(1 pt) Suppose the number of TV's in a household has a binomial distribution with parameters n = 24, and p = 10 %. Find the probability of a household having: (a) 17 or 22 TV's (b) 20 or fewer TV's (c) 17 or more TV's (d) fewer than 22 TV's (e) more than 20 TV's
Given that
n = 24
p = 0.10
(A) P(17 or 22) = P(17)+P(22)
Using binompdf
setting n=24,p= 0.10, k = 17 and 22
we get
P(17 or 22) = binompdf(24,0.10,17) + binompdf(24,0.10,22)
= 0.0000 + 0.0000
= 0.0000
(B) using binomcdf
setting n = 24,
p = 0.10 and k = 20
P(X20) = binomcdf(24,0.10,20)
= 1
(C)
using binomcdf
setting n = 24, p = 0.10 and k = 17
P(X17) = 1 - binomcdf(24,0.10,17-1)
= 1 - 1
= 0.0000
(D)
using binomcdf
setting n = 24, p = 0.10 and k = 22
P(X< 22) = binomcdf(24,0.10,22-1)
= binomcdf(24,0.1,21)
= 1
(E)
using binomcdf
setting n = 24, p = 0.10 and k = 20
P(X>20) =1- binomcdf(24,0.10,20)
= 1 - 1
= 0.0000
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