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
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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