Question

Consider the probability that no more than 19 out of 146 houses will lose power once a year. Assume the probability that a given house will lose power once a year is 8%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

n=

146

p=

0.0800

here mean of distribution=μ=np=		11.68
and standard deviation σ=sqrt(np(1-p))=		3.28
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability that no more than 19 out of 146 houses will lose power once a year :

probability =P(X<19.5)=(Z<(19.5-11.68)/3.278)=P(Z<2.39)=0.9916

(please try 0.9871 if no continuity correction is required)