This is my first year learning Prob. Stat. Please show me how to so.pdf

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This is my first year learning Prob. Stat. Please show me how to solve these questions. I have the answers, but I woullove to kndow how to get those answers. I will rate as soon as I see your comments. Thanks alot. Answers: a) k>0 b) F(x) = 1 - x^-k ; 11 Answers: a) F(x) = x^2 /9 ; 0< x <3 , zero if x<= 0, one if x>=3 b) 4/9 c) 1/4 d) 3 sqrt(2) / 2 e)2 Solution a) f(x) should be non-negative for all values hence, k>0. b) for CDF, integrate this function for x= 1 to infinity c) for k>1, it is giving some non-zero result and for k<1, it\'s zero in this case! 18) a)for cdf, integrate this function for x = 0to 3, we get : |(x^2)/9| for 0 to 3 => 2 b) integrate this function for x = 0to 2, we get : |(x^2)/9| for 0 to 2 => 4/9 c) integrate this function for x = 0 to 1.5 (as this is defined for x>0, we get : |(x^2)/9| for 0 to 1.5 => 1/4 d) intergrate this for x= 0tp m euuate it with intergation from x =m to 3.. find a m between 0 and 3 => 3 sqrt(2)/2 e) integrate xf(x) from x=0to 3 and it gives => 2.

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The Thiessen polygons method is a common method used to interpolate p.pdf

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The two-asset case The expected return for asset A is 6.25% with a standard deviation of 5.00%, and the expected return for asset B is 4.50% with a standard deviation of 5.00%. Based on your knowledge of efficient portfolis, fill in the blanks in the following table with the appropriate answers. What is the efficient proportion of asset A within the portfolio for case II? 0.50 1.00 0.00 0.43 Solution Answer Answer 1 Security A Security B Expected portfolio return Proportion Expected Return SD Proportion Expected Return SD A B C D E F (A*B)+(D*E) 1 6.25 5 0 4.5 5 6.25 0.75 6.25 5 0.25 4.5 5 5.81 0.5 6.25 5 0.5 4.5 5 5.38 0.25 6.25 5 0.75 4.5 5 4.94 0 6.25 5 1 4.5 5 4.50 CASE I Security A Security B SD of Portfolio Proportion SD Proportion SD Pab A M C N O D P F Q R S T U A*A C*C M*N D*D F*F P*Q 2*A*C*D*F*S O+R+T Squreroot of U 1.00 1.0 5.0 25.0 25.0 0.00 0.0 5.0 25.0 0.0 -0.5 0.0 25.0 5.0 0.75 0.6 5.0 25.0 14.1 0.25 0.1 5.0 25.0 1.6 -0.5 -4.7 10.9 3.3 0.50 0.3 5.0 25.0 6.3 0.50 0.3 5.0 25.0 6.3 -0.5 -6.3 6.3 2.5 0.25 0.1 5.0 25.0 1.6 0.75 0.6 5.0 25.0 14.1 -0.5 -4.7 10.9 3.3 0.00 0.0 5.0 25.0 0.0 1.00 1.0 5.0 25.0 25.0 -0.5 0.0 25.0 5.0 CASE II Security A Security B SD of Portfolio Proportion SD Proportion SD Pab A M C N O D P F Q R S T U A*A C*C M*N D*D F*F P*Q 2*A*C*D*F*S O+R+T Squreroot of U 1.00 1.0 5.0 25.0 25.0 0.00 0.0 5.0 25.0 0.0 0.4 0.0 25.0 5.0 0.75 0.6 5.0 25.0 14.1 0.25 0.1 5.0 25.0 1.6 0.4 3.8 19.4 4.4 0.50 0.3 5.0 25.0 6.3 0.50 0.3 5.0 25.0 6.3 0.4 5.0 17.5 4.2 0.25 0.1 5.0 25.0 1.6 0.75 0.6 5.0 25.0 14.1 0.4 3.8 19.4 4.4 0.00 0.0 5.0 25.0 0.0 1.00 1.0 5.0 25.0 25.0 0.4 0.0 25.0 5.0 CASE III Security A Security B SD of Portfolio Proportion SD Proportion SD Pab A M C N O D P F Q R S T U A*A C*C M*N D*D F*F P*Q 2*A*C*D*F*S O+R+T Squreroot of U 1.00 1.0 5.0 25.0 25.0 0.00 0.0 5.0 25.0 0.0 0.7 0.0 25.0 5.0 0.75 0.6 5.0 25.0 14.1 0.25 0.1 5.0 25.0 1.6 0.7 6.6 22.2 4.7 0.50 0.3 5.0 25.0 6.3 0.50 0.3 5.0 25.0 6.3 0.7 8.8 21.3 4.6 0.25 0.1 5.0 25.0 1.6 0.75 0.6 5.0 25.0 14.1 0.7 6.6 22.2 4.7 0.00 0.0 5.0 25.0 0.0 1.00 1.0 5.0 25.0 25.0 0.7 0.0 25.0 5.0 Answer 2 What is the efficient proportion of asset A within the portfolio for CASE II? Efficient portfolio offers the highest expected return for a defined level of risk or the lowest risk for a given level of expected return. Answer : efficient proportion of asset A is 0.5 ( Where standard deviation (SD) of portfolio is lowest i.e. 4.2%) Security A Security B Expected portfolio return Proportion Expected Return SD Proportion Expected Return SD A B C D E F (A*B)+(D*E) 1 6.25 5 0 4.5 5 6.25 0.75 6.25 5 0.25 4.5 5 5.81 0.5 6.25 5 0.5 4.5 5 5.38 0.25 6.25 5 0.75 4.5 5 4.94 0 6.25 5 1 4.5 5 4.50.

The two-asset case The expected return for asset A is 6.25 with a s.pdf

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This :4 pts The statement of cash flows OA OB Oc OD dlassifies a firm\'s cash fows as operating, investing, financing, and other activities shows the financial position of a firm at a given point of time summarizes all the purchase and sale of fixed assets and raw materials provides insight into a firm\'s operating, investment, and financing cash flows Solution The statement of cash flows Answer Classifies cashflows as operating, investing, financing and other activities. Which only focuses on the actual cash balance of the firm Q2 the present value of a 25000 perpetuity at a 14 percent discount rate is PV = D/r Where:- D = Dividend or coupon period r = discount rate PV = Present Value PV = 25,000/0.14 = $178,571 Q3. Which off the following is true about annuities? Answer:- An annuity is an equal stream of cash flows is paid or recieved at the beginning of each period.

This 4 pts The statement of cash flows OA OB Oc OD dlassifies a firm.pdf

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There are41students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of5min and a standard deviation of4min. (Round your answers to four decimal places.) Solution a) z=(10-5)/(4/sqrt(41))=8.00 probability that he is through grading before the11:00 P.M.TV news begins = 1 b)probability that he misses part of the report if he waits until grading is done before turning on the TV = 0.

There are41students in an elementary statistics class. On the basis.pdf

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??There does not exist a fish that can live on land.?? Write an equivalent English statement that begins with the words ??Every fish??? Let P(x) be the predicate ??x can live on land,?? where the variable x represents animals. Write both statements symbolically using P(x) and quantifiers, ( There exists). For all Solution I\'m adding a predicate N(x) to represent \"x is a animal\". 1. Every fish is unable to live on land. ?x N(x) -> ~P(x) 2.The original statement might be represented as ~?x N(x) & P(x).

There does not exist a fish that can live on land. Write an equ.pdf

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There are twelve astrological signs: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio, Sagittarius, Capricorn, Aquarius, and Pisces. For example, your sign is Pisces if your birthday falls between February 19th and March 20th. Assuming all the astrological signs are equally likely for birthdays, find the probability that among a set of n people, at least two have the same astrological sign. Hint: remember the pigeonhole principle. Find the minimum number of people needed to guarantee a greater than 50% chance that two of them share the same astrological sign. Solution a) total no of events(sample space) = 12^n Atleast two having same sign = total - no one having same sign =(12^n) - (12*11*10*...*(13-n)) if n<=12 = 0 if n>12 by pegion hole principle P(E) = 1 - ( ( 12*11*10*....*(13-n) ) / 12^n ) ----if n<=12 = 0 ------------------------------- ------if n>12 b)P(E) > 1/2 So 12*11*10...*13-n / 12^n < 1/2 Solving we get n = 5.

There are twelve astrological signs Aries, Taurus, Gemini, Cancer, L.pdf

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There are two containers. One contains the letters BUMBLEBEES the oth.pdf

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There are several documentaries about the engineering disasters associated with Hurricane Katrina There are equally as many publications on the topic. Review at least three publications (not blogs or media reports or public opinion-based sources) to address the following questions What failed Why it failed Possible corrective actions Who was at fault and why Write a 3-5 page review addressing the afore-mentioned topics and include the following as well Describe in detail what ethical issues may have been partly/wholly responsible for the failure Be sure to address lack of professionalism or conscientiousness that may have helped lead to this failure You must cite references appropriately and there must be at least three credible references. Solution Engineering Review of the Disaster caused by Hurricane Katrina Even though 5 engineering review bodies operated separately, all of them reacted to IPET’s work, the main and finest funded of the study groups. Volumes 2 through 7 of the IPET “draft final” report divide the results into six broad sections: 1. Geodetic Vertical and Water Level Datums 2. Hurricane Protection System 3. The Storm 4. The Performance of the Levees and Floodwalls 5. Performance of Interior Drainage and Pumping 6. The Consequences; and Engineering and Operational Risk and Reliability Analyses The report is more than 6,000 pages extensive so far, and the last section on risk and reliability has not yet appeared. The NAE/NRC team and the ASCE team both prepared their commentary to conform to the IPET pattern. In reviewing a study of this kind, it is significant to keep in mind that critiques may address the quality of the design and construction for facilities in place at the time of the hurricane or the adequacy of the post-hurricane investigations described in the report. For example, one might find that the wind and water levels used for the design calculations were not adequate but that the hind-casting calculations of the hurricane’s effects described in the report are excellent. The NAE/NRC team and the ASCE team had similar reactions, with some minor differences of detail. The detailed study of the geodetic levels revealed considerable room for confusion and error. Two different benchmark levels were used in creating the levee system. In some cases, the water levels were expressed against one benchmark and the height of the levees against another. Furthermore, there seemed to be no consistent effort to monitor subsidence of the levee system or its components. The focus was on building the levees to the “authorized” elevations without considering whether the corresponding water elevations were measured from the same base or whether subsequent settlement and subsidence might make the authorized levels irrelevant. Ethical Issues: All of the review teams found that the hurricane protection system was a system in name only. Planning and system-wide design were confused by political and short-term economic considerations over.

There are several documentaries about the engineering disasters assoc.pdf

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There are many famous (more properly, infamous) examples of catastrophic failures of major engineering structures and systems. Use the Internet to look up each of the following and briefly describe whether the failure involved brittle or ductile failure initially (as a cause) and brittle or ductile failure overall (macroscopically). Tacoma (WA) Narrows Bridge, known as Galloping Gertie Minnesota I-35W Bridge Space Shuttle Challenger Space Shuttle Columbia RMS Titanic Aloha Airlines Flight 243 Solution a. Tacoma bridge failure is considered a ductile failure, because the collapse due to the wind vibrations lasted for long time so that no human casualities were made. Only one dog was the casuality. 2. The failure of I-35 bridge in Minnesota was a brittle failure because it occured suddenly and caused enormous catastrophic consequences. c. Space shuttle Challenger disaster occured due to the occurence of the brittle failure cracks in O-rings which allowed the fuel leakage and thereby the explosion. No one in the shuttle surviced, therefore it is a brittle failure. O-rings are ductile in nature but, at low temperature they become brittle, so such failures come in category of ductile-to-brittle transition failures. d. Space Shuttle Columbia failure occured due to the damage to the thermal shielding tiles from impact with a piece of foam, which fell from the insulation of an external tank during the take off. Failure occured while returning to the earth, while its speed was very very high, and wing experienced a high temperature of 1540 degree celsius. Damage to the thermal shield panel was tested to be a brittle failure, and therefore it also comes in category of brittle failure. e. RMS titanic was floating at very cold temperature and when near freezing temperature was met, tests showed that steel became extremely brittle. It was also a ductile-to-brittle transition failure. Overall ship sunk in 2 hours and about 15 minutes, so overall it could be considered a ductile failure, but no help was available in that time. f. It was a gatigue failure, which progressed in 19 years, but the end failure was brittle, therefore a brittle failure. Overall (maroscopically) failure was a ductile failure, since it happened due to the corrosion and fatigue of 19 years..

There are many famous (more properly, infamous) examples of catastrop.pdf

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The Warren truss is used to support a staircase. Determine the force .pdf

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The Wind Mountain archaeological site is located in southwestern New Mexico. Wind Mountain was home to an ancient culture of prehistoric Native Americans called Anasazi. A random sample of excavations at Wind Mountain gave the following depths (in centimeters) from present-day surface grade to the location of significant archaeological artifacts%u2020.8545120807555656065959070756568 Solution (b) Given a=0.02, |t(0.01, df=n-1=14)|=2.62 So 98% CI is xbar+/- t*s/vn --> 74.2+/- 2.62*18.3/sqrt(15) --> (61.8204, 86.5796).

The Wind Mountain archaeological site is located in southwestern Ne.pdf

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The waiting time Y until delivery of a new component for an industrial operation is uniformly distributed over the interval from 1 to 5 days. The cost of this delay is given by U = 2Y2 + 3. Use the method of transformations to derive the density function of U. I have found that the distribution function of Y is P(Y <= y) = (y-1)/5. Why is this? Solution You know F_Y(y) because Y is uniformly distributed. F_Y(y)= integral_1^y 1/4 dx= (y-1)/4 when y in [1,5] = 0 when y<1 = 1 when y>5 1/4 because the length of the interval [1,5] is 4. the density function is f_y(y)=1/4 when y in [1,5]..

The waiting time Y until delivery of a new component for an industr.pdf

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