Areas Under the Normal Curve

• The probability of the random variable $X$ assuming a value between $x_1$ and $x_2$.
  $$P(x_1<X<x_2)=\int_{x_1}^{x_2}n(x;\mu,\sigma)dx=\frac{1}{\sqrt{2\pi \sigma}}\int_{x_1}^{x_2}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}dx$$

• The area under the curve between any two ordinates must also depend on the values $\mu$ and $\sigma$.

  Figure 6: $P(x_1 < X < x_2)$ : area of the shaded region.

  

  Figure 7: $P(x_1 < X < x_2)$ for different normal curves.

  

• Definition 6.1:
  
• Transformation: $Z=\frac{X-\mu}{\sigma},~z_1=\frac{x_1-\mu}{\sigma},~z_2=\frac{x_2-\mu}{\sigma}$
• Then; $E(Z)=\mu=0$ and $\sigma_Z^2=1$
  $$P(x_1< X < x_2)=\frac{1}{\sqrt{2\pi \sigma}}\int_{x_1}^{x_2}e^{-\... ...)^2}dx \Rightarrow \frac{1}{\sqrt{2\pi}}\int_{z_1}^{z_2}e^{\frac{1}{2}(z)^2}dz$$
  $$=\int_{z_1}^{z_2}n(z;0,1)=P(z_1< Z < z_2)$$

  Figure 8: The original and transformed normal distributions.

  

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

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to the right of $z = 1.84$

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

$$1 - 0.9671 = 0.0329$$

iv

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

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

$$0.8051 - 0.0244 = 0.7807$$

Figure 9: Areas for Example 6.2.

• Usage of the Table A.4;
  • The entries in the table are the areas under the standard normal curve between the mean, $z = 0$, and $z =X$.
  • The first column represents the values of $z$ from 0.0 to 3.4 by increment 0.1,
  • and the first row indicates the second digit under the decimal of the corresponding values of $z$ according to the column values.
  • Suppose we want to find the area between 0 and 1.23, then all we need to do is to read the entry where the row of 1.2 and the column of 0.03 come across.

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

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$P(Z > k) = 0.3015$

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

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

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

Figure 10: Areas for Example 6.3.

• Example 6.4: Given a random variable $X$ having a normal distribution with $\mu = 50$ and $\sigma = 10$,
• Find the probability that $X$ assumes a value between 45 and 62.
• Solution:
  $$x_1= 45~and~x=62 \xrightarrow{transformation} z_1=\frac{45-50}{10}=-0.5,~z_2=\frac{62-50}{10}=1.2$$
  $$P(45 < X < 62) = P(-0.5 < Z < 1.2)$$
  $$= P(Z< 1.2) - P(Z < -0.5 )= 0.8849 - 0.3085 = 0.5764$$

  Figure 11: Area for Example 6.4.

  

• Example 6.5 Given that a normal distribution with $\mu = 300$ and $\sigma = 50$, find the probability that $X$ assumes a value greater than 362.
• Solution:
  $$z=\frac{362-300}{50}1.24$$
  $$- P(X > 362) = P(Z > 1.24) = 1 - P(Z < 1.24 )$$
  $$= 1 - 0.8925 = 0.1075$$

  Figure 12: Area for Example 6.5.

  

• Using the Normal Curve in Reverse
• We might want to find the value of $z$ corresponding to a specified probability.
• The steps:
  1. Begin with a known area or probability.
  2. Find the $z$ values corresponding to the tabular probability that comes closest to the specified probability.
  3. Determine $x$ by rearranging the formula
     $$z=\frac{x-\mu}{\sigma}~to~give~x=\sigma z+ \mu$$

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

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

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14% of the are to the right

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

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

Figure 13: Areas for Example 6.6.