Posted by Ann on Friday, July 20, 2012 at 5:47pm.
This 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
But, I'm suppose to use the normal distribution approximation to the binomial distribution.
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*(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).
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.
You're most welcome.
I am glad things are working out.
Post if you have difficulties.