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Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Given observations:

\({\rm{47}}{\rm{.1,68}}{\rm{.1,68}}{\rm{.1,90}}{\rm{.8,103}}{\rm{.6,106,115,126,146}}{\rm{.6,229,240,240,278,278,289,289,367,385}}{\rm{.9,392,505}}\)

(a) We note that the data contains \({\rm{20}}\)data values.

We then need to determine the z-percentiles, which are the z-scores corresponding to probability \(\frac{{{\rm{i - 0}}{\rm{.5}}}}{{{\rm{20}}}}\)(or the closest probability):

\({i=1}\frac{\text{1-0}\text{.5}}{\text{20}}\text{=0}\text{.025z=-1}\text{.96}\)

\({i=2}\frac{\text{2-0}\text{.5}}{\text{20}}\text{=0}\text{.075z=-1}\text{.44}\)

\({i=3}\frac{\text{3-0}\text{.5}}{\text{20}}\text{=0}\text{.125z=-1}\text{.15}\)

\({i=4}\frac{\text{4-0}\text{.5}}{\text{20}}\text{=0}\text{.175z=-0}\text{.93}\)

\({i=5}\frac{\text{5-0}\text{.5}}{\text{20}}\text{=0}\text{.225z=-0}\text{.76}\)

\({i=6}\frac{\text{6-0}\text{.5}}{\text{20}}\text{=0}\text{.275z=-0}\text{.60}\)

\({i=7}\frac{\text{7-0}\text{.5}}{\text{20}}\text{=0}\text{.325z=-0}\text{.45}\)

\({i=8}\frac{\text{8-0}\text{.5}}{\text{20}}\text{=0}\text{.375z=-0}\text{.32}\)

\({i=9}\frac{\text{9-0}\text{.5}}{\text{20}}\text{=0}\text{.425z=-0}\text{.19}\)

\({i=10}\frac{\text{10-0}\text{.5}}{\text{20}}\text{=0}\text{.475z=-0}\text{.06}\)

\({i=11}\frac{\text{11-0}\text{.5}}{\text{20}}\text{=0}\text{.525z=0}\text{.06}\)

\({i=12}\frac{\text{12-0}\text{.5}}{\text{20}}\text{=0}\text{.575z=0}\text{.19}\)

\({i=13}\frac{\text{13-0}\text{.5}}{\text{20}}\text{=0}\text{.625z=0}\text{.32}\)

\({i=14}\frac{\text{14-0}\text{.5}}{\text{20}}\text{=0}\text{.675z=0}\text{.45}\)

\({i=15}\frac{\text{15-0}\text{.5}}{\text{20}}\text{=0}\text{.725z=0}\text{.60}\)

\({i=16}\frac{\text{16-0}\text{.5}}{\text{20}}\text{=0}\text{.775z=0}\text{.76}\)

\({i=17}\frac{\text{17-0}\text{.5}}{\text{20}}\text{=0}\text{.825z=0}\text{.93}\)

\({i=18}\frac{\text{18-0}\text{.5}}{\text{20}}\text{=0}\text{.875z=1}\text{.15}\)

\({i=19}\frac{\text{19-0}\text{.5}}{\text{20}}\text{=0}\text{.925z=1}\text{.44}\)

\({i=20}\frac{\text{20-0}\text{.5}}{\text{20}}\text{=0}\text{.975z=1}\text{.96}\)

NORMAL PROBABILITY PLOT

We must generate a normal probability plot to assess if the distribution of variables is nearly normally distributed.

A scatterplot with the observations on the horizontal axis and the z-percentiles on the vertical axis is called a normal probability plot.

It is reasonable to presume that the distribution of the observations is substantially normal if the pattern in the normal probability plot is broadly linear and does not include severe curvature.

Because the generated normal probability map is essentially linear, the observations' distribution might be normal.

(b) We note that the data contains \({\rm{20}}\) data values.

We then need to determine the Weibulm-percentiles, which are the values of \({\rm{x}}\) for which the cumulative distribution function of the standard Weibull distribution is equal to the probability \(\frac{{{\rm{i - 0}}{\rm{.5}}}}{{{\rm{20}}}}\) (or the closest probability). Moreover, the percentile is \({\rm{ln( - ln(1 - p))}}\) with \({\rm{p}}\) the probability.

\({i=1p=}\frac{\text{1-0}\text{.5}}{\text{20}}\text{=0}\text{.025z=-3}\text{.68}\)

\({i=2p=}\frac{\text{2-0}\text{.5}}{\text{20}}\text{=0}\text{.075z=-2}\text{.55}\)

\({i=3p=}\frac{\text{3-0}\text{.5}}{\text{20}}\text{=0}\text{.125z=-2}\text{.01}\)

\({i=4p=}\frac{\text{4-0}\text{.5}}{\text{20}}\text{=0}\text{.175z=-1}\text{.65}\)

\({i=5p=}\frac{\text{5-0}\text{.5}}{\text{20}}\text{=0}\text{.225z=-1}\text{.37}\)

\({i=6p=}\frac{\text{6-0}\text{.5}}{\text{20}}\text{=0}\text{.275z=-1}\text{.13}\)

\({i=7p=}\frac{\text{7-0}\text{.5}}{\text{20}}\text{=0}\text{.325z=-0}\text{.93}\)

\({i=8p=}\frac{\text{8-0}\text{.5}}{\text{20}}\text{=0}\text{.375z=-0}\text{.76}\)

\({i=9p=}\frac{\text{9-0}\text{.5}}{\text{20}}\text{=0}\text{.425z=-0}\text{.59}\)

\({i=10p=}\frac{\text{10-0}\text{.5}}{\text{20}}\text{=0}\text{.475z=-0}\text{.30}\)

\({ip=}\frac{\text{11-0}\text{.5}}{\text{20}}\text{=0}\text{.525z=0}\text{.30}\)

\({i=12p=}\frac{\text{12-0}\text{.5}}{\text{20}}\text{=0}\text{.575z=0}\text{.59}\)

\({i=13p=}\frac{\text{13-0}\text{.5}}{\text{20}}\text{=0}\text{.625z=0}\text{.76}\)

\({i=14p=}\frac{\text{14-0}\text{.5}}{\text{20}}\text{=0}\text{.675z=0}\text{.93}\)

\({i=15p=}\frac{\text{15-0}\text{.5}}{\text{20}}\text{=0}\text{.725z=1}\text{.13}\)

\({i=16p=}\frac{\text{16-0}\text{.5}}{\text{20}}\text{=0}\text{.775z=1}\text{.37}\)

\({i=17p=}\frac{\text{17-0}\text{.5}}{\text{20}}\text{=0}\text{.825z=1}\text{.65}\)

\({i=18p=}\frac{\text{18-0}\text{.5}}{\text{20}}\text{=0}\text{.875z=2}\text{.01}\)

\({i=19p=}\frac{\text{19-0}\text{.5}}{\text{20}}\text{=0}\text{.925z=2}\text{.55}\)

\({i=20p=}\frac{\text{20-0}\text{.5}}{\text{20}}\text{=0}\text{.975z=3}\text{.68}\)

WEIBULL PROBABILITY PLOT

It is reasonable to presume that the distribution of the observations is substantially normal if the pattern in the normal probability plot is broadly linear and does not include severe curvature.

Because the generated normal probability map is essentially linear, the observations' distribution might be normal.

It is reasonable to infer that the distribution of the observations is essentially Weibull distributed if the pattern in the Weibull probability plot is generally linear and does not include substantial curvature.

Because there is no noticeable curvature in the Weibull probability map, the distribution of the observations might be Weibull distributed.