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

Answer:

Percentile 5

$z=-1.64<\frac{a-24.2}{2.6}$

And if we solve for a we got

$a=24.2 -1.64*2.6=19.94$

Percentile 95

$z=1.64<\frac{a-24.2}{2.6}$

And if we solve for a we got

$a=24.2 +1.64*2.6=28.46$

Step-by-step 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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.05$  

$P(z<\frac{a-\mu}{\sigma})=0.05$

Replacing we got:

$z=-1.64<\frac{a-24.2}{2.6}$

And if we solve for a we got

$a=24.2 -1.64*2.6=19.94$

Percentile 95

$z=1.64<\frac{a-24.2}{2.6}$

And if we solve for a we got

$a=24.2 +1.64*2.6=28.46$