Answer to Question #120023 in Statistics and Probability for Arafeen

Question

(2a). According to the Ministry of Health, the height of Guinean travellers who were quarantined in Tamale
for the novel coronavirus were normally distributed about a mean of 160cm and a standard deviation
of 8cm. Find the probability that a traveller selected at random has
i) Height between 148cm and 175cm
ii) Height above 164cm
iii) Height below 179cm

Answer 1

We denote by $X$ a random variable, which is a height of a randomly selected traveler. It has a normal distribution with $\mu=160$ cm, $\sigma=8$ cm. We remind that its probability density function has the form $p(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12(\frac{x-\mu}{\sigma})^2}$ . We obtain the following probabilities (rounded to 4 decimal places):

i) $P(148\leq X\leq175)=\frac{1}{8\sqrt{2\pi}}\int_{148}^{175}e^{-\frac12(\frac{x-160}{8})^2}dx\approx0.9028$ ;

ii) $P(X\geq164)=\frac{1}{8\sqrt{2\pi}}\int_{164}^{+\infty}e^{-\frac12(\frac{x-160}{8})^2}dx\approx0.3085;$

iii) $P(X\leq179)=\frac{1}{8\sqrt{2\pi}}\int_{-\infty}^{179}e^{-\frac12(\frac{x-160}{8})^2}dx\approx0.9912;$

Anaconda (https://www.anaconda.com/) was used for computations. The following code produces numeric values:

_{from scipy import integrate}

_{import numpy as np}

_import math

_{func = lambda x:(1/(8*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-160)/8)*((x-160)/8))}

_{Pr1 = integrate.quad(func, 148, 175)}

_{Pr2 = integrate.quad(func, 164, np.Infinity)}

_{Pr3 = integrate.quad(func, -np.Infinity, 179)}

_print(Pr1)

_print(Pr2)

_print(Pr3)

Answer: 0.9028; 0.3085; 0.9912 (rounded to 4 decimal places)