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 0.99 and negative 0.32 and draw a sketch of the region.
Accepted Solution
A:
Answer: 0.2134Step-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]