We have been given that a set of average city temperatures in december are normally distributed with a mean of 16.3° C and a standard deviation of 2°C. We are asked to find the proportion of temperatures that are between 12.9°C and 14.9°C.
First of all, we will find the z-score corresponding to 12.9°C and 14.9°C using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]z=\frac{12.9-16.3}{2}[/tex]
[tex]z=\frac{-3.4}{2}=-1.7[/tex]
Similarly, we will find z-score corresponding to 14.9°C.
[tex]z=\frac{14.9-16.3}{2}[/tex]
[tex]z=\frac{-1.4}{2}=-0.7[/tex]
Now we need to find probability of z-score between [tex]-1.7\text{ and }-0.7[/tex].
[tex]P(-1.7<z<-0.7)=P(z<-0.7)-P(z<-1.7)[/tex]
Using normal distribution table, we will get:
[tex]P(-1.7<z<-0.7)=0.24196-0.04457[/tex]
[tex]P(-1.7<z<-0.7)=0.19739[/tex]
Therefore, [tex]0.19739[/tex] of temperatures are between 12.9°C and 14.9°C.
