Can someone please please please explain this problem to me. In a simple explanation without no tables or attachments! Just something you can write on a scratch piece of paper to help me understand how to solve it. Just something simple. I don't even understand where the intervals came from. Can someone please just provide me with a simple hands on solution to this problem?

Consider the sample of 65 customer satisfaction ratings given in the table (below). 45 45 44 44 39 46 42 40 44 46 46 41 46 46 45 44 46 41 43 38 48 39 38 40 42 39 43 41\\ 39 43 36 41 44 42 40 42 41 42 43 43 41 44 38 45- 45 46 40 44 44 47 45 44 43 42 44 45 45 43 Click here for the Excel Data File Carry out a chi-square goodness-of-fit test of normality for the population of all customer satisfaction ratings using a = .05. Recall that we previously calculated a = 42.95 and s = 2.6424 for the 65 ratings. Use the Empirical Rule to create the intervals. (Round your expected frequencies to 2 decimal places. Round your answer to 3 decimal places.) a normal distribution X12 = : so HO. normality. Conclude population Answer Given : *= 49.95, 5=2. 6424 and 1=65 Interval RE Fi E: ( Fi - E: ) 2 0. 0928 ( 65 ) = 1.482 0 . 15676 ex - 25 ( -00, 37.665 ) 0 .1359 ( 65 )= 8.8335 1.13508 x- 25 x + 25 ( 48.235, 00) 0 .0228 * 65 = 1.482 1. 482 Sum 10.07401 Teststatistic: 72 & (Fi - Ei ) 2 Ei = 10 . 074 D.P = 3 , 2 = 0.05 coitical value ( R'c ) = CHIINV ( 0 .09 . 3) = 7 . 815 since, Real = 10 074 7 72 c= 7-815. so, cereject Ho. Normally conclude population not a normal distribution