Question

The number of spam emails received each day follows a Poisson distribution with a mean of 50. Approximate the following probabilities. Apply the ±½ correction factor and round value of standard normal random variable to 2 decimal places.
Round your answer to four decimal places (e.g. 98.7654).

(a) More than 50 and less than 60 spam emails in a day.
(b) At least 50 spam emails in a day.
(c) Less than 50 spam emails in a day.
(d) Approximate the probability that the total number of spam emails exceeds 350 in a seven-day week.

Answer 1

here mean =50

and std deviation =sqrt(50)=7.0711

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	50
std deviation   =σ=	7.0711

More than 50 and less than 60 spam emails in a day :

probability =

P(50.5<X<59.5)

=

P(0.07<Z<1.34)=

0.9099-0.5279=

0.3820

b)

At least 50 spam emails in a day :

probability =

P(X>49.5)

=

P(Z>-0.07)=

1-P(Z<-0.07)=

1-0.4721=

0.5279

c)

Less than 50 spam emails in a day :

probability =

P(X<49.5)

=

P(Z<-0.07)=

0.4721

d)

for 7 days; expected mail =50*7=350

and std deviation =sqrt(350)=18.7083

probability that the total number of spam emails exceeds 350 in a seven-day week :

probability =

P(X>350.5)

=

P(Z>0.027)=

1-P(Z<0.03)=

1-0.5120=

0.4880