In: Statistics and Probability
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?
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