Let a random variable X be distributed normally with a mean of 100 and a standard deviation of 15.
1 answer:
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Mean number of errors in each page = 0.01
Mean number of errors in 100 pages = 0.01*100=1
It is possible to use the cumulative distribution function (CMF), but the math is a little more complex, involving the gamma-function. Tables and software are available for that purpose.
Thus it is easier to evaluate with a calculator for the individual cases of k=0,1,2 and 3.
The Poisson distribution has a PMF (probability mass function)
with λ = 1
=>
=>
or
P(k<=3)=0.9810 (to four decimal places)
Mean number of errors in 100 pages = 0.01*100=1
It is possible to use the cumulative distribution function (CMF), but the math is a little more complex, involving the gamma-function. Tables and software are available for that purpose.
Thus it is easier to evaluate with a calculator for the individual cases of k=0,1,2 and 3.
The Poisson distribution has a PMF (probability mass function)
with λ = 1
=>
=>
or
P(k<=3)=0.9810 (to four decimal places)
The given equation is,
Thus, option (C) is the correct solution.