# Homework Help: Statistics - normal distribution

1. Jun 14, 2008

### logaw

Ok, I know this problem is below everything on this forum, but I am an English major with no math skills and I'm REALLY stuck.

This question is as follows:

1)Assume that the number of items borrowed per person per year in a library
is normally distributed with a mean of 87 and a standard deviation of 4.
These values describe the distribution of borrowing for the entire
population of users in this library.

a) What is the mode for this distribution?

b) What is the median of this distribution?

c) Roughly sketch the distribution. Place the number of items borrowed on
the x axis and the number of individuals on the y axis. Indicate the
location and values of the mean and the +two and +three standard deviations
on the x axis.

d)What percent of the population borrow between 83 and 91 items per year?

e) What percent of the population borrow between 79 and 95 items per year?

f) What percent of the population borrow more than 95 items per year?

Work I've done:

Not much. I'm having trouble knowing how to get started on the equation. I don't
understand how having only the mean (87) and the standard deviation (4) is
enough information to let me figure out the mode and median. It doesn't
make sense to me.

I know how to get the standard deviation from the mean when I have all of
the data points, but there are no data points given in this problem.

Any help would be much appreciated!

2. Jun 14, 2008

### CompuChip

Don't worry, we get all kinds of questions at all different levels. We even have discussions about subjects like "1 + 1" sometimes :tongue:

You have one additional piece of information: it's a normal distribution. Normal distributions are entirely fixed by the mean and standard deviation.

Also,

3. Jun 14, 2008

### HallsofIvy

You appear to be saying that you know NOTHING at all about the "normal distribution" and I have difficulty believing that you would be given problems like this without having been introduced to the normal distribution first.
Also, since you have shown no work, I don't know exactly how "basic" you are supposed to be. To me, (1) and (2) are trivial (no calculation at all need) because of the symmetry of the normal distribution. (c) is equally simple because all normal distributions look alike- is there a graph of the "standard" normal distribution in your text? As far as (d), (e), and (f) are concerned, you can't "calculate" those; you need to convert to the "standard" normal distribution (with mean 0 and standard deviation 1) and then look them in a chart- that is probably in your textbook.
If not, a good one- and a graph of the standard distribution is at
http://www.mathsisfun.com/probability/standard-normal-distribution-table.html

4. Jun 14, 2008

### logaw

Thank you to both of you -- I appreciate the speedy replies! Ok, I understand that it is a normal distribution now, and that all three are the same. We probably did cover it in our class, but I didn't get it until now.

It is just a basic statistics class and we move through each idea fairly quickly. (Too quickly for me, obviously.) Also, we don't have any textbooks or anything to refer to; we just have to rely on our own note taking.

But I still don't entirely get how to figure out the rest of the problem. The top of the curve on the distribution is 87%, and I have the chart divided into three sections on each side -- am I supposed to find out the other percentages by looking at the chart you gave me?

I'm sorry, I know this is very basic stuff, but I am sincerely stuck.

5. Jun 14, 2008

### konthelion

To solve for for (d),(e),(f) there's basically two ways to solve for it.

1.) You can plug it in your calculator (TI-83+ is the standard and can do this) >"normalcdf" function. Of course on tests you're probably not allowed to used a calculator, so you have to do the following method...
2.) Use the formula $$z=\frac{x-\mu}{\sigma}$$ where $$\mu$$ is the mean and $$\sigma$$ is the standard deviation. Then you look on a normal distribution table.

By the way,
No, this is not true. The mean $$\mu=87$$.

Last edited: Jun 14, 2008
6. Jun 14, 2008

### logaw

EDIT: Nevermind, something finally clicked, and I have figured it out. Wow, I am very slow.

Thank you all, for your help -- it is much appreciated.

Last edited: Jun 14, 2008