The Voyage of the St. Andrew (Case Study Sample)

Statistics: Case Study 2.
Name
Institution
Question 1
HISTOGRAM SHOWING PROBABILITY DISTRIBUTION OF THE NUMBER OF FAMILIES PER PARISH OFGERMAN IMMIGRANTS.
-321013726656PROBABILITY0PROBABILITY
Number of families per parish
Relative Frequency DistributionXP(X = x)10.70620.1763040.0595060.0590

E(X) = ‑Pi.Xi = (1×0.706 + 2×0.176 + 3 ×0 + 4 × 0.059 + 5×0 + 6×0.059) = 1.648
2(X) =E(X2) – μ2
E(X2) = (1×0.706 + 4×0.176 + 9 ×0 + 16 × 0.059 + 25×0 + 36×0.059) – 1.648 = 1.762
 (X) = ‚2(X) = 1.327
HISTOGRAM SOWING PROBABILITY DISTRIBUTION OF THE KNOWN NUMBER OF FREIGHTS PURCHASED BY THE GERMAN FAMILIES
20233533072022Number of freightsNumber of freights-321972942975PROBABILITY0PROBABILITY
Relative Frequency DistributionXP(X = x)10.0751.50.02520.4252.50.1530.1253.50.140.0550.02560.0250

E(X) =‑Pi.Xi = (1×0.075 + 1.5 ×0.025 + 2×0.425 + 2.5×0.15 + 3×0.125 + 3.5×0.1 + 4×0.05 + 5×0.025 + 6×0.025 ) = 2.538
2(X) =E(X2) – μ2 = = (1×0.075 + 2.25 ×0.025 + 4×0.425 + 6.25×0.15 + 9×0.125 + 12.25×0.1 + 16×0.05 + 25×0.025 + 36×0.025 ) – 2.538 = 1.005
(X) =‚2(X) = 1.002
HISTOGRAM SHOWING PROBABILITY DISTRIBUTION OF THE KNOWN NUMBER OF PEOPLE IN A FAMILY

-371583773430PROBABILITY0PROBABILITY
Number in Family.
Relative Frequency DistributionXP(X = x)10.32220.18630.13640.10250.05160.13670.03480.01790.0160

E(X) = ‑Pi.Xi =(1×0.322 + 2×0.186 + 3×0.136 + 4×0.102 + 5×0.051 + 6×0.136 + 7×0.034 + 8×0.017 + 9×0.016 ) = 3.099
2(X)= 2(X) =E(X2) – μ2 = (1×0.322 + 4×0.186 + 9×0.136 + 16×0.102 + 25×0.051 + 36×0.136 + 49×0.034 + 64×0.017 + 81×0.016 ) – 3.099 = 4.539
 (X) = 2.131

Question 2
Yes it appears that, on average, the neuländers 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 voyage
Question 3
Average cost of the crossing for a family in pistoles and in 1998 U.S. dollars
Mean number of freights purchased = 2.538
One freight = 7.5 pistoles
2.538 = 7.5 × 2.538 = 19.035 pistoles
In dollars one freight = £2000
2.538 freights= £(2000 ×2.538) = £5,076
Question 4
It 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 5
Step 1: Sketch the curve.
The probability that X > 4 is equal to the black area under the curve.
Step 2:
Since μ=2.538 and σ=1.002 we have:
P ( X >4 )=P ( X‒ μ> 4‒2.538 ) =P ( X‒Î...