Areas Under the Normal Curve

Example 6.2: Given a standard normal distribution, find the area under the curve that lies

xxi
to the right of $ z = 1.84$

1 minus the area to the left of $ z = 1.84$ (see Table A.3)

$\displaystyle 1 - 0.9671 = 0.0329
$

xxii
between $ z = -1.97$ and $ z = 0.86$

The area to the left of $ z = 0.86$ minus the left of $ z = -1.97$

$\displaystyle 0.8051 - 0.0244 = 0.7807
$

Figure 6.9: Areas for Example 6.2.
\includegraphics[scale=0.6]{figures/06-09}

\includegraphics[scale=1]{figures/06-10}

Example 6.3: Given a standard normal distribution, find the value of k such that

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$ P(Z > k) = 0.3015$

$\displaystyle P(Z < k) = 1 - P(Z > k) = 1 - 0.3015 = 0.6985 \Rightarrow k = 0.52
$

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$ P(k < Z < -0.18) = 0.4197$

$\displaystyle P(Z < -0.18) - P(Z < k) = 0.4286 - P(Z < k) = 0.4197 \Rightarrow k = -2.37
$

Figure 6.10: Areas for Example 6.3.
\includegraphics[scale=1]{figures/06-11}

Example 6.6: Given a normal distribution with $ \mu=40$ and $ \sigma =6$, find the value of $ x$ that has

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45% of the area to the left

From Table A.3 we find $ P(Z<-0.13) =0.45$. Hence

$\displaystyle x = 6*(-0.13)+40 = 39.22.
$

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14% of the are to the right

From Table A.3, we find $ P (Z<1.08) =086$. Hence

$\displaystyle x = 6*(1.08)+40 = 46.48.
$

Figure 6.13: Areas for Example 6.6.
\includegraphics[scale=0.75]{figures/06-14}

Cem Ozdogan 2012-02-15