Homework #9 Solution3.71 A warehouse contains ten printing machines, four of which are defective. A company randomly selectsfive of the machines for purchase. What is the probability that all five of the machines are not defective?ans: There are four machines are defective. Five nondefectives must be selected from the 6 nondefectivesgroup. The probability of the selected five which are nondefective isµ65¶µ105¶=6252=142.3.75 Specifications call for for a type thermistor to test out at between 9000 and 10000 ohms at 25◦C. Tenthermistors are available, and three of these are to be selected for use. Let Y denote the number amongthree that do not conform to specifications. Find the probability distribution for Y in tabular form ifthe following conditions prevail.(a) The ten contain two thermistors not conforming to specifications.ans: All possible values of Y are 0, 1, 2. The distribution is:P (Y = 0) =µ20¶µ83¶µ103¶=715P (Y = 1) =µ21¶µ82¶µ103¶=715P (Y = 2) =µ22¶µ81¶µ103¶=115(b) The ten contain four thermistors not conforming to specifications.ans: All possible values of Y in this subproblem are Y = 0, 1, 2, 3. The distribution is:P (Y = 0) =µ40¶µ63¶µ103¶=16P (Y = 1) =µ41¶µ62¶µ103¶=12P (Y = 2) =µ42¶µ61¶µ103¶=310P (Y = 3) =µ43¶µ60¶µ103¶=1301https://www.coursehero.com/file/7336734/HW09Key/This study resource wasshared via CourseHero.com3.79 A auditor checking the accounting practices of a firm samples three accounts from an accounts receivablelist of eight. Find the probability that the auditor sees at least one past-due account under the followingconditions.ans: Let Y be the number of past due accounts found by the auditor. Then Y follows a hypergeometricdistribution with N = 8, n = 3, and suppose that k is the numb er of past due accounts among eight,the probability is:P (Y ≥ 1) = 1 − P (Y = 0) = 1 −µk0¶µ8 − k3¶µ83¶= 1 −µ8 − k3¶56(a) There are two such accounts among the eight.ans: In this case k = 2, plug in the above probability we haveP (Y ≥ 1) = 1 −µ8 − 23¶56= 1 −2056=914.(b) There are four such accounts among the eight.ans: In this case k = 4, plug in the above probability we haveP (Y ≥ 1) = 1 −µ8 − 43¶56= 1 −456=1314.(c) There are seven such accounts among the eight.ans: In this case k = 7, plug in the above probability we haveP (Y ≥ 1) = 1 −µ8 − 73¶56= 1 − 0 = 1,since there is only one non past due account and the auditor must choose at least one, i.e. the auditorwill choose 1 or 2, or 3, and so on. Once two or more accounts are chosen the auditor will see a pastdue account for certain. Therefore the probability is 1.3.84 Find the moment generating function for the Bernoulli random variable.nas: The probability of a Bernoulli random variable is p when y = 1 and 1 − p when y = 0, thusM(t) = E(etY) = et·0(1 − p) + et·1(p) = 1 − p + etp3.85 Show that the moment generating function for the binomial random variable is given byM(t) = [p(et+ (1 − p)]nUse this result to derive the mean and the variance for the binomial distribution.ans: The probability of a binomial random variable Y = y isµny¶py(1 − p)n−y.M(t) = E(etY) =nXy =0etyµny¶py(1 − p)n−y=nXy=0µny¶(etp)y(1 − p)n−y= [pet+ (1 − p)]n.2https://www.coursehero.com/file/7336734/HW09Key/This study resource wasshared via CourseHero.comHere we note that the binomial theorem indicates (a + b)n=nXy =0µny¶aybn−y. To derive the mean weneed to compute the derivative M0(t) = n[pet+ (1 − p)]n−1pet. E(Y ) = M0(0) = np. To compute thevariance we need E(Y2) which is M00(0). The second derivative of M(t) isM00(t) = (n − 1)n[pet+ (1 − p)]n−2(pet)2+ n[pet+ (1 − p)]n−1pet.Thus M00(0) = (n − 1)np2+ np = (np)2− np2+ np. The variance isV ar(Y ) = E(Y2) − (E(Y ))2= (np)2− np2+ np − (np)2= np(1 − p).3.86 Show that the moment generating function for the Poisson random variable with mean λ is given byM(t) = eλ(et−1)Use this to derive the mean and the variance of the Poisson distribution.ans: From the class we got the mgf M (t) = eλ(et−1). The first derivative is M0(t) = eλ(et−1)λet. ThusE(Y) =M0(0) =λ. The second derivative ofM(t) isM00(t) = eλ(et−1)(λet)2+ eλ(et−1)λet,then E(Y2) = M00(0) = λ2+ λ. Therefore the variance is V ar(Y ) = E(Y2) − (E(Y ))2= λ.3https://www.coursehero.com/file/7336734/HW09Key/This study resource wasshared via CourseHero.comPowered by TCPDF
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