Can anyone help with statistics?

[Word Nerd]

I have a really bad stats teacher and I'm really struggling to understand the work in his class. Can anyone help with this question?

 

20,000 men attend NDU, a university whose athletic mascot is the leprechaun. The mascot requirements demand a person no taller than five feet. If the mean height of men is 5'8" with a standard deviation of 2.5", and the student body forms a normal distribution for height,

a) how many potential leprechauns are there in the student body?

b) how many men are as tall or taller than 6'5"?

 

I don't just want an answer. I want to know how to get to the answer. If anyone can help that would be great.

[mooeypoo]

Woah, I took statistics a long time ago, but here's my input based on some internet research.. hope it helps, at least for a starter:

 

To achieve the "mean value", you add up the ages of all the students and divide by the number of students (source: http://www.mathsisfun.com/mean.html also this: http://www.sosmath.com/calculus/diff/der11/der11.html with a bit more advanced level, I am not sure which level you are in. The first should suffice for a basic level statistics question).

 

The standard deviation is, basically, the amount of values who "stray" off the mean. If you have your mean as 5'8", with most people something close to that but a few 7'1 and some 4'2" (...kindergarden? ) that's your 'deviation'.

You use the mean to calculate the deviation.

 

(this is a good example of the steps you need to do to calculate that: http://en.wikipedia.org/wiki/Standard_deviation#Example )

 

So, basically, you have the value of the standard deviation (look at the example from wikipedia on what to do with it) and you have the equation for the mean, that you can work with:

 

[math]

\frac{\text{sum of the ages of all students}}{20,000 \text{ students}} = 5'8"

[/math]

 

Working with both values, you can figure out how many potential leprechauns there are (hence, how many men that are 5'8" high are in the 20,000 total) and how many are your deviation (taller than 6'5")...

 

Hope that helped!!

 

Good luck

 

~moo

[Glider]

You need to calculate the z-score for 5' (i.e. how many standard deviations 5' is from the mean). Having done that, you can work out the proportion of the population (i.e. area under the curve) that are 5' or less (answers question a), then do the same thing for 6'8" (answers question b).