Question

An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is given.

y	0	1	2	3
p(y)	0.50	0.30	0.15	0.05

(a)

What is the probability that among 15 randomly chosen such individuals, at least 10 have no citations? (Round your answer to three decimal places.)

(b)

What is the probability that among 15 randomly chosen such individuals, fewer than half have at least one citation? (Round your answer to three decimal places.)

(c)

What is the probability that among 15 randomly chosen such individuals, the number that have at least one citation is between 5 and 10, inclusive? ("Between a and b, inclusive" is equivalent to

(a ≤ X ≤ b).

Round your answer to three decimal places.)

Answer 1

a)

X = number of people with no citation

X follow binomial distribution with n = 15 , p = 0.50

P(X >= 10)

==1-BINOMDIST(9,15,0.5,1)

= 0.1509

b)

Y = number of people with at least one citation

Y follow binomial with n = 15 , p = 0.5

P(Y < 7.5)

= P(Y <= 7)

==BINOMDIST(7,15,0.5,1)

= 0.5

c)

P(5 <= Y <= 10)

= =BINOM.DIST.RANGE(15,0.5,5,10)

0.881530762