The data in the flowing table summarize the number of one-dayindustrial absences

Question

The data in the flowing table summarize the number of one-dayindustrial absences
for each of 195 inspectors working for a company during the year1957.
number of
dominant
mice in a liter observed
frequency 0 27 1 35 2 35 3 37 4 19 5 20 6 10 7 6 8 3 9 3 10 0
(a) Find the average number of absences per inspector.
(b) Compute expected frequencies based on the Poisson distributionand compare
these with the observed frequencies. Do you think that it isreasonable to use the
Poisson distribution as the distribution for the number ofabsences?
number of
dominant
mice in a liter observed
frequency 0 27 1 35 2 35 3 37 4 19 5 20 6 10 7 6 8 3 9 3 10 0

Explanation / Answer

x f     (f*x) P(X=x) Expected frequency      =N*P(X=x) 0 27 0 0.061123 11.91892602 1 35 35 0.17083 33.3118726 2 35 70 0.238724 46.55121 3 37 111 0.222401 43.36822447 4 19 76 0.155396 30.30215906 5 20 100 0.086862 16.93813118 6 10 60 0.040461 7.889984762 7 6 42 0.016155 3.150213927 8 3 24 0.005644 1.100555587 9 3 27 0.001753 0.341768 10 0 0 0.00049 0.095519781 Totals 55 195 545 0.999839 194.9685654 Average 2.79487 Based on the above constructed table, and comparing theobserved
and expected frequencies, the sum of the observed andexpected frequencies is equivalent.
So,we think that it is reasonable to use the
Poisson distribution as the distribution for the number ofabsences. i hope this will helps you. x f     (f*x) P(X=x) Expected frequency      =N*P(X=x) 0 27 0 0.061123 11.91892602 1 35 35 0.17083 33.3118726 2 35 70 0.238724 46.55121 3 37 111 0.222401 43.36822447 4 19 76 0.155396 30.30215906 5 20 100 0.086862 16.93813118 6 10 60 0.040461 7.889984762 7 6 42 0.016155 3.150213927 8 3 24 0.005644 1.100555587 9 3 27 0.001753 0.341768 10 0 0 0.00049 0.095519781 Totals 55 195 545 0.999839 194.9685654 Average 2.79487

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