Normal Approximation-what is it?

Quwee

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Feb 2, 2006
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Suppose 20 people are randomly selected to participate in a survey. Assume that P(man)=P(woman)=.5. Use a normal approximation to estimate the probability that at least 12 women will be chosen.
 
Odd that you're asked about something you haven't learned yet.

Look in your textbook or google "normal approximation to binomial".

You'll need to apply a continuity correction (again, textbook/google), too.
 
The teacher already taught all this stuff and I am behind the rest of the class
I read the textbook but is confusing and I don't understand it.
 
I need to to do this so I can finish this page and then I can do the chapter test
 
Let X be the random variable for the number of women to be selected.

Binomial parameters:

p = 0.5
q = 1 - 0.5 = 0.5
n = 20

We are asked for P(X > 12), and we can use P(X > 12) = 1 - P(X < 12).

We're approximating a discrete distribution (binomial) with a continuous one (normal), so applying a continuity correction means we're looking for 1 - P(X < 11.5).

Normal distribution parameters:

\(\displaystyle \mbox{ \mu = np = (20)(0.5) = 10 }\) (np > 5 and nq > 5 requirements for the approximation are satisfied)
\(\displaystyle \mbox { \sigma = \sqrt{npq} = \sqrt{(20)(0.5)(0.5)} = \sqrt{5}}\)

\(\displaystyle \mbox{ Z = \frac{ X - \mu}{\sigma} = \frac{11.5 - 10}{\sqrt{5}} \approx 0.671}\)

A Z-score of 0.671 corresponds to a probability of 0.2489 in the normal tables.

Sketch the information on a bell curve to get the correct probability:

Code:
                          0.2489
                     ***   /
         0.5          :  |/_
            \   *     :      *
            _\|       :      :
           *          :      :   *
                      :      :  
    * - - - - - - - - - - - - - - - - - *   
                     10     11.5
So P(X < 11.5) = 0.5 + 0.2489 = 0.7489

And we have P(X > 12) = 1 - P(X < 12) = 1 - 0.7489 = 0.2511

If you have binomial tables which go up to n=20, although ultimately making the exercise pointless, you can compare the approximation with the actual.
 
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