Question

Ever ordered a burrito at Chipotle? If so, you’ll notice that you get a slightly different amountof filling each time due to random influences (e.g., mood of the server, fullness of the bins with thefood, etc.). Suppose you come home and weigh your same order each day for a year, discoveringthat the weight of your order is normal with mean 620 grams and SD 30 grams. You decide that a"typical day" for you at Chipotle is one where the mass of food falls in the central 60% of what’spossible. Between what two masses will you get a "typical day"? Solve without using R. [WISE,VIZ]

Answer 1

`a=620 -0.842*30=594.74`

So the value of height that separates the bottom 20% of data from the top 80% is 594.74.

`a=620 +0.842*30=645.26`

So the value of height that separates the bottom 80% of data from the top 20% is 645.26.

`(594.74,645.26)` we will have the 60% of the weigths

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 weights of a population, and for this case we know the distribution for X is given by:

`X \sim N(620,30)`

Where
`\mu=620` and
`\sigma=30`

The z score formula is given by:

`z=(x-\mu)/(\sigma)`

We want the 60% middle values so we need to have 20% of the values on each tail, and we want to find the limits so we can do this:

LOWER LIMIT

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

`P(X>a)=0.8` (a)

`P(X<a)=0.2` (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 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(z>-0.842)=0.8

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.2`

`P(z<(a-\mu)/(\sigma))=0.2`

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

`z=-0.842<(a-620)/(30)`

And if we solve for a we got

`a=620 -0.842*30=594.74`

So the value of height that separates the bottom 20% of data from the top 80% is 594.74.

UPPER LIMIT

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

`P(X>a)=0.2` (a)

`P(X<a)=0.8` (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 0.8 of the area on the left and 0.2 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.8`

`P(z<(a-\mu)/(\sigma))=0.8`

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

`z=0.842<(a-620)/(30)`

And if we solve for a we got

`a=620 +0.842*30=645.26`

So the value of height that separates the bottom 80% of data from the top 20% is 645.26.

So then the answer for this case would be:

`(594.74,645.26)` we will have the 60% of the weigths