Statistics, how do you translate Bin(n,p) to say N(,) etc

[mathpariah]

hey guys, some hopefully easy stat questions if you will

Im wondering about translating one statistical distribution to another, like going from:

Bin(n,p) to N(np,sqrt(npq)) where q=1-p

or that Po(au) is roughly equal to N(xu,sqrt(xu))

Im mostly sitting scratching my head on which method to use on each and every problem I am solving (Im studying for an exam in march). is there any chart or a really good and lightweight summary somewhere that can shed some light on this? like which method to use and how you can and why you need to translate them to another method, like in the examples above

Im also wondering when you need to use a two-sided interval, like youre going from a 95% to 97.5% by using 1-a/2 where a is the level

I could really use some help and would be grateful for any input

thanks

 

[moonman239]

hey guys, some hopefully easy stat questions if you will

Im wondering about translating one statistical distribution to another, like going from:

Bin(n,p) to N(np,sqrt(npq)) where q=1-p

or that Po(au) is roughly equal to N(xu,sqrt(xu))

Im mostly sitting scratching my head on which method to use on each and every problem I am solving (Im studying for an exam in march). is there any chart or a really good and lightweight summary somewhere that can shed some light on this? like which method to use and how you can and why you need to translate them to another method, like in the examples above

Im also wondering when you need to use a two-sided interval, like youre going from a 95% to 97.5% by using 1-a/2 where a is the level

I could really use some help and would be grateful for any input

thanks

To answer the first question, only if n*p and n*(1-p) are both 5 or greater can you use the normal distribution to approximate the binomial distribution. If these conditions are satisfied, then the binomial probability of exactly X successes in n trials approximately = the normal probability of getting a value between X-.5 and X+.5, with mean=np and sd=sqrt(npq). The binomial probability of the number of successes being less than or equal to X approximately = the normal probability of getting a value less than or equal to X+.5. The binomial probability of the number of successes being greater than or equal to X approximately = the normal probability of the number of successes being greater than or equal to X-.5.

 

[mathpariah]

To answer the first question, only if n*p and n*(1-p) are both 5 or greater can you use the normal distribution to approximate the binomial distribution. If these conditions are satisfied, then the binomial probability of exactly X successes in n trials approximately = the normal probability of getting a value between X-.5 and X+.5, with mean=np and sd=sqrt(npq). The binomial probability of the number of successes being less than or equal to X approximately = the normal probability of getting a value less than or equal to X+.5. The binomial probability of the number of successes being greater than or equal to X approximately = the normal probability of the number of successes being greater than or equal to X-.5.

thanks moonman.. I find the np and n(1-p)>5 limits curious but I can live with that.

Id be immensly appreciative if someone could look at my other questions too. does anyone know of a good summary website of basic distribution and WHEN to use which. like bin, po, N, chi square and t

thanks