All,

I am having trouble understanding this statistics problem. Please
solve showing all work. A nice tip will be given if you use Microsoft
Word and equation editor to format the answer properly, but standard
text is very acceptable too.

Let X be binomial with n=20 and p=0.3. Use the normal approximation to
approximate each of the following. Compare your results with values
obtained from Table I of App. A (which is a table of Cumulative
Binomial Distributions, basically here they just want to see how good
the approximation using the normal approximation).

You should probably use the following definitions to solve

*Normal approximation to the Binomial Distribution
Let X be binomial with parameters na dn p. For large n, X is
approximately normal with mean (n)(p) and variance (n)(p)(1-p)

*The density function for normal distribution
f(x) = (1/(sqrt(2(pi))(alpha)) times e to the (-1/2 ...etc

*and similar

I would like this tonight. Please feel free to post clarifications, I
will respond within 10 minutes.

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Clarification of Question by jgortner-ga on 26 Feb 2004 17:44 PST

Sorry, I forgot the questions!
... each of the following

a. P[X < or equal to 3]
b. P[3 < or equal to X < or equal to 6]
c. P[X > or equal to 4]
d. P[X = 4]

Thanks.

Hello jgortner!
In order to answer this question, we only need the definition you gave
about the normal approximation to binomial. You will find a Word
document with the full answer at the following address

http://www.angelfire.com/alt/elmarto/

If you have any doubts regarding this answer, please don't hesitate to
request clarification. Otherwise I await your rating and final
comments.

Best wishes!
elmarto

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Request for Answer Clarification by jgortner-ga on 26 Feb 2004 21:02 PST

Reasercher,

I have recieved your answer. I'm reviewing it now, and will have my
(most likely 5 star) review posted by morning. Thanks for you quick,
well thoughtout, and well laid out answer.

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Request for Answer Clarification by jgortner-ga on 26 Feb 2004 22:59 PST

Researcher,

In your answer you say ....

"However, in this case it?s not a very good approximation. If we look
up this probability in the binomial table, we get that this
probability is 0.573."

"If we know use the binomial table, we get that this probability is
actually 0.892. So again it appears to be a good approximation."

"The actual probability (using the binomial table) is 0.1300. Using
this method, the approximation is quite good."

Can you please explain / show your work here?

Thanks.

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Request for Answer Clarification by jgortner-ga on 26 Feb 2004 23:04 PST

Also,

Can you please show your work for

"Now we follow the same procedure as before, subtracting the mean and
dividing by the standard deviation. We get that, using the
approximation, " (number 3)

Thanks.

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Request for Answer Clarification by jgortner-ga on 26 Feb 2004 23:15 PST

I appologise for the multiple clarifications.

Also explain for 
"If we do this (following the same procedure as in the previous
questions) we get that this probability (using the normal
approximation) is 0.1208. The actual probability (using the binomial
table) "

Thanks.

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Clarification of Answer by elmarto-ga on 27 Feb 2004 18:06 PST

Hi jgortner!
First of all, sorry for the delay to answer your clarification
requests. I hope this delay didn't cause you much trouble.

If I understood your questions correctly, you had trouble
understanding how I arrived to those probability values using the
binomial distribution. I used the table available in the link I
provided. This table states P(X=x), for a binomial distribution with
n=20 and various p's. In order to find P(3<=X<=6), this is equal to
P(X=3)+P(X=4)+P(X=5)+P(X=6). So I just added the values that appeared
on the table for x=3, 4, 5 and 6 (with p=0.3). In order to find
1-P(X<4), we know that P(X<4)=P(X<=3), because the binomial
distribution is discrete. Therefore the answer to P(X<4) is the same
as the first question, which asked P(X<=3). So
P(X>=4)=1-P(X<4)=1-0.108=0.892. Finally, in order to find P(X=4), I
just copied the value that appeared at x=4 in the table (obviously,
again in column p=0.3).

Regarding your other two questions, I simply repeated the procedure
used in questions (1) and (2). In question (3), I needed P(X<4).
Therefore, this becomes

P(X<4)
=P(X-6 < 4-6)
=P( (X-6)/2.05 < -2/2.05 )
=P( Z < -0.97)

Looking up this value in the standard normal distribution table gives
the probability value 0.1646.

In question (4) I did the same:

P(3.5<X<4.5)
=P(X<4.5)-P(X<3.5)
=P( (X-6)/2.05 < (4.5-6)/2.05 ) - P( (X-6)/2.05 < (3.5-6)/2.05 )
=P( Z < -0.73 ) - P( Z < -1.21 )

And then I found those values looking in the table.

I hope this clarifies your doubts. If you have any further question
regarding my answer, please don't hesitate to request further
clarification.

Best wishes!
elmarto

Great work. Nicely formatted Microsoft Word answer. Dedicated server
for documents. Fantastic!