 F x (4 )
0.4 0.025 16
) 0.1
Why do I subtract FX(4) and FX(10) values? Because F X(.) is a cumulative function!
F X (10) 0.1(10) 8(0.025) 0.4 0.8
Range 10 x 14 CONSTANT d 2 14(0.025)x 0.5(0.025)x 82(0.025) fX (x ) 0.6 0.8 dx F X (10) 0.35 0.025x
0.025x fX (x ) 0.1 0.35 0.025x
Solutions to Exam 4
0 x 4 4 x 10 10 x 14
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