Answer: 0.2134
Step-by-step explanation:
Given : Mean readings on thermometers =[tex]\mu=0^{\circ}\ C[/tex]
Standard deviation = [tex]\sigma=1.00^{\circ}\ C[/tex]
We assume that the readings on thermometers are normally distributed.
Let x be the random variable that represents the reading on a random thermometer.
To find z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = -0.99
[tex]z=\dfrac{-0.99-0}{1}=-0.99[/tex]
For x= -0.32
[tex]z=\dfrac{-0.32-0}{1}=-0.32[/tex]
By using the standard normal distribution table , the probability that a randomly selected thermometer reads between -0.99 and -0.32 will be :-
[tex]P(-0.99<x<-0.32)=P(-0.99<z<-0.32)\\\\=P(z<-0.32)-P(z<-0.99)\\\=0.3744842-0.1610871=0.2133971\approx0.2134[/tex]
