The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. For the 10 percent who spend the most time reading national news online, how much time do they spend?

And if we solve for a we got

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:

Where and

We can solve this problem using the normal standard distribution and the z score given by:

We want to find a value a, such that we satisfy this condition:

  (a)

  (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:

But we know which value of z satisfy the previous equation so then we can do this:

And if we solve for a we got

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.

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