Question

12% of individuals insured by all-driver automobile insurance company have received at least one traffic ticket.

1) for a sample of 130 insured individuals, what is the probability that between 9 and 20 of them have received at least one ticket?

Answer 1

n = 130

p = 0.12

= n * p = 130 * 0.12 = 15.6

= sqrt(np(1 - p))

    = sqrt(130 * 0.12 * (1 - 0.12))

   = 3.705

If 9 and 20 are inclusive, then

P(9 < X < 20)

= P(8.5 < X < 20.5)

= P((8.5 - )/< (X - )/< (20.5 - )/)

= P((8.5 - 15.6)/3.705 < Z < (20.5 - 15.6)/3.705)

= P(-1.92 < Z < 1.32)

= P(Z < 1.32) - P(Z < -1.92)

= 0.9066 - 0.0274

= 0.8792

If 9 and 20 are not inclusive, then

P(9 < X < 20)

= P(10 < X < 19)

= P(9.5 < X < 19.5)

= P((9.5 - )/< (X - )/< (19.5 - )/)

= P((9.5 - 15.6)/3.705 < Z < (19.5 - 15.6)/3.705)

= P(-1.65 < Z < 1.05)

= P(Z < 1.05) - P(Z < -1.65)

= 0.8531 - 0.0495

= 0.8036