Let "X=" life of electric bulb in hours: "X\\sim N(\\mu, \\sigma^2)."
Then
(i)
Given that "\\mu=1200\\ hrs, \\sigma=200\\ hrs, n=10000"
"=1-P(Z<{1050-1200 \\over200\/\\sqrt{10000}})=1-P(Z<-75)\\approx1"
Approximately all 10,000 bulbs are expected have life 1050 hrs. or more.
(ii)
"=P(Z<150)\\approx1"
Approximately 100 percentage of bulbs are expected to fuse before 1500 hrs. of service.
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