Question 1181965: I. A sales manager receives 6 calls on average between 9:30 a.m. and 10:30 a.m. on a weekday.
Find the probability that:
a) he will receive 2 or more calls between 9:30 a.m. and 10:30 a.m. on a certain weekday.
b) he wiI1 receive exactly 2 calls between 9:30 a.m. and 9:40 a.m. on a certain weekday.
c) during a 5 day working week, there will be exactly 3 days on which he will receive no
calls between 9:30 a.m. and 9:40 a.m.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Poisson distribution with parameter 6.
b is easiest and is e^(-6)*6^2/2!
=0.0446
also 2ndVARSpoissonpdf(6,2)ENTER
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for a
probability of 0 calls is e^(-6)=0.0025
and 1 call I that *6=0.0149
That sum is 0.0174 and is not what we want. The complement is everything we do want, and that is 1-0.0174 or 0.9826, the probability.
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The number of calls he receives in a 10 minute period is a Poisson distribution with parameter 1, one-sixth of the time and one-sixth of the parameter of 6.
The probability of 0 calls in that 10 minute period is e^(-1)*1^0/0!=0.3679
The probability of not 0 calls is 1-0.3679 or 0.6321
There are 5C3, or 10 ways this can happen, and the formula is 10*0.3679^3*0.6321^2=0.1990
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