An important part of the customer services responsibilities of a telephone company relates to the speed with which reported troubles in residential services can be repaired. Suppose that past data indicate that 70% of the reported troubles in residential services can be repaired on the same day. If 100 troubles are reported on a given day, calculate the probability that at most 80 will be repaired on the same day.
This is a binomial distribution with "p=0.7,\\;n=100".
It can be approximated by the normal distribution with
"\\mu=np=100*07=70,"
"\\sigma=\\sqrt{np(1-p)}=\\sqrt{100*0.7*0.3}=4.58."
Thus, "P(X\\le80)=P(Z<\\frac{80.5-70}{4.58})=P(Z<2.29)=0.9890."
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