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Arisa [49]
3 years ago
14

Chocolate chip cookies have a distribution that is approximately normal with a mean of 24.2 chocolate chips per cookie and a sta

ndard 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?
Mathematics
1 answer:
Allisa [31]3 years ago
4 0

Answer:

Percentile 5

z=-1.64

And if we solve for a we got

a=24.2 -1.64*2.6=19.94

Percentile 95

z=1.64

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   (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  

P(z

Replacing we got:

z=-1.64

And if we solve for a we got

a=24.2 -1.64*2.6=19.94

Percentile 95

z=1.64

And if we solve for a we got

a=24.2 +1.64*2.6=28.46

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