Question

ssume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find the probability of obtaining a reading between -1.94°C and -1.5°C.

Answer 1

`P(-1.94<X<-1.5)=P((-1.94-\mu)/(\sigma)<(X-\mu)/(\sigma)<(-1.5-\mu)/(\sigma))=P((-1.94-0)/(1)<Z<(-1.5-0)/(1))=P(-1.94<z<-1.5)`

And we can find this probability with this difference:

`P(-1.94<z<-1.5)=P(z<-1.5)-P(z<-1.94)=0.0668-0.026= 0.0408`

Explanation:

Let X the random variable that represent the temperatures of a population, and for this case we know the distribution for X is given by:

`X \sim N(0,1)`

Where
`\mu=0` and
`\sigma=1`

We are interested on this probability

`P(-1.94<X<-1.5)`

And using the z score formula given by:

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

And using this formula we got:

`P(-1.94<X<-1.5)=P((-1.94-\mu)/(\sigma)<(X-\mu)/(\sigma)<(-1.5-\mu)/(\sigma))=P((-1.94-0)/(1)<Z<(-1.5-0)/(1))=P(-1.94<z<-1.5)`

And we can find this probability with this difference:

`P(-1.94<z<-1.5)=P(z<-1.5)-P(z<-1.94)=0.0668-0.026= 0.0408`