The Voyage of the St. Andrew Case Study - 1925 Words

The Voyage of the St. Andrew (Case Study Sample) Content: Statistics: Case Study 2.NameInstitutionQuestion 1HISTOGRAM SHOWING PROBABILITY DISTRIBUTION OF THE NUMBER OF FAMILIES PER PARISH OFGERMAN IMMIGRANTS.-321013726656PROBABILITY0PROBABILITYNumber of families per parishRelative Frequency DistributionXP(XÂ=Âx)10.70620.1763040.0595060.0590 E(X) = à ¢Pi.Xi = (1ÃÆ'0.706 + 2ÃÆ'0.176 + 3 ÃÆ'0 + 4 ÃÆ' 0.059 + 5ÃÆ'0 + 6ÃÆ'0.059) = 1.6482(X) =E(X2) à ¢Ã¢â€š ¬ ÃŽÂ ¼2E(X2) = (1ÃÆ'0.706 + 4ÃÆ'0.176 + 9 ÃÆ'0 + 16 ÃÆ' 0.059 + 25ÃÆ'0 + 36ÃÆ'0.059) à ¢Ã¢â€š ¬ 1.648 = 1.762Â(X) = à ¢Ã… ¡2(X) = 1.327HISTOGRAM SOWING PROBABILITY DISTRIBUTION OF THE KNOWN NUMBER OF FREIGHTS PURCHASED BY THE GERMAN FAMILIES20233533072022Number of freightsNumber of freights-321972942975PROBABILITY0PROBABILITYRelative Frequency DistributionXP(XÂ=Âx)10.0751.50.02520.4252.50.1530.1253.50.140.0550.02560.0250 E(X) =à ¢Pi.Xi = (1ÃÆ'0.075 + 1.5 ÃÆ'0.025 + 2ÃÆ'0.425 + 2.5ÃÆ'0.15 + 3ÃÆ'0.125 + 3.5ÃÆ'0.1 + 4ÃÆ'0.05 + 5 ÃÆ'0.025 + 6ÃÆ'0.025 ) = 2.5382(X) =E(X2) à ¢Ã¢â€š ¬ ÃŽÂ ¼2 = = (1ÃÆ'0.075 + 2.25 ÃÆ'0.025 + 4ÃÆ'0.425 + 6.25ÃÆ'0.15 + 9ÃÆ'0.125 + 12.25ÃÆ'0.1 + 16ÃÆ'0.05 + 25ÃÆ'0.025 + 36ÃÆ'0.025 ) à ¢Ã¢â€š ¬ 2.538 = 1.005(X) =à ¢Ã… ¡2(X) = 1.002HISTOGRAM SHOWING PROBABILITY DISTRIBUTION OF THE KNOWN NUMBER OF PEOPLE IN A FAMILY -371583773430PROBABILITY0PROBABILITYNumber in Family.Relative Frequency DistributionXP(XÂ=Âx)10.32220.18630.13640.10250.05160.13670.03480.01790.0160 E(X) = à ¢Pi.Xi =(1ÃÆ'0.322 + 2ÃÆ'0.186 + 3ÃÆ'0.136 + 4ÃÆ'0.102 + 5ÃÆ'0.051 + 6ÃÆ'0.136 + 7ÃÆ'0.034 + 8ÃÆ'0.017 + 9ÃÆ'0.016 ) = 3.0992(X)= 2(X) =E(X2) à ¢Ã¢â€š ¬ ÃŽÂ ¼2 = (1ÃÆ'0.322 + 4ÃÆ'0.186 + 9ÃÆ'0.136 + 16ÃÆ'0.102 + 25ÃÆ'0.051 + 36ÃÆ'0.136 + 49ÃÆ'0.034 + 64ÃÆ'0.017 + 81ÃÆ'0.016 ) à ¢Ã¢â€š ¬ 3.099 = 4.539Â(X) = 2.131 Question 2Yes it appears that, on average, the neulÃÆ'nders were successful in signing more than one family from a parish because the Expected of X (E(X )) is 1.648 which is more than one.The mean of 3.099 means that most of that it is likely that most of the families knew one another prior to undertaking the voyageQuestion 3Average cost of the crossing for a family in pistoles and in 1998 U.S. dollarsMean number of freights purchased = 2.538One freight = 7.5 pistoles2.538 = 7.5 ÃÆ' 2.538 = 19.035 pistolesIn dollars one freight =  £20002.538 freights=  £(2000 ÃÆ'2.538) =  £5,076Question 4It is not appropriate to estimate the average cost of the voyage from the mean family size because we are not given the components for each family i.e. age of each family members.Question 5Step 1: Sketch the curve.The probability thatÂX 4Âis equal to the black area under the curve.Step 2:SinceÂΠ¼=2.538ÂandÂà Ã†â€™=1.002Âwe have:PÂ(ÂX 4Â)=PÂ(ÂXà ¢ ÃŽÂ ¼ 4à ¢2.538Â) =PÂ(ÂXà ¢ÃƒÅ½...