Determine the sample size needed to construct a 99% confidence interval to estimate the population mean when o=91 and the margin of error equals 18.

Im having a problem find out how z0.005=2.575. i sent this same problem to a tutor about 2 hours ago and he gave me the wrong answer. Can u please explain this answer to me without using excel or any other technology? im not math savvy so can u please break this down to me in laymans terms step by step? I need this problem worked out by hand and can u explain how u got the answer to the za/2 part? I have basically tried to use every calculator and table i have. I also dont understand what the problem means when its says in this case "za/2 falls halfway between 2 values so the 2 values have been averaged" what does that even mean. I have been stuck here working on this problem for hours. Can u give me step by step answers only on those parts without using large vocabulary please?

D C M 12/1/2020 FALL 2020 STAT 3309 ONLINE CRN 23905-Arcola Mouton Student: Arcola Mouton Course: FALL 2020 STAT 3309 ONLINE CRN 23905 Instructor: Benny John Date: 12/01/20 8 5. 38 Book: Donnelly: Business Statistics, 30 Time: 10:06 Determine the sample size n needed to construct a 99% confidence interval to estimate the population mean when c = 91 and the margin of error equals 18. 1 Click the icon to view a table of standard normal cumulative probabilities. The formula for the sample size needed to estimate a population mean is given below, where Za / 2 is the critical z-score, the population standard deviation, and ME, is the margin of error. (Zux /2) 202 ( ME * ) 2 Determine the value of a that corresponds to a 99% confidence level. ( = 0.01 Now determine ? 0.01 = 0.005 2 Use this value to determine Za / 2. rounding to two decimal places. While either technology or a standard normal probability table could be used to find Za / 2. for this problem, use the table. In this case, Za / 2 falls halfway betwe values, so the two values have been averaged. Za /2 = 20.005 = 2.575 Substitute the values of Za / 2, 6, and ME- into the formula for the sample size. n = (za /2) 202 ( ME * ) 2 = (2.575)2(91)2 (18)2 plify, rounding to two decimal places. 1 = 169.47 dvised to always round up with a sample size because rounding down will not achieve the desi consider the stated margin of error as a minimum requirement. A smaller sample size result iven confidence interval. he sample size required is 170. ulative Probabilities for the Standard Normal Distribution od.pearsoncmg.com/api/v1/print/math