Let "X=" service time of equipment: "X\\sim N(\\mu, \\sigma^2\/n )."
Then "Z=\\dfrac{X-\\mu}{\\sigma\/\\sqrt{n}}\\sim N(0, 1)"
a) Given "\\mu=54\\ months, \\sigma=10 \\ months, n=40"
"\\approx P(Z<-2.529822)\\approx0.005706"
b) Given "\\mu=54\\ months, \\sigma=10 \\ months, n=40"
"=1-P(Z\\leq\\dfrac{56-54}{10\/\\sqrt{40}})\\approx"
"\\approx P(Z<1.264911)\\approx1-0.897048"
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