Respuesta :
Answer:
[tex]P(X>32)=P(\frac{X-\mu}{\sigma}>\frac{32-\mu}{\sigma})=P(Z>\frac{32-29.9}{2.31})=P(z>0.909)[/tex]
And we can find this probability with the complement rule and using the normal standard distributon table or excel we got:
[tex]P(z>0.909)=1-P(z<0.909)=1-0.818=0.182[/tex]
And if we convert this to a % we got 18.2 % of maximum temperatures higher or equal than 32 C
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the maximum monthly temperature of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(29.9,2.31)[/tex]
Where [tex]\mu=29.9[/tex] and [tex]\sigma=2.31[/tex]
We are interested on this probability
[tex]P(X>32)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>32)=P(\frac{X-\mu}{\sigma}>\frac{32-\mu}{\sigma})=P(Z>\frac{32-29.9}{2.31})=P(z>0.909)[/tex]
And we can find this probability with the complement rule and using the normal standard distributon table or excel we got:
[tex]P(z>0.909)=1-P(z<0.909)=1-0.818=0.182[/tex]
And if we convert this to a % we got 18.2 % of maximum temperatures higher or equal than 32 C
