Question

It is estimated that 0.48 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal.

What is the probability that of today's 1,000 callers at least 5 received a busy signal?

Answer 1

Use the Poisson approximation to the binomial.

0.48 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal.

P=0.48/100 =0.0048

\lambda= np=1000*0.0048=4.8

Mean/Expected number of events of interest: 4.8
POISSON.DIST Probabilities Table
	X	P(X)
	0	0.0082
	1	0.0395
	2	0.0948
	3	0.1517
	4	0.1820

P( x >=5)

=( 1- P( x <5))

= ( 1-(P( x=0)+ P( x=1)+ P( x=2)+ P( x=3)+ P( x=4))

=(1-( 0.0082+ 0.0395+ 0.0948+ 0.1517+ 0.1820)

1-0.4763

=0.5237​