STAT 400 Discussion 9 Spring 2015 1. One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value 25 in. and standard deviation 0.9 in. The length of the second piece is a normal random variable with mean and standard deviation 20 in. and 0.6 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation 0.2 in. Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between 43.45 in. and 45.65 in.? 2. Consider a 6-pack of soda. Suppose that the amount of soda in each can follows a normal distribution with mean 12.06 oz and standard deviation 0.15 oz. Assume that all cans are filled independently of each other. Find the probability of the following: a) a can is underfilled, i.e. there is less than 12 oz of soda in a can; b) all 6 cans are underfilled; c) at least one of the 6 cans is underfilled; d) exactly 2 of the 6 cans are underfilled; e) the average amount of soda in these 6 cans is less than 12 oz.3. 5.6-4 Approximate P(39.75 < X < 41.25), where X is the mean of a random sample of size 32 from a distribution with mean = 40 and variance 2 = 8. 4. 5.6-8 Let X equal the weight in grams of a miniature candy bar. Assume that = E[ X ] = 24.43 and 2 = Var[ X ] = 2.20. Find: (a) E[X] (b) Var[X] (c) P(24.17 < X < 24.82) approximately. 5. 6.4-1 Let X 1 , X 2 , … , X n be a random sample from: N ( , 2 ) unknown, known. Show that μˆ = X is the MLE for 6. 6.4-2 Let X 1 , X 2 , … , X n be a random sample from: N ( , 2 ) known, unknown. Show that niin122 μσˆX 1 is the MLE for 21. One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value 25 in. and standard deviation 0.9 in. The length of the second piece is a normal random variable with mean and standard deviation 20 in. and 0.6 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation 0.2 in. Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between 43.45 in. and 45.65 in.? Total = First + Second – Overlap. E(Total) = E(First) + E(Second) – E(Overlap) = 25 + 20 – 1 = 44. Var(Total) = Var(First) + Var(Second) + (– 1) 2 Var(Overlap) = 0.9 2 + 0.6 2 + 0.2 2 = 0.81 + 0.36 + 0.04 = 1.21. SD(Total) = 21.1 = 1.1. Total has Normal distribution. P( 43.45 < Total < 45.65 ) = 1.14465.45Z1.14445.43P = P( – 0.50 < Z < 1.50 ) = 0.9332 – 0.3085 = 0.6247.2. Consider a 6-pack of soda. Suppose that the amount of soda in each can follows a normal distribution with mean 12.06 oz and standard deviation 0.15 oz. Assume that all cans are filled independently of each other. Find the probability of the following: = 12.06, = 0.15. a) a can is underfilled, i.e. there is less than 12 oz of soda in a can; P(X < 12) = P15.006.1212Z = P(Z < – 0.40) = (– 0.40) = 0.3446. b) all 6 cans are underfilled; (0.3446) 6 = 0.001675. OR 6 C 6 (0.3446) 6 (0.6554) 0 = 0.001675. c) at least one of the 6 cans is underfilled; 1 – P(none) = 1 – (0.6554) 6 = 0.920743. d) exactly 2 of the 6 cans are underfilled; 6 C 2 (0.3446) 2 (0.6554) 4 = 0.32866. e) the average amount of soda in these 6 cans is less than 12 oz. Need P(X < 12) = ? n = 6 Normal distribution (Case 2). P(X < 12) = P615.006.1212Z = P(Z < – 0.98) = (– 0.98) = 0.1635.3. 5.6-4 4.
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