Question

Assume the readings on thermometers are normally distributed with a mean of 0degreesC and a standard deviation of 1.00degreesC. Find the probability that a randomly selected thermometer reads between negative 2.05 and negative 1.49 and draw a sketch of the region.

Answer 1

`P(-2.05<X<-1.49)=P((-2.05-\mu)/(\sigma)<(X-\mu)/(\sigma)<(-1.49-\mu)/(\sigma))=P((-2.05-0)/(1)<Z<(-1.49-0)/(1))=P(-2.05<z<-1.49)`

And we can find this probability with this difference

`P(-2.05<z<-1.49)=P(z<-1.49)-P(z<-2.05)=0.068- 0.0202= 0.0478`

And we can see the figure in the plot attached.

Explanation:

Let X the random variable that represent the redings on thermometers 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(-2.05<X<-1.49)`

We can use the z score formula given by:

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

Using this formula we got:

`P(-2.05<X<-1.49)=P((-2.05-\mu)/(\sigma)<(X-\mu)/(\sigma)<(-1.49-\mu)/(\sigma))=P((-2.05-0)/(1)<Z<(-1.49-0)/(1))=P(-2.05<z<-1.49)`

And we can find this probability with this difference

`P(-2.05<z<-1.49)=P(z<-1.49)-P(z<-2.05)=0.068- 0.0202= 0.0478`

And we can see the figure in the plot attached.