Question

Chocolate chip cookies have a distribution that is approximately normal with a mean of 23.1 chocolate chips per cookie and a standard deviation of 2.9 chocolate chips per cookie. Find Upper P 10 and Upper P 90. How might those values be helpful to the producer of the chocolate chip​ cookies?

Answer 1

`z=-1.28<(a-23.1)/(2.9)`

And if we solve for a we got

`a=23.1 -1.28*2.9=19.388`

And for the 90 percentile we can do this:

`z=1.28<(a-23.1)/(2.9)`

And if we solve for a we got

`a=23.1 +1.28*2.9=26.812`

The P10 would be 19.388 and the P90 26.812

Explanation:

Let X the random variable that represent the chocolate chip cookies of a population, and for this case we know the distribution for X is given by:

`X \sim N(23.1,2.9)`

Where
`\mu=23.1` and
`\sigma=2.9`

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

`P(X>a)=0.90` (a)

`P(X<a)=0.10` (b)

We can find a z score value who that satisfy the condition with 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.28.

Using this value we can do this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.10`

`P(z<(a-\mu)/(\sigma))=0.10`

And we can solve for the value of interest

`z=-1.28<(a-23.1)/(2.9)`

And if we solve for a we got

`a=23.1 -1.28*2.9=19.388`

And for the 90 percentile we can do this:

`z=1.28<(a-23.1)/(2.9)`

And if we solve for a we got

`a=23.1 +1.28*2.9=26.812`

The P10 would be 19.388 and the P90 26.812