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 2.06 and negative 1.43 and draw a sketch of the region.

Answer 1

`P(-2.06<X<-1.43)=P((-2.06-\mu)/(\sigma)<(X-\mu)/(\sigma)<(-1.43-\mu)/(\sigma))=P((-2.06-0)/(1)<Z<(-1.43-0)/(1))=P(-2.06<z<-1.43)`

And we can find this probability with this difference:

`P(-2.06<z<-1.43)=P(z<-1.43)-P(z<-2.06)= 0.0765 -0.0197=0.0567`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

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(-2.06<X<-1.43)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

`P(-2.06<X<-1.43)=P((-2.06-\mu)/(\sigma)<(X-\mu)/(\sigma)<(-1.43-\mu)/(\sigma))=P((-2.06-0)/(1)<Z<(-1.43-0)/(1))=P(-2.06<z<-1.43)`

And we can find this probability with this difference:

`P(-2.06<z<-1.43)=P(z<-1.43)-P(z<-2.06)= 0.0765 -0.0197=0.0567`