Respuesta :
Answer:
[tex]P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)[/tex]
And we can find this probability using the complement rule:
[tex]P(z>1.8)=1-P(z<1.8)[/tex]
And using the normal standard distirbution table or excel we got:
[tex]P(z>1.8)=1-P(z<1.8)=1-0.96407= 0.03593[/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 variable if interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
We are interested on this probability
[tex]P(X>\mu +1.8 \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]
If we apply this formula to our probability we got this:
[tex]P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)[/tex]
And we can find this probability using the complement rule:
[tex]P(z>1.8)=1-P(z<1.8)[/tex]
And using the normal standard distirbution table or excel we got:
[tex]P(z>1.8)=1-P(z<1.8)=1-0.96407= 0.03593[/tex]
