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.

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:

[tex]X \sim N(1262,118)[/tex]

Where [tex]\mu=1262[/tex] and [tex]\sigma=118[/tex]

For this part 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-1262}{118}[/tex]

And if we solve for a we got

[tex]a=1262 +1.28*118=1413.04[/tex]

So the value of height that separates the bottom 90% of data from the top 10% is 1413.04.