Posted by Ann on Friday, July 20, 2012 at 5:47pm.
The probability that a radish seed will germinate is 0.6. Estimate the probability that of 130 randomly selected seeds, exactly 90 will germinate.
Note: I keep getting the answer wrong.

Statistics  MathMate, Friday, July 20, 2012 at 6:12pm
This is a binomial distribution with parameters N=130, p=0.6, r=90
so
P(X=90)=C(N,r)p^r(1p)^(Nr)
=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 6:46pm
But, I'm suppose to use the normal distribution approximation to the binomial distribution.

Statistics  MathMate, Friday, July 20, 2012 at 7:43pm
Glad 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*(10.6))
=√(31.2)
=5.585696
Z(X=90.5)=(90.578)/5.585696=2.237859
Z(X=89.5)=(89.578)/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/onlinecalculator/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 Zvalues when looking up probabilities)
Thus
P(89.5≤X≤90.5)
=0.987380.98024
=0.00714
(approximated using normal distribution)
(compared with value of 0.00697 using the binomial distribution). 
Statistics  Ann, Friday, July 20, 2012 at 8:29pm
No 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 8:55pm
You're most welcome.
I am glad things are working out.
Post if you have difficulties.