Answer to Question #118628 in Statistics and Probability for Michael

Question

For a mean of 20 and standard deviation of 5, find P(X ≤ 17 )

Answer 1

Typically, a random variable has a discrete or continuous distribution. We shall consider these cases:

1. Suppose that $X$ has a discrete distribution. This means that $X$ takes real values $x_1\,,x_2,\,x_3,\ldots$ (countable set) with probabilities $p_1,\,p_2,\,p_3,\ldots;$ $\sum_{k=1}^{\infty}p_k=1$ . We have the following formulae for the mean and the standard deviation: $\sum_{k=1}^{\infty}x_kp_k=20$and $\sum_{k=1}^{\infty}p_k(x_k-20)^2=25$ .

$P(X\leq17)=\sum_{k\in K}p_k,$ where $\{k\in {\mathbb{N}}|x_k\leq17\}$ .

2. Assume that $X$ has a continuous distribution. Then, $X$  has a respective probability

density function $p(x)$ satisfying $\int_{-\infty}^{+\infty}p(x)dx=1$ . We have the following formulae for

the mean and the standard deviation:$\int_{-\infty}^{+\infty}xp(x)dx=20$ and

$\int_{-\infty}^{+\infty}p(x)(x-20)^2dx=25.$

$P(X\leq17)=\int_{-\infty}^{17}p(x)dx.$ In particular, in case $X$ has a normal distribution with

 $\mu=20,\,\sigma=5$ we have $P(X\leq17)=\frac{1}{5\sqrt{2\pi}}\int_{-\infty}^{17}e^{-\frac12(\frac{x-20}{5})^2}dx\approx0.274$.

The latter integral was computed with the help of Anaconda (free distribution of

the Python). The following code was used in Jupyter Notebook (a part of the distribution

for writing and running code) for calculation of the integral:

_{from scipy import integrate}

_{import numpy as np}

_import math

_{func = lambda x:(1/(5*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-20)/5)*}

_((x-20)/5))

_{Pr = integrate.quad(func, 0, 17)}

_print(Pr)