Question

In: Statistics and Probability

14% of individuals insured by all-driver automobile insurance company have at least one traffic ticket. For...

14% of individuals insured by all-driver automobile insurance company have at least one traffic ticket. For a sample of 125 insured individuals, what is the probability that between 11 and 25 of them have received at least one ticket

Expert Solution

Normal approximation for binomial distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Sample size, n = 125

P(at least one traffic ticket), p = 0.14

q = 1 - p = 0.86

Mean = np

= 125 x 0.14

= 17.5

Standard deviation =

= 3.88

P(between 11 and 25 of them have received at least one ticket) = P(10.5 < X < 25.5) (with continuity correction)

= P(X < 25.5) - P(X < 10.5)

= P(Z < (25.5 - 17.5)/3.88) - P(Z < (10.5 - 17.5)/3.88)

= P(Z < 2.06) - P(Z < -1.80)

= 0.9803 - 0.0359

= 0.9444

orchestra answered 2 years ago

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