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

Answer:

$z=-1.28<\frac{a-23.1}{2.9}$

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:

$z=1.28<\frac{a-23.1}{2.9}$

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

Step-by-step 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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.10$  

$P(z<\frac{a-\mu}{\sigma})=0.10$

And we can solve for the value of interest

$z=-1.28<\frac{a-23.1}{2.9}$

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:

$z=1.28<\frac{a-23.1}{2.9}$

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