Answer to Question #214582 in Statistics and Probability for Jceka

Question

QUESTION ONE

(a) According to the Ministry of Health, the height of travellers who were quarantined in Accra for the coronavirus were normally distributed about a mean of 170cm and a variance of 36cm. Find the probability that a traveller selected at random has i) Height between 160cm and 175cm ii) Height above 164cm. iii) Height below 180 cm

(b) The American Automobile Association reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the variable is approximately normally distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes?

Answer 1

Solution:

(a) $X\sim N(\mu,\sigma)$

$\mu=170,\sigma=\sqrt{36}=6$

(i): $P(160\le X\le175)=P(X\le 175)-P(X\le 160)$

$=P(z\le \dfrac{175-170}{6})-P(z\le \dfrac{160-170}{6}) \\=P(z\le0.83)-P(z\le-1.67) \\=P(z\le0.83)-[1-P(z\le1.67)] \\=0.79673-[1-0.95254] \\=0.74927$

(ii): $P(X>164)=1-P(X\le 164)=1-P(z\le\dfrac{164-170}{6})$

$=1-P(z\le -1)=1-[1-P(z\le 1)]=P(z\le 1) \\=0.84134$

(iii): $P(X<180)=P(z\le\dfrac{180-170}{6})=P(z\le 1.67)=0.95254$

(b) $X\sim N(\mu,\sigma)$

$\mu=25,\sigma=4.5$

$P(X<15)=P(z<\dfrac{15-25}{4.5})=P(z< -2.22) \\=1-P(z<2.22) \\=1-0.98679=0.01321$

Now, required number of calls$=0.01321\times80=1.0568\approx 1$