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

Answer:

$P(-2.05<X<-1.49)=P(\frac{-2.05-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.49-\mu}{\sigma})=P(\frac{-2.05-0}{1}<Z<\frac{-1.49-0}{1})=P(-2.05<z<-1.49)$

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.

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

Using this formula we got:

$P(-2.05<X<-1.49)=P(\frac{-2.05-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.49-\mu}{\sigma})=P(\frac{-2.05-0}{1}<Z<\frac{-1.49-0}{1})=P(-2.05<z<-1.49)$

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.