Please Help find the Z percentiles :P Construct a normal probability plot for th
ID: 3132679 • Letter: P
Question
Please Help find the Z percentiles :P
Construct a normal probability plot for the following sample of observations on coating thickness for low-viscosity paint. 0.83 0.86 0.87 1.05 1.07 1.14 1.31 1.33 1.48 1.50 1.60 1.63 1.67 1.69 1.77 1.84 Determine the z percentile associated with each sample observation. (Round your answers to two decimal places.) Sample observation 0.83 0.86 0.87 1.05 1.07 1.14 1.31 1.33 z percentile Times Sample observation 1.48 1.50 1.60 1.63 1.67 1.69 1.77 1.84 z percentile Choose the plot that best represents the normal probability plot for the sample of observations.Explanation / Answer
HERE WE NEED TO CALCULATE THE Z VALUE OF EACH DATA VALUE WHICH WILL SHOW THE PERCENTILE
FOR THAT MATTER 1ST THE MEAN OF THE DATA SET AND THE STANDARD DEVIATION NEEDS TO BE CALCULATED
=
Z SCORE WILL BE CALCULATED FOR EVERY DATA
MEAN = 1.35
STANDARD DEV = 0.34
Z SCORE = (X-MEAN)/STANDARD DEV
1) 0.83
(0.83 - 1.35)/0.34 = -1.52 = 0.0643
2) 0.86
(0.86-1.35)/0.34 = -1.44 = 0.0749
3) 0.87
(0.87-1.35)/0.34 = -1.41 = 0.0793
4)1.05
(1.05-1.35)/0.34 = -0.88 = 0.1895
5) 1.07
(1.07-1.35)/0.34 = -0.82 = 0.2061
6) 1.14
(1.14-1.35)/(0.34)= -0.6 = 0.2743
7)1.31
(1.31-1.35)/0.34 = -0.11 = 0.4562
8) 1.33
(1.33 - 1.35)/0,34= -0.05 = 0.4801
9) 1.48
(1.48-1.35)/0.34 = 0.38 = 0.6480
10)1.5
(1.5-1.35)/0.34= 0.44 = 6700
11)1.6
(1.6-1.35)/0.34 = 0.73 = 0.7673
12)1.63
(1.63-1.35)/0.34 = 0.82 = 0.7939
13)1.67
(1.67-1.35)/0.34 = 0.94 = 0.8264
14)1.69
(1.69-1.35)/0.34 = 1 =0.8413
15) 1.77
(1.77 - 1.35)/0,34 = 1.23 = 0.8907
16) (1.84-1.35)/0.34 = 1.44 = 0.9251