# Sampling Distributions and Normal Approximation

ihearyourecho

## Homework Statement

A sample survey interviews an SRS of 267 college women. Suppose (as is roughly true) that 70% of all college women have been on a diet within the past 12 months. Use a Normal approximation to find the probability that 75% or more of the women in the sample have been on a diet.

## Homework Equations

Zscore=(x-mu)/(sd/n)

## The Attempt at a Solution

I could do this if I knew how to find the standard deviation, but I don't..
Any help would be great, thanks!

## Answers and Replies

Staff Emeritus
Homework Helper
Forget the normal distribution for a moment. How would you solve this problem in theory? What distribution applies?

ihearyourecho
Well, I know it's a binomial distribution, too, but other than that, I'm not exactly sure what you're getting at...

Staff Emeritus
Homework Helper
Cool, that's what I was getting at. Just wanted to make sure you recognized the problem. Now, you want to approximate a binomial distribution with a normal distribution where the mean is given by $\mu=np$ and the variance is given by $\sigma^2=npq$.

ihearyourecho
Right right

mu=np, p=mu/n=.75/267=.00281
sigma=SQRT(np(1-p)=SQRT(267*.00281*(1-.00281)=.865

Is this right? sigma=.865? And then I just plug that into the equation I listed earlier?

Staff Emeritus
Homework Helper
No, the value you used for p is wrong. Try again.

ihearyourecho
Oh, I looked at it wrong. Mu is .70, which would make p=.00262. Is this right? Also, are my methods for the rest of the problem right, just substituting in this p value?

Staff Emeritus
Homework Helper
No, that's not right. The binomial distribution tells you the probability of k successes in n trials when p is the probability of success in an individual trial. What is a trial in this problem? Once you know that, finding the value of p should be straightforward.

ihearyourecho
I'm sorry, I just really don't understand. I don't see "one trial," all I see is 70% of all college women have been on a diet in the past 12 months. Our sample is 267 women, which means that 187 women had been on a diet.

In our book, it states "Suppose that a count X has the binomial distribution with n observations and success probability p. When n is large, the distribution of X is approximately Normal, N(np, SQRT(np(1-p))). As a rule of thumb, we will use Normal approximation when n is so large that np greaterthanequalto 10 and n(1-p) greaterthanequalto 10.

In this case, n is 267 and p is .7, which would make mu=186.9, and sigma=7.5

Staff Emeritus
Homework Helper
I'm sorry, I just really don't understand. I don't see "one trial," all I see is 70% of all college women have been on a diet in the past 12 months. Our sample is 267 women, which means that 187 women had been on a diet.
Not exactly. It means that you'd expect 187 women to have been on a diet. The actual number can be anywhere from 0 to 267.
In this case, n is 267 and p is .7, which would make mu=186.9, and sigma=7.5
Yes, this is what I was trying to get you to figure out. In your previous post, you said that mu=0.7 and p=0.00262.

Each woman in the sample corresponds to a trial/observation. The probability that the woman had been on a diet is p=0.7; this is the probability of success for each trial. The sample is then the aggregation of 267 individual trials/observations, and it's described by the binomial distribution with n=267 and p=0.7.

ihearyourecho
Alright, so now that I've got that down, can I use the equation that I had originally to finish out the problem?

Zscore=(x-mu)/(sd/SQRT(n))
?

Staff Emeritus
Homework Helper
What do you get if you use it? Is the answer reasonable?

ihearyourecho
mu=pN, mu=(.7)*(267), mu=186.9
sd=SQRT(Np(1-p)), sd=SQRT(186.9(1-.7), sd=7.488
Zscore = (x-mu)/(sd/(n^1/2), Zscore=(200.25-186.9)/(7.488/16.34), Zscore=29.13

So, no, it doesn't make sense.
Unless I did the x part wrong..

Staff Emeritus
Homework Helper
Look up what the variables in the equation you used stand for, the variable n in particular.

ihearyourecho
n (in z score according to central limit theorem) is the size of a population in an SRS.

Staff Emeritus
Homework Helper
Does the central limit theorem apply here?

ihearyourecho
It states that when n is large, the sampling distribution of the sample mean xbar is approximately normal. I'm not exactly sure what constitutes as "large," but I guess in the overall scheme of things, n is not large if we're comparing it to the population of all women.

ihearyourecho
So, if this is the case, then this would be the method:

mu=pN, m=(.7)*(267), mu=186.9
sd=SQRT(Np(1-p)), sd=SQRT(186.9(1-.7), sd=7.488
Zscore = (x-mu)/(sd), Zscore=(200.25-186.9)/(7.488), Zscore=1.78, P=.9625

How can you tell if n is "large?"

Staff Emeritus