Question

a) On average 2.5 faulty reports are made to a company’s switchboard per day.

i. Name the random variable present in this problem and state its distribution.

Calculate the probability that

ii. FOUR faulty reports will be made on Monday

iii. Less than 3 faulty reports in a 5-day work week

Answer 1

a)

i)

variable is the number of faulty report

Poisson distribution:

$P(x=k)= k! λ k e −λ ​$

mean $\lambda=2.5λ=2.5$ faulty reports per day

ii)

$P(x=4)=\frac{2.5^4e^{-2.5}}{4!}=0.1336P(x=4)= 4! 2.5 4 e −2.5 ​ =0.1336$

iii)

for a 5-day work week:

$λ=2.5⋅5=12.5$

$P(x<3)=P(x=0)+P(x=1)+P(x=2)$

$P(x=0)=e −12.5 =3.7⋅10 −6$

$P(x=1)=12.5e −12.5 =4.7⋅10 −5$

$P(x=2)=\frac{12.5^2e^{-12.5}}{2}=2.9\cdot10^{-4}P(x=2)= 2 12.5 2 e −12.5 ​ =2.9⋅10 −4$

$P(x<3)=3.7\cdot 10^{-6}+4.7\cdot 10^{-5}+2.9\cdot10^{-4}=3.407\cdot10^{-4}P(x<3)=3.7⋅10 −6 +4.7⋅10 −5 +2.9⋅10 −4 =3.407⋅10 −4$