College Statistics

#1
So let me start by saying I am not very strong in the math field, and I will be starting my summer 2 class for statistics here in a week. What Ive been doing in the mean time is reading my book trying to teach myself so that way I will be ahead before I even make it to class.

Ive read through a small portion of the book right now and the owner of the book before me left some of the sample problems the teacher gave them without answers or how to solve them I understand some of it but I was hoping I could get specific information on how to do it here.

here is the sample problem

in 1992 the journal of the American Medical Association reported that the average body temperature is 98.2 with a SD of .7

What percent of people have a body temperature between 97.3 and 98.6

o now I know that roughly 68% are within a +-1 SD range and 95% are within +-2 SDs roughly.

(average was 98.2) so the smallest within 1sd is 97.5 the largest within 1 SD is 98.9

I could do the same within 2SD's as well

How would I figure out how to get the percent of people that have a body temperature between 97.3 and 98.6?

The previous owner of this sample question said their answer was 61.5% but did not show any work nor do I know if this is correct.


Thank you for your help
 
#2
Hi PirateChris,

I think the old book owner is mostly right.

Here's one of the first thing's you'll learn in statistics: finding the area under the normal curve.

First, we assume that the data is actually normally distributed, which body temperature might (probably is) be. You had the right idea looking for standard deviations, but you need to calculate the precise number of standard deviations to get the answer.

To do this, you need to find something called the z-score, which tells you, given some value of normally distributed variable, how much does it deviate from the mean? You can then look up in a table how much area lies under the normal curve at this z-score. This corresponds to the "percentage" you're looking for.

So we're looking for the area under the curve that corresponds to the range of temperatures 97.3 to 98.6.

The formula for the z-score is this:

value - mean / standard deviation

we need (or at least I did) have to do this twice: once for 97.3 because it's the left of the mean, and once for 98.6 because it's to the right of the mean. You need to calculate the z-scores for both of these, find their values in the table and add them.

So, anyway time for calculations

z-score of 97.3 = 97.3 - 98.2 / .7 = -1.28

z-score of 98.6 = 98.6 - 98.2 / .7 = .57

area under standard normal curve from 0 to -1.28 = .3997
" " from 0 to .57 = .2157

.3997 + .2157 = .6154 = 61.5 % (this guy was careful about significant digits, I see).

I hope this helps. This will be the first thing you learn in class, so don't sweat it.