Evaluate Integral in normal distribution with probability density function. Practice Speed of Vehicles on the highway using standard normal distribution formula & cumulative distribution

- Many random phenomenon such as test scores, heights of individuals, annual rainfall at a location are modeled by Normal Distribution.
- In Normal Distribution the probability density function \(f\) for the random variable X is given by \(f(x)=\dfrac {1}{\sigma\sqrt {2\pi}}\,e\,^{-(x-\mu)^2/(2\sigma^2)}\)
- The mean of this density function is \(\mu\)
- \(\sigma=\) standard deviation of the X and it measures how spread out the values of X are.
- These are the possible graph of function \(f\).

- More the value of \(\sigma\)more the values of X are spread out.

A \(\mu=\) 15, \(\sigma=\)20

B \(\mu=\) 5, \(\sigma=\)2

C \(\mu=\) 20, \(\sigma=\)100

D \(\mu=\) 100, \(\sigma=\)20

A \(\int\limits_{70}^{75}\dfrac {1}{10}e\;^{-(x-100)^2/2}\; dx\)

B \(\int\limits_{5}^{10} e\;^{-x^2}\; dx\)

C \(\int\limits_{70}^{75}\dfrac {1}{7.02}e\;^{-(x-63)^2/15.68}\; dx\)

D \(\int\limits_{1}^{2}e\;^-x^2\; dx\)

Let \(X\) define a continuous random variable then we define cumulative distribution function of this random variable as

\(P(X\leq x)=\displaystyle\int \limits^x_{-\infty}f(x)\,dx=F(x)\)

It is basically the probability that the random variable \(X\) will take value less than \(x\).

A \(F(x)=0\;\forall x\)

B \(F(x)= \begin{cases} 0&if&x < 0\\ \dfrac{x^3}{27}&if&0\leq x<3\\ 1 &if&x \geq 3 \end{cases}\)

C \(F(x)=\dfrac{x^3}{27}\,\forall x\)

D \(F(x)=1\,\forall x\)

A \(\dfrac{5}{6}\)

B \(\dfrac{3}{4}\)

C \(\dfrac{1}{2}\)

D \(\dfrac{5}{7}\)

It is observed that speed of vehicles which run on a highway are normally distributed and if the maximum speed limits is known, the probability that a randomly chosen vehicle is traveling at a legal speed can be found.

Consider the probability density function for random variable

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

Suppose we want to know what is the probability that random variable X lie within. one standard deviation of the mean i.e.

\(P(\mu-\sigma\leq X \leq \mu+\sigma)\)

\(=\int\limits_{\mu-\sigma}^{\mu+\sigma}\;\dfrac {1}{\sigma\sqrt {2\pi}}e^{-(x-\mu)^2/2\sigma^2}\)

\(=\int\limits_{\mu-\sigma}^{\mu+\sigma}\; \dfrac {1}{\sigma\sqrt {2\pi}} \;e^{- \left (\dfrac {x-\mu}{\sigma}\right)^2 ×\dfrac {1}{2}}\;dx\) ...(i)

Put \(\dfrac {x-\mu}{\sigma}=t\Rightarrow dx=\sigma\; dt\)

\(\therefore\) Equation (i) becomes \(\int\limits_{-1}^{1}\dfrac {1}{\sigma \sqrt {2\pi}} \;e^{-t^2/2}\;\sigma\;dt \)

\(\dfrac {1}{ \sqrt {2\pi}}\int\limits_{-1}^{1} \;e^{-t^2/2}\;\;dt \) which is independent of the standard deviation \(\sigma\).

- We can similarly find the probability that random variable X lies with n standard deviation about the mean as

\(P(\mu-n\sigma\leq X\leq \mu+n\sigma)\)\(=\dfrac {1}{ \sqrt {2\pi}}\int\limits_{-n}^{n} \;e^{-t^2/2}\;\;dt \)