Question

5. (-/6 Points] DETAILS DEVORESTAT9 5.E.087.NVA MY NOTES PRACTICE ANOTHER Garbage trucks entering a particular waste management facility are weighed prior to offloading their contents. Let x = the total processing time for a randomly selected truck at this facility (waiting, weighing, and offloading). An article suggests the plausibility of a normal distribution with mean 13 min and standard deviation 4 min for X. Assume that this is in fact the correct distribution. (Round your answers to four decimal places.) (a) What is the probability that a single truck's processing time is between 12 and 14 min? (b) Consider a random sample of 16 trucks. What is the probability that the sample mean processing time is between 12 and 14 min? (c) Why is the probability in (b) much larger than the probability in (a)? The mean is u = 13. A sample mean X based on n = 16 observations is likely to be ---Select-- v the true mean than is a single observation X. That is because, while both are "centered" at u, the --Select... variability in X gives it ---Select-- ability to vary significantly from u. (d) What is the probability that the sample mean processing time for a random sample of 16 trucks will be at least 20 min? You may need to use the appropriate table in the Appendix of Tables to answer this question Need Help? Talk to Tutor

Answer 1

Sol Sol randomly selected Then, Z= x- let x = the total processing time for a truck at this facility. mean, y = 13 Standard deviation, r=4 XNN (p=13,5=4) X-13 follows Slándard Normal. a) Here, need to find P(12<x214) P(12LX L14)= P 2-13 LX-2 L 14 -13 PL-0.252EL0.25) = P(Zc0.25) - P(Z2-0.25) - 0.598706 -0.401294 (from from standard Normal table = 0.1974 12 ñ 0.1974 b) Sample size, n=16 Me= M = 13 اد According to Central Limit Theorem, ANN (ME = 13,7 = 1) Z = X-M X-13 follows standard Normal. Then -

c) The mean Here, we need to find need to find P(12 CXC 14) P(122X (14) = P 14-13 P(ALZc1) P(ZLI)- P(ZL-1) = 0.841345-0.158655 (from Sländard Normal tables perplees 0-68269 PL 0.6827 M=13. A sample mean X bared on n = 16 observations is likely to be closer to the true mean than in because, while both are single observation x. This is centered at r. the decreaned variability in it less ability to vary significantly from pe. gives d) Here, we need to find (20) PlX720)= PC ý -HÝ 20-13 Plz71) - PLZ<1) = 0.0000