E mean is supposed to be higher 66 widget weights

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Unformatted text preview: ch unusual event happens most likely because the claim is wrong (i.e mean is supposed to be higher). 66. Widget weights have mean 40 grams and standard deviation 10grams. It is also known that 60 grams is the 90th percentile of widget weights. A random sample of 100 widgets is taken. Find approximate probabilities for each of the following: a. The average weight of widgets in the random sample was greater than 39.6 grams, b. More than 10 of the widgets weighed more than 60 grams. SOLUTION: Let X indicate the weight of a random chosen widget, i.e. X ∼ (µ = 40, σ = 10). ¯ a. P(X > 39.6) C.L.T. ≈ P( X−40 > √10 100 39.6−40 √10 100 ) = P(Z > −0.4) = 1 − 0.3446 = 0.6554. b. Let Y indicate the number of widgets weighted more than 60 g, i.e. X ∼ Bin(n = 100, p = P(X > 60) = 0.1), with µ = 100(0.1) = 10 and σ 2 = 100(0.1)(0.9) = 9. np = 10 and n(1 − p) = 90 • Since np is not greater than 10, approximation could be done through CLT (why? n = 100 > 30, and Bin(100, 0.1) is a sum of 100 i.i.d. Bernoulli(0.1)). P(Y > 10) 10 − 10 ≈ P(Z > √ ) 9 = P(Z > 0) = 0.5. • (full credit is granted) Since 10 is right on the boundary of...
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This document was uploaded on 02/15/2014 for the course MATH 231 at Lehigh University .

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