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