Answer to Question #124643 in Statistics and Probability for Umair

Question

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

Answer 1

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

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

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

and $P(x\geq x_3)$ for real $x_0,x_1,x_2,x_3$ by changing the boundaries in the integral. We obtained the following answers:

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

(ii) $P(151.5\leq x\leq 182)=\frac{1}{6\sqrt{2\pi}}\int_{151.5}^{182}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.1217$

(iii)$P(x\geq 150)=\frac{1}{6\sqrt{2\pi}}\int_{150}^{+\infty}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.1797$

(iv) The aim is to find $F^{-1}(0.6)$ , where $F(y)=P(x> y)=\frac{1}{6\sqrt{2\pi}}\int_y^{+\infty}e^{-\frac12(\frac{x-144.5}{6})^2}dx$

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

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 $F(y)$ decreases and substituted different values of $y$ 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.