91影视

The function for the standard normal distribution is f(x) =$\frac{1}{\sqrt{2蟺}}{e}^{-x^2/2}dx$

Area under the bell curve is given by

Properties Used:

${鈭�}_a^{b}f\left(x\right)dx={鈭�}_a^{c}f\left(x\right)dx+{鈭�}_c^{b}f\left(x\right)dx$a < c < b

${鈭�}_a^{b}f\left(x\right)dx=-{鈭�}_b^{a}f\left(x\right)dx$

$鈭�f\left(x\right)dx=-{鈭�}_a^{-b}f\left(x\right)dx$

Calculation:

$\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{15}{e}^{-x^2/2}dx$(Given)

$\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2蟺}}({鈭�}_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+{鈭�}_{0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx$

$=\frac{1}{\sqrt{2蟺}}\left[{鈭�}_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+\left({鈭�}_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx-{鈭�}_{-1.5}^{0.5}{e}^{-\frac{x^2}{2}}dx\right)\right]$

$=\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2蟺}}\left[{鈭�}_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+\left({鈭�}_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx-{鈭�}_{0.5}^{-1.5}{e}^{-\frac{x^2}{2}}dx\right)\right]\text{(By using the property)}$

$=\frac{1}{\sqrt{2蟺}}\left[{鈭�}_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+\left({鈭�}_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx-{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx\right)\right]\text{(By using the property)}$

$=\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2蟺}}{鈭�}_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx-\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx$

$\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2蟺}}{鈭�}_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx$

We can achieve the following result by using the table values on the right side.

$2.\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=0.3829+0.8664$

$\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1.2493}{2}$

$\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=0.62465$

$\text{Hence,}\frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=0.62465$

The function for the standard normal distribution is f(x) =$\frac{1}{\sqrt{2蟺}}{e}^{-x^2/2}dx$

Area under the bell curve is given by

$\begin{array}{lc}b& \frac{1}{\sqrt{2蟺}}{鈭�}_{-b}^b{e}^{-x^2/2}dx\\ 0.5& 0.3829\\ 1& 0.6827\\ 1.5& 0.8664\\ 2& 0.9545\\ 2.5& 0.9876\end{array}$

Properties:

${鈭�}_a^{b}f\left(x\right)dx={鈭�}_a^{c}f\left(x\right)dx+{鈭�}_c^{b}f\left(x\right)dx$a < c < b

${鈭�}_a^{b}f\left(x\right)dx=-{鈭�}_b^{a}f\left(x\right)dx$

$鈭�f\left(x\right)dx=-{鈭�}_{-a}^{-b}f\left(x\right)dx$

Calculations:

$\frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-x^2/2}dx$(Given)

$$\begin{gathered} \frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2蟺}}\left({鈭�}_{-2}^2{e}^{-\frac{x^2}{2}}dx-{鈭�}_{-2}^{1.5}{e}^{-\frac{x^2}{2}}dx\right) \\ =\frac{1}{\sqrt{2蟺}}\left[{鈭�}_{-2}^2{e}^{-\frac{x^2}{2}}dx-\left({鈭�}_{-2}^{-1.5}{e}^{-\frac{x^2}{2}}dx+{鈭�}_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx\right)\right] \\ \frac{1}{\sqrt{2蟺}}{鈭�}_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2蟺}}\left[{鈭�}_{-2}^2{e}^{-\frac{x^2}{2}}dx-\left(-{鈭�}_{-1.5}^{-2}{e}^{\left.\left.-\frac{x^2}{2}dx+{鈭�}_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx\right)\right]\text{(By using the property)}}\right)\right. \end{gathered}$$

$\frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2蟺}}{鈭�}^2{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2蟺}}{鈭�}_{-1.5}^2{e}^{-\frac{x^2}{2}}dx$

We can achieve the following result by using the table values on the right side.

$2.\frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-\frac{x^2}{2}}dx=0.9545-0.8664$

$$\begin{gathered} \frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-\frac{x^2}{2}}dx=\frac{0.0881}{2} \\ \frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-\frac{x^2}{2}}dx=0.044057 \end{gathered}$$

$\text{Hence,}\frac{1}{\sqrt{2蟺}}{鈭�}_{1.5}^2{e}^{-\frac{x^2}{2}}dx=0.044057$