smc-7138-50-PDF

The probability density function for the normal distribution with mean \(\mu\) and standard deviation \(\sigma\) is

\(f(x)=\dfrac{1}{\sigma \sqrt{2 \pi}} e^{-\tfrac{(x-\mu)^2} {2 \sigma^2}}\)

The graph of \(y=e^{-\tfrac{1}{2}(x-1.5)^2}\) is shown. The point \(M\) is a local maximum.

Using a \(z\)-score table of values, calculate the area of the shaded region. Give your answer correct to three decimal places. (4 marks)

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\(\text{Comparing the graph to the normal distribution PDF:}\)

\(\mu=1.5, \ \sigma=1\)

\(e^{-\tfrac{1}{2}(x-1.5)^2} = \sqrt{2 \pi} \times f(x)\)

\(\Rightarrow\ \text{Total area under the curve} = \sqrt{2 \pi}\ \text{u}^2\)

\(\text{Convert the \(x\)-values to \(z\)-scores:}\)

\(\text{When }\ x=0:\ \ z=\dfrac{0-1.5}{1}=-1.5 \)

\(\text{When }\ x=1.5:\ \ z=\dfrac{1.5-1.5}{1}=0 \)

\(P(Z \leqslant 0)=0.5000\)

\(P(Z \leqslant 1.5)=0.9332\ \ \text{(from table)}\)

\(P(Z \leqslant -1.5)=1-0.9332=0.0668\ \ \text{(by symmetry)}\)

\(P(-1.5 \leqslant Z \leqslant 0)=0.5000-0.0668=0.4332\)

\(\text{Area under curve} = 0.4332 \times \sqrt{2\pi} \approx 1.08587\)

\(\text {Shaded area}\)	\(=\ \text{Area of rectangle}-\text{Area under curve}\)
	\(=(1.5 \times 1)-1.08587 \ldots\)
	\(=0.414 \ \text{u}^2 \ \text{(3 d.p.)}\)

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