Statistics

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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 -

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

• Statistics -

But, I'm suppose to use the normal distribution approximation to the binomial distribution.

• Statistics -

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).

• Statistics -

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 :) -

You're most welcome.
I am glad things are working out.
Post if you have difficulties.

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