Respuesta :
Answer:
[tex]z=0.842<\frac{13.16-\mu}{3}[/tex]
And if we solve for a we got
[tex]\mu=13.16 +0.842*3=15.69[/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 hourly rates of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,3)[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>13.16)=0.20[/tex] (a)
[tex]P(X<13.16)=0.80[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.20
If we use condition (b) from previous we have this:
[tex]P(X<13.16)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.80[/tex]
[tex]P(z<\frac{13.16-\mu}{\sigma})=0.80[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.842<\frac{13.16-\mu}{3}[/tex]
And if we solve for a we got
[tex]\mu=13.16 +0.842*3=15.69[/tex]
