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 (a) Height between 148cm and 175cm (b) Height above 164cm (c) Height below 179cm

Answer 1

a)
`P(148<X<175)=P((148-\mu)/(\sigma)<(X-\mu)/(\sigma)<(175-\mu)/(\sigma))=P((148-160)/(8)<Z<(175-160)/(8))=P(-1.5<z<1.875)`

`P(-1.5<z<1.875)=P(z<1.875)-P(z<-1.5)= 0.970-0.0668= 0.9032`

b)
`P(X>164)=P((X-\mu)/(\sigma)>(164-\mu)/(\sigma))=P(Z>(164-160)/(8))=P(z>0.5)`

`P(z>0.5)=1-P(z<0.5)= 1-0.691= 0.309`

c)
`P(X<179)=P((X-\mu)/(\sigma)<(179-\mu)/(\sigma))=P(Z<(179-160)/(8))=P(z<2.375)`

`P(z<2.375)= 0.991`

Explanation:

Let X the random variable that represent the heights of Guinean travels , and for this case we know the distribution for X is given by:

`X \sim N(160,8)`

Where
`\mu=160` and
`\sigma=8`

Part a

We are interested on this probability

`P(148<X<175)`

We can ue the z score formula given by:

`z=(x-\mu)/(\sigma)`

Using this formula we got:

`P(148<X<175)=P((148-\mu)/(\sigma)<(X-\mu)/(\sigma)<(175-\mu)/(\sigma))=P((148-160)/(8)<Z<(175-160)/(8))=P(-1.5<z<1.875)`

And we can find this probability with this difference using the normal standard distribution or excel:

`P(-1.5<z<1.875)=P(z<1.875)-P(z<-1.5)= 0.970-0.0668= 0.9032`

Part b

`P(X>164)=P((X-\mu)/(\sigma)>(164-\mu)/(\sigma))=P(Z>(164-160)/(8))=P(z>0.5)`

And we can find this probability with this difference using the normal standard distribution or excel:

`P(z>0.5)=1-P(z<0.5)= 1-0.691= 0.309`

Part c

`P(X<179)=P((X-\mu)/(\sigma)<(179-\mu)/(\sigma))=P(Z<(179-160)/(8))=P(z<2.375)`

And we can find this probability with this difference using the normal standard distribution or excel:

`P(z<2.375)= 0.991`