In a certain school, 6% of all students get a probation due to diverse reasons. Use the Poisson approximation to the binomial distribution to determine the probabilities that among 80 students (randomly chosen in this given school) : a) 4 will get at least one probation in any given year. b) At least 3 will get at least one probation in any given year. c) Anywhere from 3 to 6, inclusive will get at least one probation in any given year.

Step-by-step explanation:

In a certain school, 6% of all students get a probation due to diverse reasons. This means that

p = 6/100 = 0.06

Number of randomly selected students is 80. This means that

n = 80

Mean, u = np = 80*0.06 = 4.8

Using poisson probability distribution,

P (x=r) = (e^-u * u^r) / r!

a) 4 will get at least one probation in any given year. It becomes

P (x=4) = (e^-4.8 * 4.8^4) / 4! = 0.18

b) At least 3 will get at least one probation in any given year. This means

P (x greater than or equal to 3) = 1 - P (x lesser than or equal to 2)

P (x lesser than or equal to 2) = P (x = 0) + P (x = 1) + / P (x = 2)

P (x = 0) = (e^-4.8 * 4.8^0) / 0! = 0.0082

P (x = 1) = (e^-4.8 * 4.8^1) / 1! = 0.04

P (x = 2) = (e^-4.8 * 4.8^2) / 2! = 0.38

P (x lesser than or equal to 2) = 0.0082 + 0.04 + 0.38 = 0.4282

c) Anywhere from 3 to 6, inclusive will get at least one probation in any given year. This means

P (3 lesser than or equal to x lesser than or equal to 6) = P (x = 3) + P (x = 4) + P (x = 5) + P (x = 6)

P (x = 3) = (e^-4.8 * 4.8^3) / 3! = 0.15

P (x = 4) = (e^-4.8 * 4.8^4) / 4! = 0.18

P (x = 5) = (e^-4.8 * 4.8^5) / 5! = 0.17

P (x = 5) = (e^-4.8 * 4.8^6) / 6! = 0.14

P (3 lesser than or equal to x lesser than or equal to 6) = 0.15 + 0.18 + 0.17 + 0.14 = 0.64