Answer to Question #267928 in Sociology for Sadie

Question #267928

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


1
Expert's answer
2021-11-19T00:56:01-0500

a)

i)

variable is the number of faulty report

Poisson distribution:

"P(x=k)= \nk!\n\u03bb \nk\n e \n\u2212\u03bb\n \n\u200b"

mean "\\lambda=2.5\u03bb=2.5" faulty reports per day


ii)

"P(x=4)=\\frac{2.5^4e^{-2.5}}{4!}=0.1336P(x=4)= \n4!\n2.5 \n4\n e \n\u22122.5\n \n\u200b\n =0.1336"


iii)

for a 5-day work week:

"\u03bb=2.5\u22c55=12.5"

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

"P(x=0)=e \n\u221212.5\n =3.7\u22c510 \n\u22126"

"P(x=1)=12.5e \n\u221212.5\n =4.7\u22c510 \n\u22125"

"P(x=2)=\\frac{12.5^2e^{-12.5}}{2}=2.9\\cdot10^{-4}P(x=2)= \n2\n12.5 \n2\n e \n\u221212.5\n \n\u200b\n =2.9\u22c510 \n\u22124"

"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\u22c510 \n\u22126\n +4.7\u22c510 \n\u22125\n +2.9\u22c510 \n\u22124\n =3.407\u22c510 \n\u22124"



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