AMS 5 Chapter Notes - Chapter 18: Normal Distribution

• School: UCSC
• Department: Applied Math and Statistics
• Course: AMS 5
• Professor: Yonatan Katznelson

Document Summary

Four hundred draws will be made at random with replacement from the box. Estimate the chance that the sum of the draws will be more than 1,500. Estimate the chance that there will be fewer than 90 [3]s. 98. 73%; avg = 16/4 = 4 ev = 4*400 = 1600, sd = ((9+1+1+9)/4)1/2 = 51/2 se = 4001/2*51/2 ~ 44. 72 z = (1500-1600)/44. 72 ~ -2. 23 1. 27%, 100-1. 27 = 98. 73. 11. 27%; [0][1][0][0] avg = . 25 ev = . 25*400 = 100 sd = . 43 se = 20*. 43 = 8. 66 z = (89. 5-100)/8. 66 ~ -1. 2124 11. 27% Ten draws are going to be made at random with replacement from the box {[0][1][2][3]}. The chance that the sum will be in the interval from 10 to 20 inclusive equals the area under _a_ between _b_ and _c_. For the first one, your options are: the normal curve, the probability histogram for the sum.