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)