1. Section 5.6 Important Theorem in the Text: The Central Limit TheoremTheorem 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 Use the Central Limit Theorem to find a and b, both depending on n, so that the limiting distribution ofis 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 = 1/225/12. Therefore, the Central Limit Theorem tells us that the limiting distribution of is N(0, 1). Y – ———– n/2 5  n —– 2  3

2. 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 (b) X is the mean of n independent and identically distributed random variables each of which has mean  = and variance  2 = 1/225/12. Therefore, the Central Limit Theorem tells us that the limiting distribution of is N(0, 1). X – ———– 1/2 5 —–— 2  3n

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

4. 2. (a) A random sample X 1, X 2, …, X n is taken from a N(100, 64) distribution. Find each of the following: P(96 < X i < 104) for each i = 1, 2, …, n. P(96 < X i < 104) = 96 – X i – 104 – P( ———— < ———— < ———— ) = 100 888 P(– 0.50 < Z < 0.50) =  (0.50) –  (– 0.50) =  (0.50) – (1 –  (0.50)) = 0.6915 – (1 – 0.6915) = 0.3830

5. P(96 < X < 104) = 96 – X – 104 – P( ———— < ———— < ———— ) = 100 444 P(– 1.00 < Z < 1.00) =  (1.00) –  (– 1.00) =  (1.00) – (1 –  (1.00)) = 0.8413 – (1 – 0.8413) = 0.6826 2.-continued (b) P(96 < X < 104) when n = 4.

6. (c)P(96 < X < 104) when n = 16. P(96 < X < 104) = 96 – X – 104 – P( ———— < ———— < ———— ) = 100 222 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 X 1, X 2, …, X 25 is taken from a distribution defined by the p.d.f.f(x) = x / 50 if 0 < x < 10. P(X i < 6) for each i = 1, 2, …, 25. P(X i < 6) = x — dx= 50 9 / 25 0 6 x 2 —— = 100 0 6

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