Respuesta :
Answer:
[tex]P(X>110)=P(\frac{X-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{5})=P(z>2)[/tex]
And we can find this probability with the complement rule and we got:
[tex] P(Z>2) = 1-P(Z<2) [/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2) = 1-P(Z<2)=1-0.97725=0.02275[/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 prices of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,5)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=5[/tex]
We are interested on this probability
[tex]P(X>110)[/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>110)=P(\frac{X-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{5})=P(z>2)[/tex]
And we can find this probability with the complement rule and we got:
[tex] P(Z>2) = 1-P(Z<2) [/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2) = 1-P(Z<2)=1-0.97725=0.02275[/tex]
And we can find this probability on this way:
[tex]P(-0.50<z<0.65)=P(z<0.65)-P(-0.5)[/tex]