Answer to Question #120219 in Statistics and Probability for Arhin Gabriel

Question

(a) 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

Let X = the random variable denoting the height of a Guinean traveller who was quarantined in Tamale for the novel corona virus

Then by the problem,

X ~ N($\mu$ = 160, $\sigma$² = 8²)

Therefore,

Z = $\frac{X-\mu}{\sigma}$ ~ N(0, 1), Z is the standard normal variate

(i) The probability that a traveler selected at random has height between 148 cm and 175 cm

= P(148 $\leq$ X $\leq$ 175)

= P($\frac{148-160}{8}\leq\frac{X-160}{8}\leq\frac{175-160}{8}$)

= P(-1.5 $\leq$ Z $\leq$ 1.88)

= $\Phi$(1.88) - $\Phi$(-1.5)

= 0.9699 - 0.0668

= 0.9031

Answer: The probability that a traveler selected at random has height between 148 cm and 175 cm is 0.9031.

(ii) The probability that a traveler selected at random has height above 164 cm

P(X $>$ 164)

= P($\frac{X-160}{8}>\frac{164-160}{8}$)

= P(Z $>$ 0.5)

= 1 - P(Z $\leq$ 0.5)

= 1 - $\Phi$(0.5)

= 1 - 0.6915

= 0.3085

Answer: The probability that a traveler selected at random has height above 164 cm 0.3085.

(iii) The probability that a traveler selected at random has height below 179 cm

P(X < 179)

= P($\frac{X-160}{8}<\frac{179-160}{8}$)

= P(Z < 2.38)

= $\Phi$(2.38)

= 0.9913

Answer: The probability that a traveler selected at random has height below 179 cm is 0.9913.