Not sure what you mean by 'color axis is too thin`. Do you mean the thickness of the line? If so

PlotStyle -> AbsoluteThickness[5]

If you want a thicker BarLegend then

PlotLegends -> 
 BarLegend[{Hue[(# - 1)/3.1] &, {1, 3.1}}, LegendMarkerSize -> {30, 250}]

{30, 250} is {width, height}

ListLinePlot[Transpose[{vn/Tn, sumpt}],
 PlotStyle -> AbsoluteThickness[5],
 ColorFunction -> Function[{x, y}, Hue[.7 col[[x]]]],
 ColorFunctionScaling -> False,
 PlotLegends -> 
  BarLegend[{Hue[(# - 1)/3.1] &, {1, 3.1}}, LegendMarkerSize -> {30, 250}]]

Hi John,

didn't you understand? Your code is WRONG, Have you ever run it? I tried to fix the flaws and asked you if you agree to the results of the new version. But you did not answer to that question. So: please check it.

Then ((ETv)/(vnE1)) is the array of the random numbers. Check this too.

After that we can talk about colors.

Here is a 1st attempt

v = 246
E1 = 9000

Tn = {300, 500, 469, 650, 546, 389, 456, 411, 523, 700}
vn = Range[10] Tn
sumpt = Table[0, {10}]
nn = 1000
Do[
 rr = RandomReal[{1, 3.1}, 10];
 E1T = E1 vn rr/ v;
 PT = E1T/Tn - 7 Log[vn]/Tn + Log[Tn/100];
 sumpt = sumpt + PT,
 {nn}]
sumpt = sumpt/nn
col = Table[1 + (j - 1) 2.1/9, {j, 1, 10}]
ListLinePlot[
 Transpose[{vn/Tn, PT}],
 PlotStyle -> Thick,
 ColorFunction -> Function[{x, y}, Hue[col[[x]]]],
 ColorFunctionScaling -> False
 ]

There seems to be a workaround. Is it this what you want?

data = Table[{Thread[n[s]]}, {s, 20, 100, 1}] /. {a -> 1, 
     Tn -> 0.0005, g -> 10.75, f -> 5.6*10^-9} // Flatten[#, 3] &;
data11 = Table[{Thread[A1[s]]}, {s, 20, 100, 1}] /. {a -> 1, 
     Tn -> 0.0005, g -> 10.75, f -> 5.6*10^-9} // Flatten[#, 3] &;
data = {-3.44558, -3.41357, -3.37737, -3.33768, -3.29502, -3.24983, \
-3.20244, -3.15314, -3.10215, -3.04968, -2.99589, -2.94094, -2.88496, \
-2.82808, -2.7704, -2.71203, -2.65306, -2.59358, -2.53367, -2.47341, \
-2.41287, -2.35212, -2.29122, -2.23024, -2.16923, -2.10824, -2.04733, \
-1.98654, -1.92592, -1.86551, -1.80535, -1.74547, -1.68592, -1.62673, \
-1.56792, -1.50952, -1.45155, -1.39405, -1.33704, -1.28053, -1.22454, \
-1.16909, -1.11419, -1.05987, -1.00612, -0.952961, -0.900404, \
-0.848456, -0.797122, -0.746411, -0.696325, -0.64687, -0.598048, \
-0.549861, -0.502311, -0.455397, -0.409121, -0.363481, -0.318476, \
-0.274104, -0.230361, -0.187246, -0.144754, -0.102882, -0.0616244, \
-0.0209778, 0.0190633, 0.0585042, 0.0973505, 0.135608, 0.173283, 
   0.210381, 0.246909, 0.282874, 0.318282, 0.35314, 0.387454, 
   0.421233, 0.454483, 0.487211, 0.519424};

data11 = {1.01252*10^-9, 1.13838*10^-9, 1.27117*10^-9, 1.41033*10^-9, 
   1.55528*10^-9, 1.70539*10^-9, 1.85999*10^-9, 2.01842*10^-9, 
   2.17996*10^-9, 2.34391*10^-9, 2.50956*10^-9, 2.6762*10^-9, 
   2.84314*10^-9, 3.00971*10^-9, 3.17523*10^-9, 3.33909*10^-9, 
   3.50068*10^-9, 3.65943*10^-9, 3.81481*10^-9, 3.96633*10^-9, 
   4.11354*10^-9, 4.25601*10^-9, 4.39339*10^-9, 4.52533*10^-9, 
   4.65156*10^-9, 4.77183*10^-9, 4.88592*10^-9, 4.99367*10^-9, 
   5.09495*10^-9, 5.18966*10^-9, 5.27775*10^-9, 5.35918*10^-9, 
   5.43394*10^-9, 5.50208*10^-9, 5.56364*10^-9, 5.61871*10^-9, 
   5.66737*10^-9, 5.70975*10^-9, 5.74598*10^-9, 5.77621*10^-9, 
   5.8006*10^-9, 5.81934*10^-9, 5.83259*10^-9, 5.84056*10^-9, 
   5.84344*10^-9, 5.84144*10^-9, 5.83476*10^-9, 5.82362*10^-9, 
   5.80821*10^-9, 5.78876*10^-9, 5.76547*10^-9, 5.73855*10^-9, 
   5.70821*10^-9, 5.67464*10^-9, 5.63805*10^-9, 5.59864*10^-9, 
   5.55659*10^-9, 5.51208*10^-9, 5.4653*10^-9, 5.41643*10^-9, 
   5.36562*10^-9, 5.31305*10^-9, 5.25886*10^-9, 5.20321*10^-9, 
   5.14625*10^-9, 5.0881*10^-9, 5.02891*10^-9, 4.96879*10^-9, 
   4.90787*10^-9, 4.84627*10^-9, 4.78408*10^-9, 4.72141*10^-9, 
   4.65836*10^-9, 4.59502*10^-9, 4.53147*10^-9, 4.4678*10^-9, 
   4.40409*10^-9, 4.34039*10^-9, 4.27679*10^-9, 4.21333*10^-9, 
   4.15009*10^-9};

col = (# - 20)/80 & /@ Table[20 + (j - 1) 80/99, {j, 1, 100}];
dat3 = Transpose[{data, data11}];
rest = {Hue[col[[Position[dat3, #][[1, 1]]]]], Point[#]} & /@ dat3;
rest = rest /. Point[{x_, y_}] -> Point[{x, Log[y]}];


ListLogPlot[{dat3[[1]], dat3[[-1]]}, PlotStyle -> PointSize[.02], 
 PlotRange -> {{-1.5, 1.0}, {10^-12, 10^-8}}, Frame -> True,
 Epilog -> {PointSize[.02], rest}]

Hi John,

your code gives error messages because of synthax.mistakes. I had a close look at it and concluded that you want to calculate PT and do it like this

v = 246
E1 = 9000

Tn = {300, 500, 469, 650, 546, 389, 456, 411, 523, 700}
vn = Range[10] Tn
E1T = E1 vn RandomReal[{1, 3.1}, 10]/ v
PT = E1T/Tn - 7 Log[vn]/Tn + Log[Tn/100]

Is this right?

But a Plot doesn't look like yours

ListLinePlot[Transpose[{vn/Tn, PT}]]

Any comments?

Hello Rohit, that looks great,

but on my system that doesn't work. I don't know why, there is a Part - Error. Ok, I am using a hand-axe material out of the stone - age.

Part::partw: Part 1 of {} does not exist. >>
Part::pspec: Part specification {}[[1,1]] is neither an integer nor a list of integers. >>

Seems my workaround is not necesaary

Thanks Hans, I think now it is perfect

The function takes arguments x and y but they are not used in the body of the function. What is s?

In the previous example the x values were integers Range[10] so col[[x]] is a valid expression. In this example the x values are reals so col[[x]] is invalid. Rewrite col as an interpolating function that maps the x values to the corresponding value to use for Hue, or perhaps use Rescale.

And still another method (be sure to calculate sumpt in advance)

ShowLegend[
 ListLinePlot[Transpose[{vn/Tn, sumpt}], PlotStyle -> Thick, 
  ColorFunction -> Function[{x, y}, Hue[.7 col[[x]]]], 
  ColorFunctionScaling -> False, ImageSize -> Large, 
  PlotRange -> {{0, 12}, {0, ub}}, AxesOrigin -> {0, 0}],
 {Hue[.7 (1 - #)] &, 8, "   \!\(\*
StyleBox[\"3.1\",\nFontSize->14]\)", "   \!\(\*
StyleBox[\"1.0\",\nFontSize->14]\)", LegendShadow -> None, 
  LegendSize -> {.18, .8}, LegendPosition -> {.7, -.5}
  }]

Is there a reason why you have not upgraded to v12?

Well, for me it is simply too expensive. I am retired. And even "if"", I don't like the idea to have to have a download instead of a CD, and that only for one only machine. I use Mathematica on my desktop, and on my laptop, but at different times. So essentially once.

And after all there are lots of functionalities in v7 about which I have no knowledge. I get the impression that the higher versions of Mathematica contain functionalities which are not "really" necessary, and on the other hand , those which are really cool.

And I admire your knowledge of Mathematica and your problem-solving-ability.

Hi Rohit: Big Smile! Cool!!

But why does the Legend goes only up to 3.0? (As you may remember I use(still) Mma v7)

There in the posted image there is a color axis corresponding to the ratio, It is important since it makes the graphics more detailed, the color axis shown in above figure shows how the colors corresponding to each points related to the values of ((ET*v)/(vn*E1))

Secondly, I only want simply the values of PT (for each random number i.e. one point for each set of data, don't need any avg like thing) plotted against vn/Tn

What do you mean by " the correct color axis " ? The colors are coded accroding to the random number used in the calculation fpr the point in question , your ((ETv)/(vnE1)) .

Thanks for giving your valueable time, I think now it is exactly what I needed, but how to assign the correct color axis, and also guide me how to draw the color line in case of log color data(i.e. a log color axis)

OK. But you are merely repeating your words. For me it is not all all clear what you want to do. And: I suggest that you eliminate all these subscript-expressions. They may be a reason for a lot of mistakes

?????????????????????????

What do you want to do?

Perhaps you can give a handmade example?

I still don't understand the question? What are you asking for? Kind of lost here.

Like this? I think you will have to adapt the Hue-expression

Plot[
 Cos[x] Exp[- .1 x^2], {x, -5, 5},
 PlotStyle -> Thick,
 ColorFunction -> Function[{x, y}, Hue[ y]]
 ]