COMM 215 Midterm Midterm Revie Review w
Basics of Probability Gilles Cazelais
Lots of problems that come up in applications tions invo involv lvee phenom phenomena ena for which which exact exact pred predic ictio tions ns are are eith either er impos impossi sibl blee or very ery difficult. The best we can do in such cases is to determine the probability of the possible outcomes. outcomes. We can think of probability probability as a number between 0 and 1 that measures how likely likely an event is to happen. An event event with prob probab abil ilit ity y zero zero woul would d be an impos impossi sibl blee even eventt and and an even eventt with with prob probab abili ility ty one one would be certain to happen. An experiment is an acti activ vity ity with with an observ observable able resul result. t. A sample sample space of an experiment is a set of outcomes such that in each trial of the experiment, one one and only one of of these outcomes occurs. Each element of a sample space is called a simple event . An event is a subset of a sample space.
N times times and the event E event E occurs occurs n n((E ) times, then the ratio n ratio n((E )/N is is called the relative frequency of E . In genera general, l, if the number number of trials N N is high enough, the relative frequency is a good approximation of P ( P (E ). ) . In fact, we could have defined P defined P ((E ) by n(E ) . N N Anothe Anotherr method method com common monly ly used used to assign assign a probability distribution to a sample space is by simply simply assuming assuming it. For instance instance,, in many situations it is reasonable to assume that all outcomes in the sample space are occur. If you you have have a coin coin equally equally likely likely to occur. and have no reason to suspect that it is not fair, it is then reasonable to assume that both outcomes heads and a nd tails are are equally likely to occur. In this case we would assign P ( P (E ) = lim
→∞
P ( P (heads) heads) = P ( P (tails) tails) = 1/2. Given a sample space S = {E 1 , E 2 , . . . , En } Statist Statistica icall methods methods could then then be used used to with n with n simple events, a probability distri- test if the assumption that the coin is fair is bution for S S is an assignment of a number acceptable or not. P ( P (E i ) to each simple event E event E i so that 1. 0 ≤ P ( P (E i ) ≤ 1, for all i = 1, . . . , n 2. P ( P (E 1 ) + P + P ((E 2 ) + · · · + P + P ((E n ) = 1. The probabi probabilit lity y of an even eventt E , denoted by P ( P (E ), ), is the sum of the probabilities of all simple events contained in E . In practice, probability distributions are often ten esti estima mate ted d by exper experim imen ent. t. For exam exam-ple, ple, if you have have a coin coin and and want ant to know know what P ( P (heads) heads) is, you could flip it one million times times and collect collect statistic statistics. s. The numnumber of heads heads obtained divided by one million could then be used as a very good estimate of P ( P (heads). heads). If an experim experimen entt is performed performed
If S = {E 1 , E 2 , . . . , En } and if we assume that all simple events are equally likely to occur, then 1 P ( P (E 1 ) = P ( P (E 2 ) = · · · = P = P ((E n ) = . n In this case we can calculate the probability of any event E as as follows number of simple events in E in E P ( P (E ) = . number of simple events in S in S I The above formula works only when the simple events in the sample space are equally likely likely.. For example, if a die is weighte weighted d and the outcomes are not equally likely, then the above formula would not apply. Typeset with LATEX on April 20, 2006.
Basic Counting Principles Multiplication Principle
Consider a multistep process in which Step 1 has n 1 possible outcomes, Step 2 has n 2 possible outcomes, ...
Step r has n possible outcomes. r
Then, the entire process has n 1 × n2 × · · · × n possible outcomes. r
Inclusion-Exclusion Principle
If A A and B are two finite sets, then |A ∪ B | = | = | A| + | + |B | − |A ∩ B |. A permutation of n distinct items taken r at a time is an ordered list of r r distinct items chosen from a set of n n distinct items. The number of such permutations is given by P (n, r ) = n (n − 1)(n − 2) · · · (n − r + 1) =
n!
(n − r )!
.
Observe that P (n, n) = n !. A combination of n distinct items taken r at a time is an unordered set of r distinct items chosen from a set of n n distinct items. The number of such combinations is given by C (n, r ) =
Another notation for C (n, r ) is
n r
n! . r!(n − r )!
.
Problems
1. (a) How many many bit bit strings strings of of length length 8 are are possible? possible? (b) How many many of these start with a 1 or end with 00? Answers: (a)
256 (b) 160
2. How many many functions functions are there from a set with 6 element elementss to a set with 4 elements. elements. Answer: 4096
3. How many many one-to-one one-to-one functions functions are there from a set with 6 element elementss to a set with (a) (a) 4 elem elemen ents ts (b) (b) 6 elem elemen ents ts (c) (c) 10 10 ele eleme ment nts. s. Answers:
(a) 0 (b) 720 (c) (c) 151 200
4. Each user on a computer system has a password, which is six to eight characters long, where each character character is a letter (case sensitive) or a digit. If each password must contain at least one digit and at least one letter, how many possible passwords are there? Answer: 167410838583040 ≈
1.67 × 1014
5. (a) (a) Ho How w man many y di diff erent erent 7-place license plates are possible if the first 3 places are for capital letters and the other 4 for digits? (b) What if no letter and digit can be repeated repeated in a single license license plate? (c) What if adjacent adjacent letters letters and digits have to be di ff erent? erent? Answers:
(a) (a) 175 760 760 000 000 (b) 78 624 624 000 000 (c) 118 118 462 462 500 500
6. The 10 letters ABCDEFGHIJ ABCDEFGHIJ are used to form strings of length 7 (order matters). matters). (a) How many many possible possible strings are there if we do not allow repetition? repetition?
(b) (c) (d) (e)
How many many strings strings begin with the letter G if repetitions repetitions are allowed? allowed? How many strings strings contain contain the substring substring GAB if repetitions repetitions are not allowed? allowed? How many many strings strings begin or end with the substring substring GAB if repetition repetitionss is allowed? allowed? How many many strings contain contain the letters letters A and B, with A somewhere somewhere to the left of B, if repetitions repetitions are not allowed?
Answers: (a)
604800 (b) 1 000 000 (c) 4200 (d) (d) 19990 (e) 141120 141120
7. How many di ff erent erent subsets of 5 letters are possible from the the 26 letters of the alphabet? Answer: 65780 8. In how many ways can we select select a chairpers chairperson, on, secretary secretary,, and treasurer treasurer form a group of 25 persons? persons? Answer: 13800 9. How many many strings of five decimal decimal digits (a) do not contain contain the same digit five times? (b) begin with with an odd digit? (c) have have exactly three three digits that are 5s? Answers:
(a) 99990 (b) 50 000 (c) (c) 810
10. A biologist biologist is attemptin attemptingg to classify 20,000 species of insects by assigning assigning 3 letter letter initials (not necessarily necessarily distinct) to each species. Is it possible to classify all the species in this way? Answer: No 11. We have 25 lightbul lightbulbs bs distributed distributed as follows follows and we select select 5 at random (order does not matter). matter). 40-W 9 (a) (b) (c) (d)
60-W 6
75-W 10
In how many many ways can this be done. How many selections will contain contain exactly two 75-W bulbs. How many many selection selectionss will contain contain at least one 75-W bulbs. How many many selection selectionss will contain contain two 40-W and three 60-W bulbs
Answers: (a)
53130 (b) 20475 (c) 50127 (d) 720
12. How many many 5-card hands are possible from a standard standard deck of 52 cards? cards? How many many of these hands hands will have exactly three hearts and two spades? How many hands will have 3 kings and a pair? Answers: 2598960,
22308, 288
13. Consider Consider the following following 10 × 10 grid. C r
B r
r
A
(a) How many paths joining joining the two two points points A and C are possible if we start at point A and are only allowed to travel right and up on the grid? (b) Of all the paths paths from A to C found in (a), how many pass through point B ? Answers: (a)
184756 (b) 58212
14. How many 10-bit strings contain five consecutive consecutive zeros or five consecutive ones? Answer:
222
15. How many diff erent erent ways are there to choose a dozen donuts from the five varieties at a donut shop? Answer: 1820 16. The number 5 can be expressed as a sum of 3 p ositive integers, integers, taking order into account, in 6 ways, namely 1+1+3= 1+3+1= 3+1+1= 1+2+2= 2+1+2= 2+2+1. Let m and n be positive integers such that m ≤ n . In howmany many ways ways can n be expressed as a sum of m −1 positive integers, taking order into account? Answer: −1 n
m
Binomial Theorem For any set S with n elements, the number of subsets of S with r elements is given by
n r
=
n!
(n − r)! r!
.
A useful combinatorial formula is Pascal’s identity
n + 1 = r
n
n , r
+
r−1
1 r n.
It can be proved by the following following combinatorial argument. Consider a set S set S with with n n + 1 elements n and fix attention on a particular element in the set, call it element a. There are r− subsets 1 n of S with r elements that contain a, with r elements that do not S with r a , and there are r subsets of S S with r n+1 contain a contain a.. Since there are a total of r subsets of S with r elements, elements, Pascal’s identity holds. S with r
Binomial Binomial Theorem. Theorem. For
any n
∈ N and
any x, y
∈ R,
n
n
(x + y) =
n
k=0
xk yn−k .
k
Proof. We prove prove it by induction. For n = 1 we have
1
(x + y)1 =
x0 y1 +
0
1 1
Assume the formula holds for some fixed arbitrary n
x1 y0 = x + y. y.
∈ N.
Then,
(x + y)n+1 = (x + y)(x + y)n n
n
= (x + y)
k=0 n
=
xk yn−k
k
k=0
n k+1 n−k + x y k
n
n
n
set ℓ =k+1
n+1
=
ℓ=1
ℓ
n
=
−1
n
n
ℓ
x y
n+1−ℓ
+
n ℓ=0
ℓ
1
ℓ
x y
n+1−ℓ
n
= x
n+1
+
ℓ=1 n
= x
n+1
+
ℓ=1
n+1
ℓ=0
set ℓ=k
− − ℓ=1
=
k
k=0
xk yn+1−k
n + 1 ℓ
n
1
ℓ
n + 1 ℓ
+
n ℓ
xℓ yn+1−ℓ n
+ y
n+1
+
n ℓ=1
ℓ
xℓ yn+1−ℓ + yn+1
xℓ yn+1−ℓ + yn+1
xℓ yn+1−ℓ
which completes the proof by induction. AT X on April 22, 2006. Gilles Gilles Cazelais. Typeset Typeset with L E
+ x
n+1
ℓ
xℓ yn+1−ℓ
Permut Permutations ations vs. Combinatio Combinations ns
It is very important to make the distinction between permutations and combinations binations.. In permutat permutations, ions, order matters matters and in com combinat binations ions order does not matter. The important information can be summarized by: Perm Permuta utatio tion n Combin Combinati ation on
Order matters matte rs does not matter matter
Number P (n, k ) = C (n, k ) =
n! (n−k)! n! (n−k)!k!
Examples:
A company has to select 3 o fficers cers from a pool of 6 candid candidate ates. s. Ho How w many diff erent erent ways can this be done if: (a) The officers are distinct? (b) The officers are not distinct? distinct? It is very important whether or not these o fficers are distinct. (a) If the o fficers are distinct, we are picking a triple ( s1 , s2 , s3 ) with each s i be b eing a candidate, and order matters. This means we are finding a 3-permutation 6! 6! from from a set of 6 elemen elements. ts. So there there are: P(6, 3) = (6− 4 = 120 · 5 · · 4 (6−3)! = 3! = 6 · 5 distinct ways to pick these o fficers. (b) If the o fficers are not distinct, the triples ( s1 , s2 , s3 ), (s1 , s3 , s2 ), (s3 , s2 , s1 ), etc. etc. are the the same since since the positio positions ns are the same. same. So, we are findin findingg a 3combinations from a set of 6 elements. So there are: 6! 6! 6·5·4 C(6, 3) = (6− 5 · 4 4 = 20. (6−3)!3! = 3!3! = 3·2·1 = 5 ·
1
Standard Normal Distribution Table
0
z
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5
z
.0 .00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.00 .0000 .03 .0398 .07 .0793 .11 .1179 .15 .1554 .19 .1915 .22 .2257 .25 .2580 .28 .2881 .31 .3159 .34 .3413 .36 .3643 .38 .3849 .40 .4032 .41 .4192 .43 .4332 .44 .4452 .45 .4554 .46 .4641 .47 .4713 .47 .4772 .48 .4821 .48 .4861 .48 .4893 .49 .4918 .49 .4938 .49 .4953 .49 .4965 .49 .4974 .49 .4981 .49 .4987 .49 .4990 .49 .4993 .49 .4995 .49 .4997 .49 .4998
.0040 .0438 .0832 .1217 .1591 .1950 .2291 .2611 .2910 .3186 .3438 .3665 .3869 .4049 .4207 .4345 .4463 .4564 .4649 .4719 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .4987 .4991 .4993 .4995 .4997 .4998
.0080 .0478 .0871 .1255 .1628 .1985 .2324 .2642 .2939 .3212 .3461 .3686 .3888 .4066 .4222 .4357 .4474 .4573 .4656 .4726 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .4987 .4991 .4994 .4995 .4997 .4998
.0120 .0517 .0910 .1293 .1664 .2019 .2357 .2673 .2967 .3238 .3485 .3708 .3907 .4082 .4236 .4370 .4484 .4582 .4664 .4732 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .4988 .4991 .4994 .4996 .4997 .4998
.0160 .0557 .0948 .1331 .1700 .2054 .2389 .2704 .2995 .3264 .3508 .3729 .3925 .4099 .4251 .4382 .4495 .4591 .4671 .4738 .4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 .4988 .4992 .4994 .4996 .4997 .4998
.01 .0199 .05 .0596 .09 .0987 .13 .1368 .17 .1736 .20 .2088 .24 .2422 .27 .2734 .30 .3023 .32 .3289 .35 .3531 .37 .3749 .39 .3944 .41 .4115 .42 .4265 .43 .4394 .45 .4505 .45 .4599 .46 .4678 .47 .4744 .47 .4798 .48 .4842 .48 .4878 .49 .4906 .49 .4929 .49 .4946 .49 .4960 .49 .4970 .49 .4978 .49 .4984 .49 .4989 .49 .4992 .49 .4994 .49 .4996 .49 .4997 .49 .4998
.02 .0239 .06 .0636 .10 .1026 .14 .1406 .17 .1772 .21 .2123 .24 .2454 .27 .2764 .30 .3051 .33 .3315 .35 .3554 .37 .3770 .39 .3962 .41 .4131 .42 .4279 .44 .4406 .45 .4515 .46 .4608 .46 .4686 .47 .4750 .48 .4803 .48 .4846 .48 .4881 .49 .4909 .49 .4931 .49 .4948 .49 .4961 .49 .4971 .49 .4979 .49 .4985 .49 .4989 .49 .4992 .49 .4994 .49 .4996 .49 .4997 .49 .4998
.0279 .0675 .1064 .1443 .1808 .2157 .2486 .2794 .3078 .3340 .3577 .3790 .3980 .4147 .4292 .4418 .4525 .4616 .4693 .4756 .4808 .4850 .4884 .4911 .4932 .4949 .4962 .4972 .4979 .4985 .4989 .4992 .4995 .4996 .4997 .4998
.0319 .0714 .1103 .1480 .1844 .2190 .2517 .2823 .3106 .3365 .3599 .3810 .3997 .4162 .4306 .4429 .4535 .4625 .4699 .4761 .4812 .4854 .4887 .4913 .4934 .4951 .4963 .4973 .4980 .4986 .4990 .4993 .4995 .4996 .4997 .4998
.0359 .0753 .1141 .1517 .1879 .2224 .2549 .2852 .3133 .3389 .3621 .3830 .4015 .4177 .4319 .4441 .4545 .4633 .4706 .4767 .4817 .4857 .4890 .4916 .4936 .4952 .4964 .4974 .4981 .4986 .4990 .4993 .4995 .4997 .4998 .4998
Gilles Cazelais. Typeset with LATEX on April 20, 2006.
Midterm Review
W. Nawfal M.Sc .
Winter 2016
1. Using the data below, complete the table, create a histogram, and draw an ogive graph. Categor y
Frequency
1-5
5
6-10
2
11-15
5
16-20
1
21-25
4
26-30
1
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Winter 2016
2. Create a Stem & Leaf Diagram using the data below. !" !$" %$" %!" &&" &'" (&" ((" ('" )*" )+" )%" )(" ))" ))" )'" '*" '*" '&" '&" ')" !$)
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
3. Using the data below create a Box Plot !" !$" %$" %!" &&" &'" (&" ((" ('" )*" )+" )%" )(" ))" ))" )'" '*" '*" '&" '&" ')" !$)
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Winter 2016
measurement for each variable var iable in the data set. 4. Indicate the level of data measurement a. Airline Week b. Day of Week c. Time of Day d. Male/Female e. Business/Pleasure f.
Length of Trip Trip
g. Hours to Destination h. Children < 10 years (yes / no) i.
Pieces of Luggage
j.
Pieces Carried On
k. Times Flown l.
Satisfaction Level ( 1 = bad, 10 = great)
4 of 18
Midterm Review
W. Nawfal M.Sc .
Winter 2016
5. A store manager tracks the number of customer complaints each week. The following data reflect a random sample of ten weeks. 11
19 19
4
6
8
9
6
4
0
3
a. Calculate the mean, median, variance, and standard deviation
b. If one week is selected at random, what is the probability that it is the the bottom Quartile of complaints?
c. if 6 weeks weeks are picked picked at at random (no (no replacem replacement) ent) what what is the the probabili probability ty that each each week picked has more complaints than the previous week? (the 2nd has more complaints than the 1st, the 3rd has more complaints than the 2nd, and so on.)
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
6. RLW is a Montreal based firm that specializes in high-end running shoes, and has two factories in Montreal (one in the East-End and one in Lachine) that produce its shoes. The East-End factory produces an average of 335 units per day with a standard deviation equal to 11 units. The factory in Lachine produces an average of 145 units per day with a standard deviation equal to 8 units. Based on this information, which line is relatively more consistent?
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
7. A Montreal real-estate company owns six office buildings in the Greater Montreal area that it leases to businesses. The lease price per square foot differs by building due to location and building amenities. Currently, all six buildings are fully leased at the prices shown here. $ / Ft2
Ft2
Building 1
$75
125,000
Building 2
$85
37,500
Building 3
$90
77,500
Building 4
$45
35,000
Building 5
$55
60,000
Building 6
$110
130,000
average (mean) price per square foot for these buildings a. Compute the weighted average
b. Compute the standard deviation of the pr ice per square foot for these buildings
7 of 18
Midterm Review
W. Nawfal M.Sc .
Winter 2016
8. Assume that a standard deck of 52 playing cards is randomly shuffled and the first 2 cards are dealt to you. What is the probability that you have a blackjack ? A blackjack is where one card is an ace and the other card is worth 10 points. The 10-point cards are kings, queens, jacks and 10’s. 10’s.
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
9. Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources depar tment has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. Based on this information, what is the probability that an employee employee will have have less less than 20 minutes minutes of unused sick time? time?
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
10. Three events occur with probabilities of P(E1) = 0.35, P(E2) = 0.25, P(E3) = 0.40. Other probabilities are: P(B | E1) = 0.25, P(B | E2) = 0.15, P(B | E3) = 0.60.
a. Compute P(E1 | B)
Need to find P(B) first!
b. Compute P(E2 | B)
c. Compute P(E3 | B).
10 of 18
Midterm Review
W. Nawfal M.Sc .
Winter 2016
11. The personnel manager of a large corporation gives an aptitude test to all newly-hired employees employees in an attempt to determine their potential for a management position. Based on past results, the personnel manager knows that 70% of the new employees pass the test (P) and the remaining 30% of the new employees fail the test (F). Among the new employees who pass the test, 25% will end up in a management position (M) within one year, while the other 75% will never reach a management position. Among the new employees who fail the test, 10% will end up in a management position within one year, and the remaining remaining 90% will never never reach a management management position. position. a. Among Among all newly-hir newly-hired ed employee employeess who take the aptitud aptitude e test, what percenta percentage ge will fail the test and reach a management position within one year?
b. If an employee who has been at the company for more than one year is selected at random and the employee is not in a management position, what is the probability that the employee employee passed the aptitude test? test?
c. What percentage of all employees employees will end up in a management management position within one year?
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W. Nawfal M.Sc .
Winter 2016
12. Until the summer of 2008, the real estate market in Fresno, California, had been booming, with prices skyrocketing. Recently, a study showed the sales patterns in Fresno for singlefamily homes. One chart char t presented in the commission’s repor t is reproduced here. It shows the number of homes sold by price range and number of days the home was on the market. Days on the Market Price Range ($000)
1-7
8-30
Over 30
Under $200
125
15
30
$200-$500
200
150
100
$501-1000
400
525
175
Over $1000
125
140
35
850
830
340
170 450 110 300 2020
a. Using the relative frequency approach to probability assessment, what is the probability that a house will be on the market more than 7 days?
b. Is the event 1–7 days on the market independent of the price $2 00–$500?
c. Suppose a home has just sold in Fresno and was on the market less than 8 days, what is the most likely price range for that home?
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
13. A company wants to know how important higher education is when selecting a new CEO. Prior studies have shown that 60% of all CEOs will be successful, 85% of successful CEOs have a university degree, while 70% of those who don’t have a University degree turn out to be unsuccessful CEOs CEOs a. What is the probability probability of a CEO having having a university university degree
b. What is the probability of a CEO not having a university degree and being successful
c. What is the probab probability ility of a CEO havi having ng a unive university rsity degree degree or being being successful successful
Need to complete the table first! Recognize that since P(S'|D') = 70% then P(S|D') must = 30% ~ then solve for D'
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Midterm Review
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Winter 2016
14. Radio Shack stocks four alarm clock radios. If it has fewer than four clock radios available at the end of a week, the store restocks the item to bring the in-stock level up to four. four. If If weekly demand is greater than the four units in stock, the store loses the sale. The radio sells for $25 and costs the store $15. The Radio Shack manager estimates that the probability distribution of weekly demand for the radio is as follows: X (Weekly Demand)
P(x)
0
0.05
1
0.05
2
0.10
3
0.20
4
0.40
5
0.1
6
0.05
7
0.05
a. What is the expected expected weekly weekly deman demand d for the alarm alarm clock clock radio? radio?
E(x) = Sum(Xi)(Pi) = (0 x 0.05) + (1 x 0.05) + (2 x 0.10) ... = 3.6 radios b. What is the probability that weekly demand will be greater than the number of available available radios? r adios?
P(x > 4) = P(5) + P(6) + P(7) = 40% c. What is the expected weekly profit from the sale of the alarm clock radio? (Remember: There are only four clock radios available in any week to meet demand.)
d. On average, how much profit is lost each week because the radio is not available when demanded?
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
15. Use the binomial formula to calculate the following probabilities for an experiment in which n = 5 and p = 0.4: that x is at most 1 a. the probability that
b. the probability that that x is at least 4
that x is less than 1 c. the probability that
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
16. A manufacturing firm produces a product that has a ceramic coating. The coating is baked on to the product, and the baking process is known to produce 15% defective items (for example, cracked or chipped finishes). Every hour, 20 products from the thousands that are baked hourly are sampled from the ceramic coating process and inspected. a. a. What is the probability that 5 defective items will be found in the next sample of 20?
b. On average, how many defective items would be expected to occur in each sample of 20?
c. How likely is it that 15 or more non-defective (good) items would occur in a sample due to chance alone?
Where X = the number of "good" items thus use P = 85% Alternatively
Where X = the number of "bad" items thus use P = 15%
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
17. For a standardized normal distribution, calculate the following probabilities: a. P(z < 1.5)
b. P(z
" 0.85)
c. P(-1.28 < z < 1.75)
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Midterm Review
W. Nawfal M.Sc .
Winter 2016
year, a 21% drop 18. The average sales price of tablets was $386 during the first quarter of this year, compared to the same period per iod last year, year, according according to a repor t from IMS Research. The price decline is a result of intense competition in the tablet market Assume that the standard deviation of prices was $25, and that tablet pricing follows a normal nor mal distribution. a. If a local Best Buy outlet sold s old 80 tablets at or below $314, how many tablets where sold this year?
b. Compute the third quartile of the tablet prices sold by this Best Buy outlet
c. What is the probability of selling a tablet that’s priced more than the price of the average price of a tablet + tax? (assume the tax is 15%)
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