Answer :
Answer:
[tex]z=1.28<\frac{a-49}{16}[/tex]
And if we solve for a we got
[tex]a=49 +1.28*16=69.48[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 69.48.
Conclusion: the 10% who spend the most time reading national news online spend 69.48 minutes or more.
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 average time an individual reads online national news of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(49,16)[/tex]
Where [tex]\mu=49[/tex] and [tex]\sigma=16[/tex]
We can solve this problem using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
We want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.9[/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 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-49}{16}[/tex]
And if we solve for a we got
[tex]a=49 +1.28*16=69.48[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 69.48.
Conclusion: the 10% who spend the most time reading national news online spend 69.48 minutes or more.