The number of chocolate chips in a bag of Chips Ahoy! Chocolate chip cookies are approximately normally distributed, with a mean of 1262 chips and a standard deviation of 118 chips. Find the number of chocolate chips that makes the 90th percentile of the number of chocolate chips in the bag.

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 1413.04.  

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 number of chocolate chips of a population, and for this case we know the distribution for X is given by:

Where and

For this part 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 1413.04.  

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.