Question

Suppose the number of children in a household has a binomial distribution with parameters n = 24 and p= 40%. Find the probability of a household having :

(a) 17 or 23 children

(b) 21 or fewer children

(c) 20 or more children

(d) fewer than 23 children

(e) more than 21 children

Answer 1

Solution-a:

n=24

p=0.40

P(X=17)+P(X=23)

24C17*0.40^17*0.60^24-17+24C23*0.40^23*0.60^24-23

=0.0017+0.0000000101331

=0.0017

Solution-B:

P(X<=21)

1-P(X>21)

1-{P(X=22)+P(X=23)+P(X=24)}

=1-{ 1.74796e-07+ 1.01331e-08+ 2.81475e-10}

= 0.9999998

Solution-c;

P(X>=20)

P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)

=1.51417e-05+ 1.922756e-06+1.74796e-07+ 1.01331e-08+ 2.81475e-10

= 1.724967e-05

=0.000017

Solution-d:

P(X<23)

1-P(X>23)

=1-P(X=24)

=1-2.81475e-10

=1

Solution-e:

P(X>21)

=P(X=22)+P(X=23)+P(X=24)

=1.74796e-07+ 1.01331e-08+ 2.81475e-10}

=1.852106e-07

=0.0000001852