Question

Suppose the number of children in a household has a binomial distribution with parameters n-8 and p-50%. Find the probability of a household having (a) 3 or 7 children (b) 5 or fewer children (c) 3 or more children (d) fewer than 7 children (e) more than 5 children Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 0 times. You have 4 attempts remaining. Email instructor Page generated at 07/27/2018 at 03:31am EDT

Answer 1

Answer:

Step-by-step explanation:

Given that the number of children in a household has a binomial distribution with parameters n-8 and p-50%

As per binomial definition we have

$P(X=r) = nCr p^r q^{n-r}$

=$8Cr (0.5)^r (0.5)^{n-r} \\=8Cr(0.5)^8$

a) P(x=3 or 7) = P(3)+P(7) = 0.2188+0.0313=0.2501

b) P(X≤5)=0.8555

c) P(X≥3)=0.8555

d) P(x<7) =0.9648