Question

Suppose that 40% of all home buyers will do some remodeling to their home within the first five years of home ownership. Assuming this is true, use the binomial distribution to determine the probability that in a random sample of 20 homeowners, 4 or fewer will remodel their homes. Use the binomial table.

1.) The probability that 4 or fewer people in the same indicate that they will remodel their homes is ____.

Thank you!!!!

Answer 1

Solution:
Given in the question
P(all home buyers will do some remodeling to their home within the first five years of homeownership) = 0.4
Number of sample = 20
We need to calculate P(X<=4), here we will use a binomial distribution which can be calculated as
P(X=n |N,p) = NCn*(p^n)*(1-p)^(N-n)
P(X<=4) = P(X=0) + P(X=1) + P(X=2) +P(X=3) + P(X=4) = 20C0*(0.4^0)*(1-0.4)^(20-0) + 20C1*(0.4^1)*(1-0.4)^(20-1) + 20C2*(0.4^2)*(1-0.4)^(20-2) + 20C3*(0.4^3)*(1-0.4)^(20-3) + 20C4*(0.4^4)*(1-0.4)^(20-4) = 0.00004 + 0.0005 + 0.0031 + 0.0123 + 0.035 = 0.051
The probability that 4 or fewer people in the same indicate that they will remodel their homes is 5.1%.