Question

Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 542, of which only 1 became a repeat offender.

(a) Suppose that of 1074 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still p = 1/542. Explain why the Poisson approximation to the binomial would be a good choice for r = number of repeat offenders out of 1074 convicted drunk drivers who toured the morgue.

The Poisson approximation is good because n is large, p is small, and np < 10.The Poisson approximation is good because n is large, p is small, and np > 10.     The Poisson approximation is good because n is large, p is large, and np < 10.The Poisson approximation is good because n is small, p is small, and np < 10.

What is λ to the nearest tenth?


(b) What is the probability that r = 0? (Use 4 decimal places.)


(c) What is the probability that r > 1? (Use 4 decimal places.)


(d) What is the probability that r > 2? (Use 4 decimal places.)


(e) What is the probability that r > 3? (Use 4 decimal places.)

Answer 1

Here N = 1074 and p = 1/542

For Binomial Distribution with large n, calculating the mass function is pretty complex. So for those complex “large” Binomials (n ≥100) and for small p (usually ≤0.01), we can use a Poisson with λ = nπ (≤20) to approximate it!

What is λ to the nearest tenth?

λ = 1074 * ( 1/542) = 1.98 = 2.0

(b) What is the probability that r = 0?

P( r = 0; 2.0) = (2.0)⁰ e^-2/ 0! = 0.1353

(c) What is the probability that r > 1?

P( r > 1; 2.0) = = 1 - P( r = 0; 2.0) - P( r = 1; 2.0) = 1 - (2.0)¹ e^{- 2.0}/ 1! - (2.0)⁰ e^-2.0/ 0!

= 1 - 0.2707 - 0.1353 = 0.5940

(d) What is the probability that r > 2?

P( r > 2; 2.0) = = 1 - P( r = 0; 2.0) - P( r = 1; 2.0) - P( r = 2; 2.0)

= 1 - (2.0)¹ e^{- 2.0}/ 1! - (2.0)⁰ e^-2.0/ 0! -  (2.0)² e^-2.0/ 2!

= 1 - 0.2707 - 0.1353 - 0.2707 = 0.3233

(e) What is the probability that r > 3?

P( r > 3 ; 2) = = 1 - P( r = 0; 2) - P( r = 1; 2) - P( r = 2; 2) - P( r = 3; 2)

= 1 - (2)¹ e^{- 2}/ 1! - (2)⁰ e^-2/ 0! -  (2)² e^-2/ 2! -  (2)³ e^-2/ 3!

= 1 - 0.2707 - 0.1353 - 0.2705 - 0.1804 = 0.1429