Respuesta :
Answer:
[tex]P(X>\mu + 2\sigma)[/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]
Using the z score formula we want this probability:
[tex] P(Z>2)[/tex]
And using the complement rule we got:
[tex] P(Z>2) = 1-P(Z<2) = 1-0.97725= 0.023[/tex]
And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:
No. This event happens only 2.5% of the time
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 variable of interst of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex] the deviation
We are interested on this probability
[tex]P(X>\mu + 2\sigma)[/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]
Using the z score formula we want this probability:
[tex] P(Z>2)[/tex]
And using the complement rule we got:
[tex] P(Z>2) = 1-P(Z<2) = 1-0.97725= 0.023[/tex]
And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:
No. This event happens only 2.5% of the time
