Question

Suppose the number of TV's in a household has a binomial distribution with parameters n = 17, and p = 80 %. Find the probability of a household having:

(a) 12 or 14 TV's

(b) 12 or fewer TV's

(c) 15 or more TV's

(d) fewer than 14 TV's

(e) more than 12 TV's

Answer 1

Binomial distribution: P(x) = nCx p​​​​​​x q​​​​​​n-x

n = 17

p = 0.8

q = 1 - p = 0.2

a) P(12 or 14 TVs) = P(12) + P(14)

= 17C12x0.812x0.25 + 17C14x0.814x0.23

= 0.1361 + 0.2393

= 0.3754

b) P(12 or fewer TVs) = 1 - P(X > 12)

= 1 - [P(13) + P(14) + P(15) + P(16) + P(17)]

= 1 - (0.2093 + 0.2393 + 0.1914 + 0.0957 + 0.0225)

= 0.2418

P(15 or more TVs) = P(15) + P(16) + P(17)

= 0.1914 + 0.0957 + 0.0225

= 0.3096

d) P(fewer than 14 TVs) = 1 - P(14 or more TVs)

= 1 - [P(14) + P(15) + P(16) + P(17)]

= 1 - (0.2393 + 0.1914 + 0.0957 + 0.0225)

= 0.4511

e) P(more than 12 TVs) = 1 - P(12 or fewer)

= 1 - 0.2418

= 0.7582