Question

Suppose the number of cars in a household has a binomial distribution
with parameters n = 20, and p = 10 %.
Find the probability of a household having:
(a) 7 or 9 cars
(b) 7 or fewer cars
(c) 10 or more cars
(d) fewer than 9 cars
(e) more than 7 cars

Answer 1

X ~ Bin ( n , p)

Where n = 20 , p = 0.10

a)

Using EXCEL , BINOM.DIST ( number_S, trials, probability_s , cumulative )

P(7 OR 9) = P(7) + P(9)

= BINOM.DIST ( 7 , 20 , 0.10 , FALSE) + BINOM.DIST ( 9 , 20 , 0.10 , FALSE)

= 0.0020 + 0.0001

= 0.0021

b)

P(X <= 7) = BINOM.DIST ( 7 , 20 , 0.10 , TRUE )

= 0.9996

c)

P(X >= 10) = 1 - P(X <= 9)

= 1 - BINOM.DIST ( 9 , 20 , 0.10 , TRUE)

= 1 - 1.0000

= 0

d)

P(X < 9) = P(X <= 8)

= BINOM.DIST ( 8 , 20 , 0.10 , TRUE)

= 0.9999

e)

P(X > 7) = 1 - P(X <= 7)

= 1 - BINOM.DIST ( 7 , 20 , 0.10 , TRUE)

= 1 - 0.9996

= 0.0004