77.7k views
5 votes
The number of chocolate chips in a bag of Chips Ahoy! Chocolate chip cookies are approximately normally distributed, with a mean of 1262 chips and a standard deviation of 118 chips. Find the number of chocolate chips that makes the 90th percentile of the number of chocolate chips in the bag.

User Ferus
by
7.2k points

1 Answer

2 votes

Answer:


z=1.28<(a-1262)/(118)

And if we solve for a we got


a=1262 +1.28*118=1413.04

So the value of height that separates the bottom 90% of data from the top 10% is 1413.04.

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

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


X \sim N(1262,118)

Where
\mu=1262 and
\sigma=118

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


P(X>a)=0.1 (a)


P(X<a)=0.9 (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:


P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.9


P(z<(a-\mu)/(\sigma))=0.9

But we know which value of z satisfy the previous equation so then we can do this:


z=1.28<(a-1262)/(118)

And if we solve for a we got


a=1262 +1.28*118=1413.04

So the value of height that separates the bottom 90% of data from the top 10% is 1413.04.

User Mloskot
by
7.1k points