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?
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)
10.4763
=0.5237
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