We remind that for the normally distributed random variable "X" with a mean of 20 and a standard deviation of 4 one has (see e.g., https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm): "P(a\\leq X\\leq b)=\\frac{1}{4\\sqrt{2\\pi}}\\int_a^be^{-\\frac12(\\frac{x-20}{4})^2}dx" . For computations we used the following program in Anaconda (https://www.anaconda.com/):
from scipy import integrate
import numpy as np
import math
func = lambda x:(1/(4*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-20)/4)*((x-20)/4))
e = integrate.quad(func, a,b)
Instead of a and b we inserted values that we are interested in.
(a) The aim is to find "\\alpha" : "P(\\alpha<X)=0.25". We substitute different values of "\\alpha" in:
"\\frac{1}{4\\sqrt{2\\pi}}\\int_{\\alpha}^{+\\infty}e^{-\\frac12(\\frac{x-20}{4})^2}dx" and get: "\\alpha\\approx22.70" .
(b) We look for "\\alpha" such that: "P(\\alpha<X)=0.85". By substituting different choices of "\\alpha" we get: "\\alpha\\approx15.85" .
(c) The aim is to find such "\\alpha" that: "P(\\alpha<X)=0.9" . We get: "\\alpha\\approx14.87" . Now we look for "\\beta" satisfying "P(X<\\beta)=0.9". We find that "\\beta\\approx25.13".
(d) The aim is to find "\\alpha" : "P(X<\\alpha)=0.15" . We get: "\\alpha\\approx15.85".
Comments
Leave a comment