有没有办法使用pstricks \lput*{:U}{68.3\%透明背景？
这是我的 MWE：
\documentclass[pstricks]{standalone}
\usepackage{pst-func,pst-node,amsmath,xfp}
\begin{document}
\definecolor{fillColor}{RGB}{125, 185, 250}
\psset{yunit=10cm,xunit=1.5}
\begin{pspicture}(-4.7,-1)(4.5,0.7)
\pscustom[fillstyle=solid,fillcolor=fillColor!25,linestyle=none]{%
\psline(-3,0)\psGauss[mue=0,sigma=1]{-3}{3}\psline(3,0)}
\pscustom[fillstyle=solid,fillcolor=fillColor!50,linestyle=none]{%
\psline(-2,0)\psGauss[mue=0,sigma=1]{-2}{2}\psline(2,0)}
\pscustom[fillstyle=solid,fillcolor=fillColor!75,linestyle=none]{%
\psline(-1,0)\psGauss[mue=0,sigma=1]{-1}{1}\psline(1,0)}
\psaxes[xAxis=false, yAxis=true,Dy=0.1,showorigin=true,tickstyle=bottom]{->}(-3.7,0)(-3.5,0)(3.5,0.6)[$$,0][$f(x)$,-140]
\psaxes[yAxis=false,Dy=0.25,labels=none,ticks=none]{<->}(0,0)(-3.5,-1)(3.5,0.6)
\psGauss[linecolor=fillColor, linewidth=1.5pt,mue=0,sigma=1]{-3.2}{3.2}%
\psline[linestyle=dashed, linecolor=gray!50](-3,0)(-3,0.1)
\psline[linestyle=dashed, linecolor=gray!50](-2,0)(-2,0.2)
\psline[linestyle=dashed, linecolor=gray!50](-1,0)(-1,0.3)
\psline[linestyle=dashed, linecolor=gray!50](0,0)(-0,0.45)
\psline[linestyle=dashed, linecolor=gray!50](1,0)(1,0.3)
\psline[linestyle=dashed, linecolor=gray!50](2,0)(2,0.2)
\psline[linestyle=dashed, linecolor=gray!50](3,0)(3,0.1)
\multido{\iA=-3+1,\iB=-3+1}{3}{\psxTick(\iA){\mu \iB $\sigma$}}
\psxTick(-0){$\mu$}
\psxTick(1){$\mu+1\sigma$}
\psxTick(2){$\mu+2\sigma$}
\psxTick(3){$\mu+3\sigma$}
\rput[rt](4,-0.04){$x$}
\rput[rt](4,-0.125){$z$}
\psaxes[yAxis=false,Dy=0.25, linecolor=white,ticksize=8pt]{}(0,-0.1)(-3.5,0)(3.5,0.6)
\pcline[fillcolor=fillColor]{<->}(-1,0.24)(1,0.24){\lput*{:U}{68.3\%}}
\end{pspicture}
\end{document}
我想要获取 68.3% 文本框周围的背景颜色。
谢谢！
答案1
\pcline{<->}(-1,0.24)(1,0.24)\ncput{\colorbox{fillColor!75}{68.3\%}}
