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!

Stats: Approximating a binomial with a normal distribution

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A multiple choice test consists of a series of questions, each with four possible answers.

    How many questions are needed in order to be 99% confident that a student who guesses blindly at each question scores no more than 35% on the test?

    2. Relevant equations

    So I know that this is a binomial setting with p=0.25 and 'n' is what we are trying to solve for.
    for binomial, μ=n*p, σ=sqrt(n*p(1-p))
    P(B(n,0.25)≤0.35*n)=0.99
    And because of the binomial setting, we must use a correction factor, in this case '+0.5'
    Z= (x - μ)/σ

    3. The attempt at a solution

    First I should say I know the answer is suppose to be n≥92

    So how I start this problem is I use the standardizing formula

    Z= (x - μ)/σ

    which in this case would be

    Z= (0.35*n + 0.5 - n*0.25)/(sqrt(n*0.25*0.75)

    This simplifies to

    Z=(0.1*n + 0.5)/(sqrt(n)*sqrt(0.1875))

    I think what I have done so far is correct. But where I get confused is finding a value for Z,

    I thought what I have to do is something like :

    P(B(n,0.25)≤0.35*n)=Φ((0.1*n + 0.5)/(sqrt(n)*sqrt(0.1875))) = 0.99

    so, Φ((0.1*n + 0.5)/(sqrt(n)*sqrt(0.1875))) = 0.99

    look up 0.01 on the table which gives me Φ(2.33)=Φ((0.1*n + 0.5)/(sqrt(n)*sqrt(0.1875)))

    and so I should be able to set those equal and solve for n:

    2.33 = ((0.1*n + 0.5)/(sqrt(n)*sqrt(0.1875))

    When I solve for n I get a quadratic formula but neither answers I get is the correct answer.

    Any help would be appreciated.

    Thanks!
     
    Last edited: Mar 16, 2016
  2. jcsd
  3. Mar 16, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    When you write P(B(n,0.25)≥0.35*n)=0.99 you are asking to be 99% sure that the student scores at least 35% (i.e., 35% or better). That was not what you were asked!
     
  4. Mar 16, 2016 #3
    Sorry I made a typo just in that sentence, I meant P(B(n,0.25)≤0.35*n)=0.99.
     
  5. Mar 16, 2016 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In the context of this particular problem, just blindly using the "1/2 correction" is a mistake. What would be correct would be to ask for the solution of
    [tex] \Phi \left( \frac{.5 + \lfloor 0.35 n \rfloor -.25 n}{.433 \sqrt{n}} \right) = 0.99, [/tex]
    where "##\lfloor 0.35 n \rfloor##" is the largest integer ##\leq 0.35 n##.

    Personally, I would not try to solve that exactly as written; instead (if I were doing the question) I would drop rounding-down and the 1/2-correction altogether and just solve the resulting very simple problem. Then, if I really wanted to be sure of the solution, I would check manually one or two values of ##n## surrounding the solution, either using the exact binomial or the normal approximation with the more involved form of 1/2-correction indicated above.
     
    Last edited: Mar 16, 2016
  6. Mar 16, 2016 #5
    It was made clear by my instructor that we should "apply a normal approximation with a continuity correction" for this problem. But even if I ignore all that and do what you suggest (as it should work that way) I get:



    [tex] \Phi \left( \frac{ 0.1 n }{.433 \sqrt{n}} \right) = \Phi \left( 2.33 \right) = 0.99, [/tex]

    so,

    [tex]\frac{ 0.1 n }{.433 \sqrt{n}} = 2.33[/tex]

    with all this I get [tex]n= 101.786[/tex]

    and if I actually include the correction I get [tex]n_1=101.539[/tex] and [tex]n_2=0.2453[/tex] from the quadratic.

    None of these are the correct answer so I'm wondering what it is I am doing wrong.

    Thanks
     
  7. Mar 16, 2016 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Further to my response in #4: for ##X_n \sim \text{Binom}(n, .25)## the probabilities ##P(X_n \leq .35 n)## are not monotone in ##n## over short intervals of ##n##, so if you plot ##P(X_n \leq .35 n)## vs. ##n## you get a graph with a "sawtooth" behavior, which rises over the long run but wiggles in the short run. The normal approximation with modified 1/2- correction behaves that way too. The reason is that as ##n## increases the integer values ##N_n## in the events ##\{ X_n \leq .35 n \} = \{ X_n \leq N_n \}## are non-decreasing but sometimes remain constant for a few neighboring values of ##n##. If ##N_n = N_{n+1}## the probability ##P(X_k \leq N_k)## can go down as ##k## increases from ##n## to ##n+1##. That happens because the distribution of ##X_{n+1}## is shifted to the right, but is narrower than that of ##X_n##, so the end result could be a decrease or an increase in the probability for "##\leq N_n ##". For example, for ##n## going from 79 to 86 the values of ##N_n##, P_exact = ##P(\text{Binom}(n, .25n) \leq N_n)## and P_normal = ##\Phi((.5 + \lfloor .35 n \rfloor -.25 n)/\sqrt{.1875n})## are
    [tex] \begin{array}{cccc}
    n & N_n & \text{P_exact} & \text{ P_normal} \\
    79 & 27 & 0.975007 & 0.977978 \\
    80 & 28 & 0.983370 & 0.985907 \\
    81 & 28 & 0.980154 & 0.982868 \\
    82 & 28 & 0.976467 & 0.979337 \\
    83 & 29 & 0.984286 & 0.986724 \\
    84 & 29 & 0.981281 & 0.983895 \\
    85 & 29 & 0.977840 & 0.980611 \\
    86 & 30 & 0.985153 & 0.987495
    \end{array}
    [/tex]
     
    Last edited: Mar 16, 2016
  8. Mar 16, 2016 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Without the continuity correction: if you use z = 2.33 you get your n = 101.786, but if you use the more accurate value z = 2.326 you get n = 101.436. Ok, they are not that different, but one of them rounds to 102 while the other rounds to 101.

    More seriously, though, is the non-monotone behavior of the probability, as explained in post #6. That means that you can have several nearby solutions to the required inequality ##P(\text{Binom}(n,.25n) \leq .35 n) = P(\text{Binom}(n,.25n) \leq \lfloor .35 n \rfloor) \geq 0.99##. This happens in both the exact analysis and in the normal approximation (with 1/2-correction included after the rounding down operation).

    By the way: your statement "First I should say I know the answer is suppose to be n≥92" is misleading: n around 92 is too small to achieve the 99% probability.
     
    Last edited: Mar 16, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Stats: Approximating a binomial with a normal distribution
Loading...