Question

Consider the probability that at least 97 out of 146 houses will not lose power once a year. Assume the probability that a given house will not lose power once a year is 61% approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Solution :

Given that,

p = 0.61

q = 1 - p =1-0.61=0.39

n = 146

Using binomial distribution,

= n * p = 146*0.61=89.06

= n * p * q = 146*0.61*0.39=5.8935

Using continuity correction

,P(x >97) = 1 - P(x <96.5 )

= 1 - P((x - ) / < (96.5-89.06) / 5.8935)

= 1 - P(z < 1.26)   

Using z table   

= 1-0.8962

probability= 0.1038