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April 23, 2014

April 23, 2014

Posted by **Ann** on Friday, July 20, 2012 at 5:47pm.

Note: I keep getting the answer wrong.

- Statistics -
**MathMate**, Friday, July 20, 2012 at 6:12pmThis is a binomial distribution with parameters N=130, p=0.6, r=90

so

P(X=90)=C(N,r)p^r(1-p)^(N-r)

=C(130,90)*0.6^90*0.4^40

=5.3347282074*10^33*1.0804695562359849*10^-20 * 1.2089258196146345*10^-16

=0.00697

- Statistics -
**Ann**, Friday, July 20, 2012 at 6:46pmBut, I'm suppose to use the normal distribution approximation to the binomial distribution.

- Statistics -
**MathMate**, Friday, July 20, 2012 at 7:43pmGlad that you mentioned that an approximation is required. The question asks for exactly 90 seeds, which is discrete.

Here's how I would proceed to approximate a discrete random variable from a continuous distribution.

"Exactly 90" is approximately equal to the random variable X=89.5 to 90.5.

We can generally approximate a binomial distribution by a normal distribution when np>5. Here np=130*0.6=78 > 5, so approximation will be reasonable.

The equivalent μ=np=78

σ

=√(npq)

=√(130*.6*(1-0.6))

=√(31.2)

=5.585696

Z(X=90.5)=(90.5-78)/5.585696=2.237859

Z(X=89.5)=(89.5-78)/5.585696=2.058830

Here, we are dealing with a small difference of two probabilities, so normal tables (on paper) by interpolation may or may not be adequate. I suggest you use a calculator with a Z function, or use a normal distribution calculator online, such as:

http://stattrek.com/online-calculator/normal.aspx

Using 5 digits, I get

P(X=90.5)=0.98738, and

P(X=89.5)=0.98024

(remember to use the respective Z-values when looking up probabilities)

Thus

P(89.5≤X≤90.5)

=0.98738-0.98024

=0.00714

(approximated using normal distribution)

(compared with value of 0.00697 using the binomial distribution).

- Statistics -
**Ann**, Friday, July 20, 2012 at 8:29pmNo wonder I keep getting the wrong answer. Thank you so much for helping me how to get the answer. Now I know how to do the next problem which is similar to this problem.

- Statistics :) -
**MathMate**, Friday, July 20, 2012 at 8:55pmYou're most welcome.

I am glad things are working out.

Post if you have difficulties.

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