QMB-210 Lecture Notes - Lecture 15: Exponential Distribution, Normal Distribution, Standard Deviation

• School: University of Tampa
• Department: Quantitative Methods in Business
• Course: QMB-210
• Professor: Shaista Nisar

Document Summary

Welsh 2: probability of the whole curve is always 100, half of the curve is 50% which is 0. 5. The time customers spend on the phone for service follows the normal distribution with a mean of 12 minutes and a standard deviation of 3 minutes. = median: known: = 12 and = 3, find the z-score for x = 14: The area under the normal curve equals 1. 0, so. P(z > 0. 67) = 1 p(z 0. 67) For a standard normal distribution, determine the following (cid:1005) (cid:1004). (cid:1011)(cid:1008)(cid:1012)(cid:1010) = (cid:1004). (cid:1006)(cid:1009)(cid:1005)(cid:1008) (cid:1005)(cid:1008) (cid:1004). (cid:1010)(cid:1011) (cid:1005)(cid:1006) (cid:1004) (cid:1004). (cid:1011)(cid:1008)(cid:1012)(cid:1010) Using the table find 1. 50 >> it is 0. 9332. The upper tail is, therefore, 1 0. 9332: p(z< -1. 22) If the question asked for greater than this part of the curve then you would subtract 0. 1112 from 1: p(-0. 86< z < 1. 76) ***we want the middle section, so, 0. 9608 0. 1949 = 0. 7659: p(0. 32 < z < 2. 15)