cltexamples

cltexamples - Distribution of the Sample Mean and Linear...

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Distribution of the Sample Mean and Linear Combinations – Examples Example 1 Let X 1 ;X 2 ;:::;X 100 denote the actual net weights of 100 randomly selected 50-pound bags of fertilizer. a. If the expected weight of each bag is 50 pounds and the standard deviation is 1 pound, approx- imate P (49 : 75 · ¹ X · 50 : 25) using the CLT. Solution Since ¹ ¹ X = ¹ = 50 and ¾ ¹ X = ¾ p n = 1 p 100 = 0 : 10 : P (49 : 75 · ¹ X · 50 : 25) = P μ 49 : 75 ¡ 50 0 : 10 · Z · 50 : 25 ¡ 50 0 : 10 = ©(2 : 5) ¡ ©( ¡ 2 : 5) = 0 : 9876 : b. If the expected weight is 49.8 pounds rather than 50 pounds, so that on average bags are under…lled, approximate P (49 : 75 · ¹ X · 50 : 25) . Solution The desired probability is: P (49 : 75 · ¹ X · 50 : 25) = P μ 49 : 75 ¡ 49 : 8 0 : 10 · Z · 50 : 25 ¡ 49 : 8 0 : 10 = ©(4 : 5) ¡ ©( ¡ 0 : 5) = 0 : 6915 : Example 2 The breaking strength of a rivet has a mean value of 10,000 psi and a standard devi- ation of 500 psi. a. What is the approximate probability that the sample mean breaking strength for a random sample of 40 rivets is between 9,900 psi and 10,200 psi? Solution The desired probability is: P (9900 · ¹ X · 10200) = P 0 @ 9900 ¡ 10000 500 p 40 · Z · 10200 ¡ 10000 500 p 40 1 A = ©(2 : 53) ¡ ©( ¡ 1 : 26) = 0 : 8905 : b. If the sample size had been 15 rivets rather than 40 rivets, could the probability requested in part a be approximated from the given information? Why or why not? Solution
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cltexamples - Distribution of the Sample Mean and Linear...

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