Thread: Another stat prob

Thanks

Need some help guys I really don't think I'm doing this right

right now I have:

a. p(x=3)

20!/3!1! (.05)^3 (.95)^17 = .24

b. p (x > 19) = p(x=19) + p(x=20) = .27

c. Mx = 1 s.d.= .97

Binomial

p = 0.05

q = 1-p = 0.95

n = 20

1) Pr(3)

2) Pr(19) + Pr(20)

3) np

These definitions must be in your text. Find them and you will be able to identify similar information for other distributions.

Originally Posted by Smokey007

Need some help guys I really don't think I'm doing this right

right now I have:

a. p(x=3)

20!/3!1! (.05)^3 (.95)^17 = .24

b. p (x > 19) = p(x=19) + p(x=20) = .27

c. Mx = 1 s.d.= .97

Sorry about notation differences. It happens.

Pr(3) = p(x=3)

You have the right idea, but something is going whacky.

In part a, you have a 1! in the denominator where there should be a 17! Even so, what you have written comes nowhere near 0.24. Must be more arithmetic errors going on. [(20*19*18)/(3*2)]*(0.05^3)*(0.95^17) = 0.0596 or so.

In part b, that probability is WAY to big. I'm thinking maybe you made the same kind of denominator error.

20*(.05^19)*0.95 + 0.05^20 is a very small number.

Good call on Part c.