Question

It is estimated that 0.44 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 1,200 callers at least 5 received a busy signal? (Round your answer to 4 decimal places.)

Answer 1

Ans:

Normal approximation:

mean=1200*0.44=528

standard deviation=sqrt(1200*0.44*(1-0.44))=17.195

P(x>=5)=1-P(x<=4)

z=(4.5-528)/17.195

z=-30.44

P(x>=5)=1-P(x<=-30.44)

P(x>=5)=1-0.0000=1.0000

or using binomial distribution:

P(x>=5)=1-binomdist(4,1200,0.44,true)=1.0000