A website experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution
Exponential distribution
"rate = \\frac{12}{60}=0.2 \\\\\n\n1. \\;P(x>10) = 1-P(x<10) \\\\\n\n= 1- (1- e^{-0.2 \\times 10}) \\\\\n\n= e^{-0.2 \\times 10} \\\\\n\n= 0.1353 \\\\\n\n2. \\; P(20<x<25) =P(x<25) -P(x<20) \\\\\n\n= 1- e^{-0.2 \\times 25}- (1- e^{-0.2 \\times 20}) \\\\\n\n= 0.0183-0.0067 \\\\\n\n= 0.0116 \\\\\n\n3. \\; P(x<k) = 1-e^{-0.2 \\times k} \\\\\n\n0.25= 1 -e^{-0.2 \\times k} \\\\\n\ne^{-0.2 \\times k}= 0.75 \\\\\n\nln(-0.2 \\times k) = ln(0.75) \\\\\n\n-0.2k= -0.2877 \\\\\n\nk= 1.4385 \\;min"
4. Required probability will follow Poisson distribution with mean 12
P(x<7) = P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4) +P(x=5) +P(x=6)
By using Excel
=POISSON(6,12,1)
P(x<7) = 0.0458
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