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!

 

[vela]

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...

 

[vela]

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?

 

[vela]

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?

 

[vela]

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

 

[vela]

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))
?

 

[vela]

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..

 

[vela]

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.

 

[vela]

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?"

 

[vela]

You're missing the point. The central limit theorem doesn't apply to this problem because you're not looking at a sample mean. Therefore, the equation you used doesn't apply to this problem, and the largeness of n, at least in the context of the CLT, is irrelevant.

The question was simply: if X is normally distributed with mean 186.9 and standard deviation 7.488, what is the probability that X>200.25?

 

[ihearyourecho]

Oh, sorry. There isn't an example of CLT in the book, and even when looking up one online right now, I'm still not exactly sure what it's used for. Thanks for your help.