On average, 3 customers make purchases every 2 hours at a certain hardware. Assume that the number of customers making purchases follows a Poisson distribution. If a customer has just made a purchase, calculate (to 4 decimal places) the probability that the time until the next customer arrives to make a purchase is less than 7 minutes:
"\\begin{aligned}\n\n\\lambda &=\\frac{3}{2 \\times 60} \\\\\n\n&=3 \/ 120 \\\\\n\n\\lambda &=0.025\n\n\\end{aligned}"
By using Poisson distribution "P(X=x)=\\frac{e^{-\\lambda} \\cdot \\lambda^{x}}{x!}" ,
"\\begin{aligned}\n\nP(X<7) &=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6) \\\\\n\n&=0.95731+0.02438+0.00330+0.0000+00000+0.000+0.0000 \\\\\n\nP(X<7) &=1.0000\n\n\\end{aligned}"
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