I am having trouble understanding this probability/statistics problem. 

A survey shows that 66% of all families have pets. If 500 families are
randomly select, what is the probability that at least 317 of them
have pets. (use the normal approximation  to the binormal with the
appropiate correction).

***please explain step by step and use Microsoft word if possible to
show symbols...thanks

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Clarification of Question by redcapelliragazo-ga on 05 May 2004 18:40 PDT

P.S. I really need an answer in 24 hours

Hi redcapelliragazo!!


When n*p and n*(1-p) are both large, say, greater than 5 (some texts
say 10 or 15), the binomial distribution is well approximated by the
normal distribution.
If you have Mean = n*p and Standard Deviation (STD) = sqrt(n*p*(1-p)), then
Z(X) = (X - n*p)/sqrt(n*p*(1-p)) is the standard normal variable.
Note on notation: sqrt means square root.
 
"Continuity Correction: when we use a continuous curve(normal) to
approximate a discrete distribution (binomial), we must spread out the
value of the discrete random variable over the continuous scale.
Thus the whole number 2 corresponds to [1.5 , 2.5) on the continous scale. 
If we wish to say P(r greater than or equal to 5), this would
correspond to P(X>4.5), since the bar for 5 actually stretches from
4.5 to 5.5, and we need to include the entire bar of 5 and move right.
If we wished to say P(r less than or equal to 5), this would
correspond to P(X<5.5) , since again the bar ends at 5.5, and we need
to include the whole bar for 5 and move left."
From "Lecture 14 Normal Distribution(Continuous)" at the website of
the Dept. of Applied Mathematics & Statistics, Stony Brook University.
www.ams.sunysb.edu/~bakoohag/lectures/Lecture14.html 
 

Note on notation: sqrt means square root.


The problem:
"A survey shows that 66% of all families have pets. If 500 families are
randomly select, what is the probability that at least 317 of them
have pets?"

n = 500
p = 0.66

STD = sqrt(500*0.66*0.34) =
    = sqrt(112.2) =
    = 10.5925

n*p = 330 > 5
n*(1-p) = 170 > 5
We can use the normal approximation to the binomial distribution.
 
We need to approximate the following value from binomial:
P(X > 317)

Using the continuous correction, we need to find using the normal distribution:
P(X > 316.5) , so X = 316.5

Then:
Z(316.5) = (316.5 - 330) / 10.5925 =
         = -13.5 / 10.5925 =
         = -1.27

We must find using a table the following:
P(Z > -1.27)
Because the normal distribution is symmetrical 
P(Z > -1.27) = P(Z < 1.27) =
             = 0.8980 

The probability that at least 317 of the selected families have pets is 89.8%.

I found this value by using this table:
"Tables for a Normal Distribution":
http://math.uprm.edu/~wrolke/esma3101/ztable.htm

For help in the use of normal distribution tables:
"Using the Normal Table":
http://www.mis.coventry.ac.uk/~styrrell/pages/ntabexam.htm 

------------------------------------------------------------

For additional reference visit the following pages:

"The Normal Distribution and Related Continuos Probability
Distribution - NORMAL DISTRIBUTION - NORMAL APPROXIMATION TO THE
BINOMIAL DISTRIBUTION":
Press the 'Visit site' button if necessary.
http://library.thinkquest.org/10030/7ndndna.htm

"THE NORMAL APPROXIMATION TO THE BINOMIAL AND POISSON DISTRIBUTIONS":
http://myphliputil.pearsoncmg.com/student/bp_berenson_bbs_9/section6_5.pdf

"Continuity Correction":
http://stat-www.berkeley.edu/~stark/SticiGui/Text/gloss.htm#continuity_correction

"The Binomial Distribution and Experiments":
http://www.andrews.edu/~calkins/math/webtexts/prod09.htm#BNOR

-------------------------------------------------------------
Unfortunately I cannot upload a Word file here, but I think that the
notation for this problem are easy enough to make it easy to
understand it. If you really need the Word file, let me know and I
will try to find a way to put it online to you. Note that we cannot
e-mail to askers due editor's policy.

Search strategy:
"normal approximation  to the binomial"
"normal approximation  to the binomial" correction
"normal distribution" tables


I hope this helps you. Before rate this answer, if you find something
unclear and/or incomplete, please let me know by using the
clarification request, I will gladly respond to your requests to
improve the answer if needed.

Best regards.
livioflores-ga

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Request for Answer Clarification by redcapelliragazo-ga on 06 May 2004 10:30 PDT

could you please clarify:

what sigma is 
what mu is ( is this the average/mean)?

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Request for Answer Clarification by redcapelliragazo-ga on 06 May 2004 11:04 PDT

I used a diffrent table from yours while doing my work. If possible
could you explain why our numbers dont match up...heres the
link...http://www.math2.org/math/stat/distributions/z-dist.htm

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Clarification of Answer by livioflores-ga on 07 May 2004 00:55 PDT

Hi!! 

First clarification:

What sigma is?
Sigma is the symbol used to denote the Standard Deviation.
See some definitions here:
"Standard deviation" From Wikipedia, the free encyclopedia:
http://en.wikipedia.org/wiki/Standard_deviation

"STANDARD DEVIATION: Dictionary Entry and Meaning" at Hiperdictionary:
http://www.hyperdictionary.com/dictionary/standard+deviation


What mu is?
mu is the symbol used in statistics to denote the mean.
See a definition here:
"Mean", a four pages definition at HyperStat:
http://www.ruf.rice.edu/~lane/hyperstat/A15885.html

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Second clarification:

The table used by you gives values for the area under the normal curve
from -oo to Z < 0 (note that Z is negative). The table that I used
gives values for the area under the normal curve from -oo to Z > 0
(note that Z is positive for this table).

So when you search the value for Z = 1.27 , the table gives you the
following result:
P(X < -1.27) = 0.10204

But you must find P(X > -1.27) that is different to P(X < -1.27) !!!

Note that you can continue using this table, but you must apply a
little  trick (remember that I used one trick to refer the result to a
value in the table):
P(X > -1.27) = 1 - P(X < -1.27) (both sets are complementary!!)
             = 1 - 0.10204 (by using your table)
             = 0.89796

Now you can see that both results match. We have only a difference
between 0.8980 and 0.89796 because your table is more accurate (it
uses more decimals, so it gives more accurate results).

I hope this helps you. 

Regards.
livioflores-ga

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Request for Answer Clarification by redcapelliragazo-ga on 07 May 2004 11:13 PDT

Thank you so much that cleared a lot up.

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Clarification of Answer by livioflores-ga on 07 May 2004 12:29 PDT

Thank you for the good rating and the generous tip. I will be waiting
here for answer your future questions.

Best regards.
livioflores-ga