Question #124643
The heights of 500 students are normally distributed with a mean of 144.5 centimeters and a variance of 36 centimeters. Assuming that the heights are recorded to the nearest half-centimeter, find the Probability.
(a) Less than 130.0 centimeters?
(b) Between 151.5 and 182.0 centimeters inclusive?
(c) Greater than or equal to 150.0 centimeters?
(d) P(x > x0) = 0.6
1
Expert's answer
2020-06-30T18:42:16-0400

Since heights are normally distributed, the probability P(xx0)P(x\leq x_0) is given by (see https://en.wikipedia.org/wiki/Normal_distribution#Standard_normal_distribution):

P(xx0)=1σ2πx0e12(xμσ)2dxP(x\leq x_0)=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{x_0}e^{-\frac12(\frac{x-\mu}{\sigma})^2}dx

with μ=144.5\mu=144.5 , σ=6\sigma=6 . In a similar way, we can express the probabilities P(x1xx2)P(x_1\leq x\leq x_2)

and P(xx3)P(x\geq x_3) for real x0,x1,x2,x3x_0,x_1,x_2,x_3 by changing the boundaries in the integral. We obtained the following answers:

(i) P(x130)=162π130e12(x144.56)2dx0.0078P(x\leq 130)=\frac{1}{6\sqrt{2\pi}}\int_{-\infty}^{130}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.0078

(ii) P(151.5x182)=162π151.5182e12(x144.56)2dx0.1217P(151.5\leq x\leq 182)=\frac{1}{6\sqrt{2\pi}}\int_{151.5}^{182}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.1217

(iii)P(x150)=162π150+e12(x144.56)2dx0.1797P(x\geq 150)=\frac{1}{6\sqrt{2\pi}}\int_{150}^{+\infty}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.1797

(iv) The aim is to find F1(0.6)F^{-1}(0.6) , where F(y)=P(x>y)=162πy+e12(x144.56)2dxF(y)=P(x> y)=\frac{1}{6\sqrt{2\pi}}\int_y^{+\infty}e^{-\frac12(\frac{x-144.5}{6})^2}dx

For y143.0y\approx143.0 (it was rounded to 1 decimal place) we get that F(143.0)0.5987F(143.0)\approx0.5987

For the computations we used the following program written in Anaconda (https://www.anaconda.com/products/individual):


from scipy import integrate

import numpy as np

import math


func = lambda x:(1/(6*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-144.5)/6)*((x-144.5)/6))

Pr1 = integrate.quad(func, -np.Infinity, 130)

Pr2 = integrate.quad(func, 151.5, 182)

Pr3 = integrate.quad(func, 150, +np.Infinity)


print(Pr1)

print(Pr2)

print(Pr3)


For the last task we used the fact that F(y)F(y) decreases and substituted different values of yy until we got the values near 0.6.

Answers: (i) 0.0078 (ii) 0.1217 (iii) 0.1797 (iv) 143.0; the answers (i)-(iii) are rounded to 4 decimal places; (iv) is rounded to 1 decimal place.


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