STAT150 Lecture Notes - Lecture 5: Central Limit Theorem, Sample Size Determination, Triangular Distribution

• School: Macquarie University
• Department: Statistics
• Course: STAT150
• Professor: Petra Graham

Document Summary

=(cid:1005)(cid:1013)(cid:1004) (cid:373)i(cid:374)s, =(cid:1009)(cid:1007). (cid:1008) (cid:373)i(cid:374)s: probability >3 hours (i. e. >180 mins) = 1 norm. dist (180, 190, 53. 4, true) = 0. 5753, on the 25th percentile of distribution = norm. inv (0. 25, 190, 53. 4) = 153. 98. 1=(cid:1006)(cid:1005), 1=4. 7, y1 = 30 (cid:2869)=(cid:2871)(cid:2868) (cid:2870)(cid:2869)(cid:2872). (cid:2875) =(cid:883). 9(cid:883) > performed better overall. Norm. inv (0. 95, 0, 1) = 1. 6449 > therefore, y1 is in the top 5% =(cid:2874)(cid:2873)(cid:2868) (cid:2873)(cid:2868)(cid:2868: sample proportion p p =xn=number of successes number of trials. If x is normally distributed then p will also be normally distributed, with mean p (the true proportion) and standard error (cid:2926)(cid:2927)n. 85% of customers will use a credit card. X = 200, n = 250, p = 0. 85, p = 200/250 = 0. 8. = norm. dist (0. 8, 0. 85, sqrt (0. 85*0. 15/250), true) > norm. dist (p , p, se(p ), true) z=p p pqn = (cid:884). (cid:884)(cid:883)(cid:886) = 0. 0134 > there is only a 1. 3% chance that <200 customers would use their credit card.