Using the normal distribution, it is found that the third quartile is of Q3 = 0.67, hence option B is correct.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of 0ºC, hence [tex]\mu = 0[/tex].
- The standard deviation is of 1ºC, hence [tex]\sigma = 1[/tex].
The third quartile is X when Z has a p-value of 0.75, as [tex]100\frac{3}{4} = 75[/tex], hence X when Z = 0.67.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.67 = \frac{X - 0}{1}[/tex]
[tex]X = 0.67[/tex]
To learn more about the normal distribution, you can take a look at https://brainly.com/question/24663213
