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#### karush

##### Well-known member

- Jan 31, 2012

- 3,119

(i) Represent this information on a standard normal curve diagram, indicating clearly the area representing 90\%

(ii) Find the value of \textbf{t}. $P(Z\le t) =0.9\quad Z = 1.282\quad t=57+(4.4(1.282))=62.64$ hours

\begin{tikzpicture}[scale=0.6]

%preamble \usepackage{pgfplots}

\newcommand\gauss[2]{1/(#2*sqrt(2*pi))*exp(-((x-#1)^2)/(2*#2^2))} % Gauss function, parameters mu and sigma

\begin{axis}[every axis plot post/.append style={

mark=none,samples=50,smooth}, % All plots: 50 samples, smooth, no marks

axis x line*=bottom, % no box around the plot, only x axis

axis y line=none, % the * suppresses the arrow tips

enlargelimits=upper, % extend the axes a bit to the right and top

domain=-2:3, % Default for all plots: from -4:4

xtick={-.455,.9},

xticklabels={$-.455$,$.9$},

width=10cm,

height=4cm]

\addplot [fill=gray!30, draw=none, domain=-0.45:0.655] {\gauss{-0.455}{0.9}} \closedcycle;

\addplot {\gauss{-0.455}{0.9}};

\end{axis}

\end{tikzpicture}

ok I could't get this tikz to match the information