Question

Consider the probability that fewer than 1919 out of 146146 houses will lose power once a year. Assume the probability that a given house will lose power once a year is 8%8%.

Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Solution:

We are given n = 146, p = 8% = 0.08, q = 1 – p = 1 – 0.08 = 0.92

We have to find the probability that fewer than 19 out of 146 houses will lose power once a year.

Here, we have to use normal approximation to binomial distribution.

Mean = np = 146*0.08 = 11.68

SD = sqrt(npq) = sqrt(146*0.08*0.92) = 3.2780482

We have to find P(X<19) ≈ P(X<18.5)

(By subtracting continuity correction factor 0.5)

Z = (X – mean) / SD

Z = (18.5 - 11.68) / 3.2780482

Z = 6.82/3.2780482

Z = 2.080506

P(Z<2.080506) = 0.98126

(By using z-table)

Required probability = 0.9813