Q 54 - Calculus

Question

The function for the standard normal distribution is

                                         $f\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2\beta }$

Its graph is that of the bell curve. Probability and statistics

books often have tables like the one following, which

lists some approximate areas under the bell curve:

                Areas under the bell curve

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

Use the information given in the table, properties of definite

integrals, and symmetry to find

(a) $\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{15}{e}^{-x^2/2}dx$ (b) $\frac{1}{\sqrt{2\pi }}{\int }_{1.5}^2{e}^{-x^2/2}dx$

Step-by-Step Solution

Answer

(a) $\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=0.62465$

(b)$\frac{1}{\sqrt{2\pi }}{\int }_{1.5}^2{e}^{-\frac{x^2}{2}}dx=0.044057$

Part a: Step 1: Given Information

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

Area under the bell curve is given by

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

Step 2: Properties and Calculation

Properties Used:

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

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

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

Calculation:

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

$\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2\pi }}({\int }_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+{\int }_{0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx$

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

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

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

$=\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2\pi }}{\int }_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx-\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx$

$\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{0.5}{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2\pi }}{\int }_{-1.5}^{1.5}{e}^{-\frac{x^2}{2}}dx$

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

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

$\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=\frac{1.2493}{2}$

$\frac{1}{\sqrt{2\pi }}{\int }_{-0.5}^{1.5}{e}^{-\frac{x^2}{2}}dx=0.62465$

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

Part b: Step 1:

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

Area under the bell curve is given by

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

Step 2: Properties and Calculations

Properties:

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

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

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

Calculations:

$\frac{1}{\sqrt{2\pi }}{\int }_{1.5}^2{e}^{-x^2/2}dx$ (Given)

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

$\frac{1}{\sqrt{2\pi }}{\int }_{1.5}^2{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2\pi }}{\int }_{1.5}^2{e}^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2\pi }}{\int }^2{e}^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2\pi }}{\int }_{-1.5}^2{e}^{-\frac{x^2}{2}}dx$

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

$2.\frac{1}{\sqrt{2\pi }}{\int }_{1.5}^2{e}^{-\frac{x^2}{2}}dx=0.9545-0.8664$

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

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