Frequencies and Histogramsrestart;We load the Statistics and Plots packages.with(Statistics);with(plots);We enter our data as a list with measurements in \316\274g/ml of protein.X:=[76.33,77.63,149.49,54.38,55.47,51.70,78.15,85.40,41.98,69.91,128.40,88.17,58.50,84.70,44.40,57.73,88.78,86.24,54.07,95.06,114.79,53.07,72.30,59.36,59.20,67.10,109.30,82.60,62.80,61.90,74.78,77.40,57.90,91.47,71.50,61.70,106.00,61.10,63.96,54.41,83.82,79.55,153.56,70.17,55.05,100.36,51.16,72.10,62.32, 73.53,47.23,35.90,72.20,66.60,59.76,95.33,73.50,62.20,67.20,44.73,57.68]; We order our measurements from smallest to largest.Y:=sort(X,`<`);We create a frequency table extending from 30 to 170 with 7 class intervals of width 20 by using the FrequencyTable command. The five columns are class interval, frequency, relative frequency, cumulative frequency, and relative cumulative frequency.FrequencyTable(X,range=30..170,bins=7);We use the Histogram and FrequencyPlot to first create a frequency histogram relative to our table, and then to superimpose a frequency polygon over it to show the relationship of the two graphs. Note the bincount for FrequncyPlot is always two larger than for Histogram.p1:=Histogram(X,bincount=7,range=30..170,frequencyscale=absolute,color=red):p2:=FrequencyPlot(Y,bincount=9,range=10..190,frequencyscale=absolute,color=blue,thickness=3):display(p1,p2);A small adjustment gives us a relative frequency histogram and a relative frequency polygon.p3:=Histogram(X,bincount=7,range=30..170,frequencyscale=relative,color=red):p4:=FrequencyPlot(Y,bincount=9,range=10..190,frequencyscale=relative,color=blue,thickness=3):display(p3,p4);We now see what we get by cutting the class interval width in half. We use 13 class intervals of size 10 extending from 30 to 160.FrequencyTable(X,range=30..160,bins=13);When you get a result like this, you can double click the output to get the data in a spreadsheet like display, or you can use the interface command to increase the array size.interface(rtablesize=1000);We try again.FrequencyTable(X,range=30..160,bins=13);We create a frequency histogram from this table.p5:=Histogram(X,bincount=13,range=30..160,frequencyscale=absolute,color=red):display(p5);This histogram gives evidence that the two largest measurements can be viewed as outliers.