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

Answer:

$P(-1.94<X<-1.5)=P(\frac{-1.94-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.5-\mu}{\sigma})=P(\frac{-1.94-0}{1}<Z<\frac{-1.5-0}{1})=P(-1.94<z<-1.5)$

And we can find this probability with this difference:

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

Step-by-step 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:

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

And using this formula we got:

$P(-1.94<X<-1.5)=P(\frac{-1.94-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.5-\mu}{\sigma})=P(\frac{-1.94-0}{1}<Z<\frac{-1.5-0}{1})=P(-1.94<z<-1.5)$

And we can find this probability with this difference:

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