I'm sure this is simple but I don't know...

I'm sure this is simple but I don't know...
Message posted by Chris (via 24.136.50.126) on December 12, 2001 at 8:58 PM (ET)

What is the mean and stardard deviation for a binomial distribution with 450 trials and a probability of success on any one trial of 0.15?

Any help would be greatly appreciated!

READERS RESPOND:
(In chronological order. Most recent at the bottom.)

Re: I'm sure this is simple but I don't know...
Message posted by Tomi (via 213.133.214.2) on December 13, 2001 at 4:55 AM (ET)

Yes, it is very simple.

What is the mean and stardard deviation for a binomial distribution with 450 trials and a probability of success on any one trial of 0.15?

If no. of trials is n and prob. of success is p, then mean = np and variance = np(1-p). Standard deviation is then the square root of the variance.

In your case, n = 450, p = 0.15.

I think you can do the rest!

Re: I'm sure this is simple but I don't know...
Message posted by JG (via 128.8.22.34) on December 13, 2001 at 4:58 AM (ET)

Mean=450*.15 and Variance=mean*.85 .

Thanks Tomi
Message posted by chris (via 24.136.50.126) on December 13, 2001 at 9:10 AM (ET)

Thank you so much Tomi the mean=67.5 and the SD is something about 7.6. i don't have my calculator handy but i get the idea thats a million.

Re: I'm sure this is simple but I don't know...
Message posted by Tomi (via 154.32.143.54) on December 13, 2001 at 7:24 PM (ET)

You're welcome!

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