1. Consider the number of emails in any 10 minute period from panicked students

Question

1. Consider the number of emails in any 10 minute period from panicked students the day before homework is due; the average number of emails in any 10 minute period is 3 with a standard deviation of 2.1. Suppose a random sample of 36, non-overlapping, independent 10 minute periods is selected.
a. What’s the approximate distribution of the average number of emails received in a 10 minute period?
b. What’s the probability the average number of emails received in a 10 minute period is over 2.5 emails?
c. What’s the probability the average number of emails received in a 10 minute period is between 2.5 and 3 emails?

Explanation / Answer

a) average number of emails received in a 10 minute period is normal with mean =3 and

std error of mean =std deviation/(n)1/2 =2.1/(36)1/2 =0.35

b) probability the average number of emails received in a 10 minute period is over 2.5 emails =P(X>2.5)

=1-P(X<2.5)=1-P(Z<(2.5-3)/0.35)=1-P(Z<-1.4286)=1-0.0766 =0.9234

c)

probability the average number of emails received in a 10 minute period is between 2.5 and 3 emails

=P(2.5<X<3) =P((2.5-3)/0.35<Z<(3-3)/0.35) =P(-1.4286<Z<0)=0.5-0.0766 =0.4234

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