# Beer

#### Beer

##### New Member
Hi,

I am preparing for a test and I need help in this question:

Tree # of Individuals # of Trees S.D. of Diameter

Oak 64 58 11.2
Maple 50 42 7.4
Hickory 12 17 9.6
Ash 8 39 14.5
Birch 7 25 12.0

Question: How many Maple trees had a diameter greater than 49.4 cm?

#### trinker

##### ggplot2orBust
What a disappointment this thread has nothing to do with beer

I'm confused what number of trees and number of individuals is. I would expect a mean of some sort but you don't appear to have that.

#### Beer

##### New Member
What a disappointment this thread has nothing to do with beer

I'm confused what number of trees and number of individuals is. I would expect a mean of some sort but you don't appear to have that.
Hi,

The answer says "49.4 cm is exactly one standard deviation above the mean maple tree diameter. Therefore, 16% of the maple trees wil have a larger diameter larger than 49.4. 16% x 50 trees = 8 trees."

I don't understand how he arrived at the fact that "49.4 cm is exactly one standard deviation above the mean maple tree diameter" And what is the mean maple tree diameter?

Thanks.

#### trinker

##### ggplot2orBust
Beer said:
# of Individuals # of Trees
Can you tell us what these 2 columns mean? What is # of individuals and what is number of trees? Using your calculations 49.9 ([TEX]\bar{x}[/TEX] + 1 sd) and 7.4 (sd) we can figure out the mean. 49.4 - 7.4 = 42. This is the same number you have in your middle column so I'd assume this is actually your mean column not #of trees as you have it labeled. The answer uses 50 for n, which is your first column. I'd assume then the first column is your n.

also are we assuming this data is normally distributed?

Here is a link about standard deviations. This will help you understand where the 16% came from. From there the .16 x 50 is pretty straight forward.

Let us know how your understanding is coming (ie I get it or I need more help because I don't understand ...[blank]....).

Tyler

#### Beer

##### New Member
Thanks Trinker,

I am beginning to understand. Clearly the mistake is in the labelling. This is from a GRE practice book.
By the way, I couldn't see the link about the standard deviations.

Thanks again,

Beer.

#### jrai

##### New Member
According to the Empirical rule, if the data is approximately normally distributed (not highly skewed) then 68% of the observations lie within 1 standard deviation & 95% lie within 2 SDs. So we know that 32% of the obs are outside mean +- SD with half in the upper tail & other half in the lower tail. Therefore, 16% obs will be more than 49.4 & 16% will be less than 34.6.

If we know for sure that the data is not normally distributed then apply Chebeyshev's rule (can be used for any distribution) which states that at least 75% obs will fall within the 2 SDs & 89% will fall within 3 SDs. The general rule is 1-(1/k^2) where k is the # of standard deviations.

#### trinker

##### ggplot2orBust
I added the wiki link to my post above (AND HERE). Look at the link (particularly the Gaussian graph) I sent and you should see how the book derived 16% (or close to that).

This could also be calculated by finding a standard score and then using a look up table or a computer software to find the probability (I actually get .1586553 which rounds to 16)

#### Beer

##### New Member
According to the Empirical rule, if the data is approximately normally distributed (not highly skewed) then 68% of the observations lie within 1 standard deviation & 95% lie within 2 SDs. So we know that 32% of the obs are outside mean +- SD with half in the upper tail & other half in the lower tail. Therefore, 16% obs will be more than 49.4 & 16% will be less than 34.6.

If we know for sure that the data is not normally distributed then apply Chebeyshev's rule (can be used for any distribution) which states that at least 75% obs will fall within the 2 SDs & 89% will fall within 3 SDs. The general rule is 1-(1/k^2) where k is the # of standard deviations.
Thanks Jrai,

How does the general rule work when the SD is 1?

#### Beer

##### New Member
I added the wiki link to my post above (AND HERE). Look at the link (particularly the Gaussian graph) I sent and you should see how the book derived 16% (or close to that).

This could also be calculated by finding a standard score and then using a look up table or a computer software to find the probability (I actually get .1586553 which rounds to 16)
Thanks Trinker.

##### Ninja say what!?!
sigh...tricked with the impression Beer was going to be offered by the OP. Good one.