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Section 5.6 Important Theorem in the Text : The Central Limit Theorem Theorem 5.6-1

Section 5.6 Important Theorem in the Text : The Central Limit Theorem Theorem 5.6-1. 1. (a). Let X 1 , X 2 , … , X n be a random sample from a U ( –2, 3) distribution. Define the random variable Y = X 1 + X 2 + … + X n =. n  X i . i = 1.

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Section 5.6 Important Theorem in the Text : The Central Limit Theorem Theorem 5.6-1

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  1. Section 5.6 Important Theorem in the Text: The Central Limit Theorem Theorem 5.6-1 1. (a) Let X1 , X2 , … , Xn be a random sample from a U(–2, 3) distribution. Define the random variable Y = X1 + X2 + … + Xn = n  Xi . i = 1 Use the Central Limit Theorem to find a and b, both depending on n, so that the limiting distribution of is N(0, 1). Y– a —— b Y is the sum of n independent and identically distributed random variables each of which has mean  = and variance 2 = 1/2 25/12. Therefore, the Central Limit Theorem tells us that the limiting distribution of is N(0, 1). n/2 Y– ———– 5n —– 23

  2. (b) Use the Central Limit Theorem to find a and b, with only b depending on n, so that the limiting distribution of is N(0, 1). X– a —— b X is the mean of n independent and identically distributed random variables each of which has mean  = and variance 2 = 1/2 25/12. Therefore, the Central Limit Theorem tells us that the limiting distribution of is N(0, 1). 1/2 X– ———– 5 —–— 23n

  3. 1.-continued (c) Suppose n = 25. Use the Central Limit Theorem to approximate P(Y 12). Y– 12 – ———  ———— = 25/2 25/2 P(Y 12) = P P(Z ) = – 0.07 25 —– 23 25 —– 23 (– 0.07) = 1 – (0.07) = 0.4721

  4. 2. (a) A random sample X1 , X2 , … , Xnis taken from a N(100, 64) distribution. Find each of the following: P(96 < Xi < 104) for each i = 1, 2, …, n . 96 – Xi– 104 – P( ———— < ———— < ———— ) = 100 100 100 P(96 < Xi < 104) = 8 8 8 P(– 0.50 < Z < 0.50) = (0.50)– (– 0.50) = (0.50)– (1 – (0.50)) = 0.6915 – (1 – 0.6915) = 0.3830

  5. 2.-continued (b) P(96 < X < 104) when n = 4. 96 – X– 104 – P( ———— < ———— < ———— ) = 100 100 100 P(96 < X < 104) = 4 4 4 P(– 1.00 < Z < 1.00) = (1.00)– (– 1.00) = (1.00)– (1 – (1.00)) = 0.8413 – (1 – 0.8413) = 0.6826

  6. (c) P(96 < X < 104) when n = 16. 96 – X– 104 – P( ———— < ———— < ———— ) = 100 100 100 P(96 < X < 104) = 2 2 2 P(– 2.00 < Z < 2.00) = (2.00)– (– 2.00) = (2.00)– (1 – (2.00)) = 0.9772 – (1 – 0.9772) = 0.9544

  7. 3. (a) A random sample X1 , X2 , … , X25is taken from a distribution defined by the p.d.f. f(x) = x / 50 if 0 < x < 10 . P(Xi < 6) for each i = 1, 2, …, 25 . 6 6 x — dx = 50 x2 —— = 100 P(Xi < 6) = 9 / 25 0 0

  8. (b) Use the Central Limit Theorem to approximate P(X < 6).  = 20 / 3 2 = 50 – 400 / 9 = 50 / 9 X– 6 – P( ———— < ———— ) = 20/3 20/3 P(X < 6) = P(Z < – 1.41) = 2 / 3 2 / 3 (– 1.41) = 1– (1.41) = 1 – 0.9207 = 0.0793

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