Question

Construct a normal probability plot for the following sample of observations on coating thickness for low-viscosity paint.

0.83	0.87	0.89	1.02	1.09	1.14	1.27	1.29
1.46	1.51	1.60	1.64	1.67	1.70	1.78	1.83

Determine the z percentile associated with each sample observation. (Round your answers to two decimal places.)

Sample observation	0.83	0.87	0.89	1.02	1.09	1.14	1.27	1.29
z percentile
Sample observation	1.46	1.51	1.60	1.64	1.67	1.70	1.78	1.83
z percentile

Answer 1

Answer:

Given that,

Construct a normal probability plot for the following sample of observations on coating thickness for low-viscosity paint.

0.83	0.87	0.89	1.02	1.09	1.14	1.27	1.29
1.46	1.51	1.60	1.64	1.67	1.70	1.78	1.83

Count: N=16

Sum, Σx=21.59

Mean ():

=Σx/N

=21.59/16

=1.349375

Therefore, the mean=1.3494.

Standard Deviation ():

=Σ(xi - μ)2/N

=1.76749375/16

=√0.110468359375

=0.332367

Therefore, the standard deviation ()=0.3324

Determine the z percentile associated with each sample observation:

Then,

Z-value:

Z-percentile:

By using Z-percentile calculator or Using excel Norm.Inv(T, 0, 1).

Calculation table:

Sample Observation	0.83	0.87	0.89	1.02	1.09	1.14	1.27	1.29
Z-Value	-1.56	-1.44	-1.38	-0.99	-0.78	-0.62	-0.24	-0.18
Z-Percentile	5.94	7.49	8.38	16.11	21.77	26.76	40.52	42.86	Mean=1.3494.
Sample Observation	1.46	1.51	1.60	1.64	1.67	1.70	1.78	1.83	Standard deviation () =0.3324
Z-Value	0.33	0.48	0.75	0.87	0.97	1.06	1.29	1.45
Z-Percentile	62.93	68.44	77.34	80.78	83.4	85.54	90.15	92.65