91影视

The probability of an event occurring is defined by probability. There are many scenarios in which we must forecast the outcome of an event in real life. The outcome of an event may be certain or uncertain. In such instances, we say that the event has a chance of happening or not happening.

Given observations:

\(\begin{array}{l}0.77,\;1.2,\;3,\;1.62,\;2.81,\;2.48,\;1.74,\;0.47,\;3.09,\;1.31,\;1.87,\\0.96,\;0.81,\;1.43,\;1.51,\;0.32,\;1.18,\;1.89,\;1.2,\;3.37,\;2.1,\;0.59,\\1.35,\;0.9,1.95,\;2.2,\;0.52,\;0.81,\;4.75,2.05\end{array}\)

We note that the data contains 30 data values.

We then need to determine the z-percentiles, which are the z-scores corresponding to the probability

\(\frac{{i - 0.5}}{{24}}\)(or the closest probability):

\(\begin{array}{l}i = 1 \Rightarrow \frac{{1 - 0.5}}{{30}} = 0.0167 \Rightarrow z = - 2.13\\i = 2 \Rightarrow \frac{{2 - 0.5}}{{30}} = 0.05 \Rightarrow z = - 1.645\\i = 3 \Rightarrow \frac{{3 - 0.5}}{{30}} = 0.0833 \Rightarrow z = - 1.38\\i = 4 \Rightarrow \frac{{4 - 0.5}}{{30}} = 0.1167 \Rightarrow z = - 1.19\\i = 5 \Rightarrow \frac{{5 - 0.5}}{{30}} = 0.15 \Rightarrow z = - 1.04\\i = 6 \Rightarrow \frac{{6 - 0.5}}{{30}} = 0.1833 \Rightarrow z = - 0.90\end{array}\)

\(\begin{array}{l}i = 7 \Rightarrow \frac{{7 - 0.5}}{{30}} = 0.2167 \Rightarrow z = - 0.78\\i = 8 \Rightarrow \frac{{8 - 0.5}}{{30}} = 0.25 \Rightarrow z = - 0.67\\i = 9 \Rightarrow \frac{{9 - 0.5}}{{30}} = 0.2833 \Rightarrow z = - 0.57\\i = 10 \Rightarrow \frac{{10 - 0.5}}{{30}} = 0.3167 \Rightarrow z = - 0.48\\i = 11 \Rightarrow \frac{{11 - 0.5}}{{30}} = 0.35 \Rightarrow z = - 0.39\\i = 12 \Rightarrow \frac{{12 - 0.5}}{{30}} = 0.3833 \Rightarrow z = - 0.30\\i = 13 \Rightarrow \frac{{13 - 0.5}}{{30}} = 0.4167 \Rightarrow z = - 0.22\\i = 14 \Rightarrow \frac{{14 - 0.5}}{{30}} = 0.45 \Rightarrow z = - 0.13\\i = 15 \Rightarrow \frac{{15 - 0.5}}{{30}} = 0.4833 \Rightarrow z = - 0.04\\i = 16 \Rightarrow \frac{{16 - 0.5}}{{30}} = 0.5167 \Rightarrow z = 0.04\end{array}\)

\(\begin{array}{l}i = 17 \Rightarrow \frac{{17 - 0.5}}{{30}} = 0.55 \Rightarrow z = 0.13\\i = 18 \Rightarrow \frac{{18 - 0.5}}{{30}} = 0.5833 \Rightarrow z = 0.22\\i = 19 \Rightarrow \frac{{19 - 0.5}}{{30}} = 0.6167 \Rightarrow z = 0.30\\i = 20 \Rightarrow \frac{{20 - 0.5}}{{30}} = 0.65 \Rightarrow z = 0.39\\i = 21 \Rightarrow \frac{{21 - 0.5}}{{30}} = 0.6833 \Rightarrow z = 0.48\end{array}\)

\(\begin{array}{l}i = 22 \Rightarrow \frac{{22 - 0.5}}{{30}} = 0.7167 \Rightarrow z = 0.57\\i = 23 \Rightarrow \frac{{23 - 0.5}}{{30}} = 0.75 \Rightarrow z = 0.67\\i = 24 \Rightarrow \frac{{24 - 0.5}}{{30}} = 0.7833 \Rightarrow z = 0.78\\i = 25 \Rightarrow \frac{{25 - 0.5}}{{30}} = 0.8167 \Rightarrow z = 0.90\\i = 26 \Rightarrow \frac{{26 - 0.5}}{{30}} = 0.85 \Rightarrow z = 1.04\\i = 27 \Rightarrow \frac{{27 - 0.5}}{{30}} = 0.8833 \Rightarrow z = 1.19\\i = 28 \Rightarrow \frac{{28 - 0.5}}{{30}} = 0.9167 \Rightarrow z = 1.38\end{array}\)

\(\begin{array}{l}i = 29 \Rightarrow \frac{{29 - 0.5}}{{30}} = 0.95 \Rightarrow z = 1.645\\i = 30 \Rightarrow \frac{{30 - 0.5}}{{30}} = 0.9833 \Rightarrow z = 2.13\end{array}\)

(a) NORMAL PROBABILITY PLOT

To determine if the distribution of values is approximately normally distributed, we have to create a normal probability plot.

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

If the pattern in the normal probability plot is roughly linear and does not contain strong curvature, then it is safe to assume that the distribution of the observations is approximately normal.

The created normal probability plot contains strong curvature, thus the distributions of the observations it not normally distributed.

(b) Determine the square root of every data value:

\(\begin{array}{l}0.8775,1.0954,1.7321,1.2728,1.6763,1.5748,1.3191,0.6856,1.7578,1.1446,1.3675,0.09798,\\0.9,1.1958,1.2288,0.5657,1.0863,1.3748,\\1.0954,1.8358,1.4491,0.7681,1.1619,\\0.9487,1.3964,1.4832,0.7211,0.9,2.1794,1.4318\end{array}\)

NORMAL PROBABILITY PLOT

To determine if the distribution of values is approximately normally distributed, we have to create a normal probability plot.

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

If the pattern in the normal probability plot is roughly linear and does not contain strong curvature, then it is safe to assume that the distribution of the observations is approximately normal.

The created normal probability plot contains no strong curvature Thus, the distributions of the squared observations are approximately normally distributed. Thus, it is plausible that the square root of precipitation is normally distributed.

(c) Take the cubic root of every data value:

\(\begin{array}{l}0.9166,1.0627,1.4422,1.1745,1.4111,1.3536,1.2028,0.7775,\\1.4565,1.0942,1.232,0.9865,0.9322,1.1266,1.1473,0.684,1.0567,1.2364\\1.0627,1.4993,1.2806,0.8387,1.1052,0.9655,1.2493,1.3006,0.8041,0.9322,1.681,1.2703\end{array}\)

NORMAL PROBABILITY PLOT

To determine if the distribution of values is approximately normally distributed, we have to create a normal probability plot.

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

If the pattern in the normal probability plot is roughly linear and does not contain strong curvature, then it is safe to assume that the distribution of the observations is approximately normal.

The created normal probability plot contains no strong curvature, thus the distributions of the squared observations are approximately normally distributed, and thus it is plausible that the cubic root of precipitation is normally distributed.