Question

Y~B(16,0.8)

P(Y>15.36) Anyone???

Answer 1

Normal distribution?

Answer 2

yup

Answer 3

Do you have a table of distributions of z vs probability?
z = (15.36-16)/0.8 = - 0.8

Answer 4

Yup, I do, but it doesn't go to decimal point. So, if I were to look for n=16, there wouldn't be any for p=15.36

Answer 5

The answer is 0.1407, not sure at all how to get that if the table doesn't have it...

Answer 6

You'd want to convert the number in terms of mean and sd, kind of to standardize it.
The mean is 16, the value is 15.36, so the number of sd from the mean is then
z=(16-15.36) divided by the sd = 0.64/0.8 = 0.8
The tables give probabilities in terms of the z value, which is 0.7881 corresponding to z=0.8

Answer 7

Yes, it does give it to 0.8 as P. Sorry, it's the X value that isn't shown. 15.36 is the x value, n=16, P=0.8. I have 16 and 0.8 but not 15.36... It doesn't go in decimals... And the answer is 0.1407... again, not sure how to solve it...

Answer 8

Is 16 the mean, and 0.8 the sd?

Answer 9

I mean 0.8 is the probability

Answer 10

Ah.. let me see..

Answer 11

15.36 is the mean + the variation

Answer 12

There's something I do not understand, so the mean is 16. 15.36 is an observation, say x. and 0.8 is a probability... of what?

Answer 13

This is the full question:

For the random variable Y, for which Y ~ B(16,0.8), calculate P(Y>mean+variation)

Answer 14

So, 0.8 is just the probability of something occuring... (it could be anything)

Answer 15

So the variable does not have to be normally distributed?

Answer 16

It's binomial distribution... Which is the same as normal distribution, isn't it? I'm just doing statistics 1, so it should be normally distributed

Answer 17

Sorry, internet connection problems.

Answer 18

No, binomial distribution has a different set of values of probabilities.

Answer 19

Ah Ok. So, How would you do that here?

Answer 20

So 16 is the number of trials, and 15.36 is the minimum success required, and 0.8 is the probability of success, right?

Answer 21

Yup. That's correct.

Answer 22

Sorry, electricity went.

Answer 23

Have you used the formula to calculate the probability, instead of looking up tables?

Answer 24

If
n=no. of trials
k=min. success required, and
p=probability of success,
then
P(k,n,p)=(n,k)(p^k)(1-p)^k
Does that sound familiar to you?

Answer 25

No, it isn't familiar to me, but it does make a lot more sense. Do I just plug in the values?

Answer 26

I am not too familiar with this distribution, but try it anyway.

Answer 27

OK. I'll tell you if it works

Answer 28

Thks.

Answer 29

but... how do I do (n,k)?

Answer 30

That's probably where they require an integer.
(n,k) is n!/((n!)(n-k)!)
where n! means the factorial of n.

Answer 31

Ah Ok.

Answer 32

I made a mistake with the formula I gave, it should read
P(k,n,p)=(n,k)(p^k)(1-p)^(n-k)
I plugged in the formula with k=15.36 and got 0.0766.

Answer 33

If you are in a rush, you may want to submit a new thread and specify explicitly it's a binomial distribution. I should have caught it when you put B(16,0.8), but didn't.

Answer 34

Ah Ok. Yes, it was wrong when I did the first equation... I'm not in a rush. It's just something I have to understand. It's my internet which keeps falling away that's the problem.

Answer 35

Yes, it's not that answer, but I give you a medal for trying :)

Answer 36

Sorry, I wish I were more familiar with cumulative binomial distributions.
If I found anything later on, I'll post it here.

Answer 37

I'll repost... but last time I did that, no one went to answer it... I guess Binomial Distribution scares people, and they'd rather go for the more simpler Distribution or equations... so now I tend not to write it as Binomial distribution.. but I guess I will :)

Answer 38

Are you good at normal probability distributions?