Answer to Question #120125 in Statistics and Probability for Popcy

Question

QUESTION 2
(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
b) The average time taken for the nearest Covid-19 ambulance team to convey a patient showing
symptoms to the nearest designated hospital follows the normal distribution about a mean of 60 seconds and a standard deviation of 8 seconds. If 200 patients were selected at random, how many of them could be taken to the hospital in
i) Less than 50 seconds (3 Marks)
ii) More than 64 seconds (3 Marks)
c) The Shelf life of a particular dairy product is normally distributed with a mean of 12 days and a variance of 9 days. About what percentage of the products last between 13 and 16 days

Answer 1

a) Let $X=$ the height of the traveller: $X\sim N(\mu, \sigma^2).$ Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)$

Given $\mu=160\ cm,\sigma=8\ cm$

(i)

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

$$=P(Z<{175-160\over 8})-P(Z\leq{148-160\over 8})=$$

$$=P(Z<1.875)-P(Z\leq-1.5)\approx$$

$$\approx0.9696-0.0668=0.9028$$

(ii)

$$P(X>164)=1-P(X\leq 164)=1-P(Z\leq{164-160\over 8})=$$

$$=1-P(Z\leq0.5)\approx1-0.6915=0.3085$$

(iii)

$$P(X<179)=P(Z<{179-160\over 8})=P(Z<2.375)\approx$$

$$\approx0.9912$$

b) Let $X=$ the time taken for the nearest Covid-19 ambulance team to convey a patient:

$X\sim N(\mu, \sigma^2).$ Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1).$

Given $\mu=60\ s, \sigma=8\ s.$

(i)

$$P(X<50)=P(Z<{50-60\over 8})=P(Z<-1.25)\approx$$ $$\approx0.1056$$

$200\cdot0.1056\approx21$

(ii)

$$P(X>64)=1-P(Z<{64-60\over 8})=$$ $$=1-P(Z<0.5)\approx0.3085$$

$0.3085\cdot 200=62$

c) Let $X=$ the Shelf life of a particular dairy product: $X\sim N(\mu, \sigma^2).$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1).$

Given $\mu=12, \sigma^2=9.$

$$P(13<X<16)=P(X<16)-P(X\leq13)=$$

$$=P(Z<{16-12\over 3})-P(Z\leq{13-12\over 3})\approx$$

$$\approx P(Z<1.3333)-P(Z\leq0.3333)\approx$$

$$\approx0.9088-0.6306=0.2782$$

About $27.82\ \%$ 27.82 of the products last between 13 and 16 days.