Question

Chocolate chip cookies have a distribution that is approximately normal with a mean of 24.2 chocolate chips per cookie and a standard deviation of 2.6 chocolate chips per cookie. Find Upper P 5 and Upper P 95. How might those values be helpful to the producer of the chocolate chip​ cookies?

Answer 1

Percentile 5

`z=-1.64<(a-24.2)/(2.6)`

And if we solve for a we got

`a=24.2 -1.64*2.6=19.94`

Percentile 95

`z=1.64<(a-24.2)/(2.6)`

And if we solve for a we got

`a=24.2 +1.64*2.6=28.46`

Explanation:

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

`X \sim N(24.2,2.6)`

Where
`\mu=24.2` and
`\sigma=2.6`

We want to find the percentiles 5 and 95 for this case.

Percentile 5

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

`P(X>a)=0.95` (a)

`P(X<a)=0.05` (b)

We want to find a percentile with 0.95 of the area on the left and 0.05 of the area on the right it's z=-1.64. On this case P(Z<-1.64)=0.05 and P(z>-1.64)=0.05

Using this condition we got:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.05`

`P(z<(a-\mu)/(\sigma))=0.05`

Replacing we got:

`z=-1.64<(a-24.2)/(2.6)`

And if we solve for a we got

`a=24.2 -1.64*2.6=19.94`

Percentile 95

`z=1.64<(a-24.2)/(2.6)`

And if we solve for a we got

`a=24.2 +1.64*2.6=28.46`