SOLUTION: A binomial distribution has a mean of 12 and a standard deviation of 2.683. Find n and p. {{{np=12}}} {{{sqrt of npq is 2.683}}} {{{2.683/12=.22358333}}} {{{1-.22358333=.7764}}} {

Algebra ->  Probability-and-statistics -> SOLUTION: A binomial distribution has a mean of 12 and a standard deviation of 2.683. Find n and p. {{{np=12}}} {{{sqrt of npq is 2.683}}} {{{2.683/12=.22358333}}} {{{1-.22358333=.7764}}} {      Log On


   

Question 362858: A binomial distribution has a mean of 12 and a standard deviation of 2.683. Find n and p.
np=12 sqrt+of+npq+is+2.683 2.683%2F12=.22358333 1-.22358333=.7764 p=+.7764 np=12 n%28.7764%29=12 12-.7764=11.2236 n=11.2236 I'm not sure if this is right.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let mu be the mean and sigma be the standard deviation.


So mu=np and sigma=sqrt%28np%281-p%29%29 by definition. Since we're given that the distribution has "a mean of 12 and a standard deviation of 2.683", we know that mu=12 and sigma=2.683.


So this means that 12=np and 2.683=sqrt%28np%281-p%29%29. Now solve for 'p' in 12=np to get p=12%2Fn and plug this into the second equation to get 2.683=sqrt%28n%2812%2Fn%29%281-12%2Fn%29%29


From there, you can simplify to get 2.683=sqrt%2812%281-12%2Fn%29%29


I'll let you take it from here.