Question

Homework: Distribution of Sample Means (Vveek 12, O. HW Score: 71.21 % , 7.83 of 11 pts 7 of 11 (9 complete) Score: 0.33 of 1 pt Question Help 8.1.25 The shape of the distribution of the Sme required to get an oil change at a 10-minute ol-change faciity is unknown, However, records indicate that the mean time is 11.5 minutes, and the standard deviation is 3.9 minutes. Complete parts (a) theough (c) below Click here to view the standard noemal distribution table (page 1) Cick here to view the standard norma distribuson table (page 2) (a) To compute probabines regarding the sample mean using the normal model what size sample would be required? Chocse the requred sample size beiow. A The nomal model cannot be used f the shape of the distnbution is unknown The sample size needs to be greater than 30 C. Any sample size could be used. D. The sample size needs to be less than 30. (b) What is the probabilty that a random sample of n 35 oil changes results in a sample mean tme less han 10 minutes? The probabity is approximately L (Round to four decimal places as needed.) Enter your answer in the answer boox and then oick Check Answer 1 part remaining Claar Al Check Answar

Please help with this question, but NOT with the standard distribution table. My professor uses **normalcdf** or some formula. NO STANDARD NORMAL DISTIVUTION TABLE please. I also need step by step answer. Thank you

Answer 1

TI-84 Plus Silver Edition TI-84 Plus Silver Edition TEXAS INSTRUMENTS TEXAS INSTRUMENTS normalcdf -1 E99 .0114529307 Ormaicdf lower: -1 E99 UPPer: -2.275 g: 1 Paste 10-11.5 =-2.275 Z = 3.9 V35 usin g nomalcdf (-1E99,-2.275,0,1) in ti -84 calculator The probability is approximately 0.01145