Let "X=" the number of successes, "X\\sim B(n, p)."
Given that "n=529, p=0.02." Then
"n(1-p)=529(1-0.02)=518.42>10"
We can use the normal model
"=N(0.02, 0.0061)"
to approximate the sampling distribution of "\\hat{p}".
"\\approx P(z<2.4590)-P(z<1.3115)\\approx"
"\\approx0.99303-0.90516\\approx0.0879"
The probability that the sample proportion of items that scan innacurately is between 2.8% and 3.5% is "0.0879."
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