Question

If a random variable X follows Normal distribution such that P (9,6<X<13.8)=0.07008 and P(X>9.6)= 0.8159. Find the mean & variance of X and find P(X> 13.8/X>9.6).

Answer 1

We remind that the probability density function for the normal distribution has the form: $p(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12(\frac{x-\mu}{\sigma})^2}$ . We have: $\int_{9.6}^{13.8}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12(\frac{x-\mu}{\sigma})^2}dx=0.07008$ and $\int_{9.6}^{+\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12(\frac{x-\mu}{\sigma})^2}dx=0.8159$ . We make the change: $z=\frac{x}{\sigma}-\frac{\mu}{\sigma}=\frac{x}{\sigma}-\alpha$ and receive: $\int_{\frac{9.6}{\sigma}-\alpha}^{\frac{13.8}{\sigma}-\alpha}\frac{1}{\sqrt{2\pi}}e^{-\frac12z^2}dz=0.07008$ and $\int_{\frac{9.6}{\sigma}-\alpha}^{+\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12z^2}dz=0.8159$ with $\alpha=\frac{\sigma}{\mu}$ .

From the last integral we receive that: $\frac{9.6}{\sigma}-\alpha=-0.9$. We substitute that in the first one and get: $-0.663=\frac{13.8}{\sigma}-\alpha$ . We receive: $\frac{4,2}{\sigma}=0,237$. Thus, we get: $\sigma=17,722$ and $\mu=1.442.$ $P(X>13.8)=\int_{13.8}^{+\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12(\frac{x-\mu}{\sigma})^2}dx\approx0.2428$ and

$P(X>9.6)=\int_{9.6}^{+\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12(\frac{x-\mu}{\sigma})^2}dx\approx0.3226$.

$P(X>13.8/X>9.6)=\frac{P((X>13.8)\, \cap\, (X>9.6))}{P(X>9.6)}=\frac{P(X>13.8)}{P(X>9.6)}=\frac{0.2428}{0.3226}\approx0.7526$

We used the following code in Anaconda for computations:

_{from scipy import integrate}

_{import numpy as np}

_import math

_{func = lambda x:(1/(17.722*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-1.442)/17.722)*((x-1.442)/17.722))}

_{e = integrate.quad(func, 9.6,+np.inf)}

_print(e)

We subsituted different values of $\mu$ and $\sigma$ in it.