1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Sampling Distributions and Normal Approximation

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant equations

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



    3. 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!
     
  2. jcsd
  3. Apr 5, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Forget the normal distribution for a moment. How would you solve this problem in theory? What distribution applies?
     
  4. Apr 5, 2010 #3
    Well, I know it's a binomial distribution, too, but other than that, I'm not exactly sure what you're getting at...
     
  5. Apr 5, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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 [itex]\mu=np[/itex] and the variance is given by [itex]\sigma^2=npq[/itex].
     
  6. Apr 6, 2010 #5
    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?
     
  7. Apr 6, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, the value you used for p is wrong. Try again.
     
  8. Apr 6, 2010 #7
    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?
     
  9. Apr 6, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  10. Apr 6, 2010 #9
    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
     
  11. Apr 6, 2010 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
    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.
     
  12. Apr 6, 2010 #11
    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))
    ?
     
  13. Apr 6, 2010 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What do you get if you use it? Is the answer reasonable?
     
  14. Apr 7, 2010 #13
    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..
     
  15. Apr 7, 2010 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Look up what the variables in the equation you used stand for, the variable n in particular.
     
  16. Apr 7, 2010 #15
    n (in z score according to central limit theorem) is the size of a population in an SRS.
     
  17. Apr 7, 2010 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Does the central limit theorem apply here?
     
  18. Apr 7, 2010 #17
    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.
     
  19. Apr 7, 2010 #18
    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?"
     
  20. Apr 7, 2010 #19

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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?
     
  21. Apr 7, 2010 #20
    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook