Question

AlwaysGreen currently serves 20% of the single-family homes in the US. Assuming binomial distribution, what is the probability that AlwaysGreen serve at least 5 homes in a random sample of 18 single-family home?

Answer 1

Solution

Given that ,

p = 0.20

1 - p = 0.80

n = 18

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * px * (1 - p)n - x

P(X 5) = 1 - P(X < 5)

= 1 - ( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) )

= 1 - ( ((18! / 0! (18 - 0)!) * 0.200 * (0.80)18 - 0 + ((18! / 1! (18 - 1)!) * 0.201 * (0.80)18 - 1 + ((18! / 2! (18 - 2)!) * 0.202 * (0.80)18 - 2 + ((18! / 3! (18 - 3)!) * 0.203 * (0.80)18 - 3 + ((18! / 4! (18 - 4)!) * 0.204 * (0.80)18 - 4 )

= 1 - ( 0.018 + 0.0811 + 0.1723 + 0.2297 + 0.2153 )

Probability = 0.2836