Respuesta :
Answer:
[tex]P(X<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-123.8}{3.1})=P(z<-1.226)[/tex]
And we can find this probability using excel or the normal standard tabe and we got:
[tex]P(z<-1.226)=0.110[/tex]
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 temperatures of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(123.8,3.1)[/tex]
Where [tex]\mu=123.8[/tex] and [tex]\sigma=3.1[/tex]
We are interested on this probability
[tex]P(X<120)[/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<120)=P(\frac{X-\mu}{\sigma}<\frac{120-\mu}{\sigma})=P(Z<\frac{120-123.8}{3.1})=P(z<-1.226)[/tex]
And we can find this probability using excel or the normal standard tabe and we got:
[tex]P(z<-1.226)=0.110[/tex]
Using the normal distribution, it is found that there is a 0.111 probability for a given year that the temperature never exceeds 120 deg. F in a given July in Death Valley.
Normal Probability Distribution
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 and the standard deviation are given by, respectively, [tex]\mu = 123.8[/tex] and [tex]\sigma = 3.1[/tex]
The probability that on a single day the probability does not exceed 120 ºF in the death valley is the p-value of Z when X = 120, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 123.8}{3.1}[/tex]
[tex]Z = -1.23[/tex]
[tex]Z = -1.23[/tex] has a p-value of 0.1111.
Hence, there is a 0.111 probability for a given year that the temperature never exceeds 120 deg. F in a given July in Death Valley.
More can be learned about the normal distribution at https://brainly.com/question/24663213
