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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?

User Suneetha
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Answer:

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

User Hossein Asadi
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