# Standard deviation and proportion

• September 4th 2009, 10:24 AM
Standard deviation and proportion
I have never taken a single stat class and I'm almost done reading a textbook on Probability and Statistics. So I decided to help tutor STAT at work, but umm.... It turns out that understanding probability/stat theorems with calculus proof doesn't necessary means that I can also do the basic stat... Please help me help the students...

Question:

The distribution of the body temperatures of adults is roughly bell-shaped, with a mean of 98.6 degrees Fahrenheit and a standard deviation of 0.8 degrees Fahrenheit.

What is the proportion of adults whose body temperature is above 98.6 degrees?

What is the proportion of adults whose body temperature is between 97 degrees and 100.2 degrees?

So I understand how to find standard deviation, that is the square root of the different of the expected value and the mean, but how do I go about this? Thanks.
• September 4th 2009, 11:33 AM
awkward
Quote:

I have never taken a single stat class and I'm almost done reading a textbook on Probability and Statistics. So I decided to help tutor STAT at work, but umm.... It turns out that understanding probability/stat theorems with calculus proof doesn't necessary means that I can also do the basic stat... Please help me help the students...

Question:

The distribution of the body temperatures of adults is roughly bell-shaped, with a mean of 98.6 degrees Fahrenheit and a standard deviation of 0.8 degrees Fahrenheit.

What is the proportion of adults whose body temperature is above 98.6 degrees?

What is the proportion of adults whose body temperature is between 97 degrees and 100.2 degrees?

So I understand how to find standard deviation, that is the square root of the different of the expected value and the mean, but how do I go about this? Thanks.

The phrase "bell-shaped" is meant to tip you off that you are supposed to assume the distribution of temperatures is Normal.

So the question of what proportion of adults have a body temperature above 98.6 degrees translates into

"What is the probability that X > 98.6, where X has a Normal distribution with mean 98.6 and standard deviation 0.8? "

To answer this by looking up probabilities in a table of the cumulative Normal(0,1) distribution, you would consider $(X - \mu) / \sigma$, which has a Normal(0,1) distribution.

(Actually, since 98.6 is the mean, you don't even need a table or knowledge of the standard deviation for this part of the question , but you will need it for the next.)

Can you take it from there?
• September 4th 2009, 02:17 PM