anonymous
  • anonymous
Help statistics question!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
Are you referring to np > 5 and n(1-p) > 5 ?
jim_thompson5910
  • jim_thompson5910
I also found this
1 Attachment
jim_thompson5910
  • jim_thompson5910
oh N = 10n N = population size n = sample size that sounds familiar in a way

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jim_thompson5910
  • jim_thompson5910
sorry N > 10n
jim_thompson5910
  • jim_thompson5910
alright
jim_thompson5910
  • jim_thompson5910
i found this too http://www2.fiu.edu/~tardanic/size.pdf it says `According to Moore/McCabe, this is true, strictly speaking, as long as the population is at least 100 times larger than the sample.`
jim_thompson5910
  • jim_thompson5910
ok I guess the rule isn't 100% solid
jim_thompson5910
  • jim_thompson5910
I would go with your book since that's what the teacher will use
jim_thompson5910
  • jim_thompson5910
sure
jim_thompson5910
  • jim_thompson5910
sigma = standard deviation sigma = sqrt(n*p*(1-p)) we don't know the value of p, but we do know n and sigma n = 50 sigma = 0.07 are you able to solve for p?
jim_thompson5910
  • jim_thompson5910
hmm there may be another way to do this
jim_thompson5910
  • jim_thompson5910
ok I figured out a shorter way
jim_thompson5910
  • jim_thompson5910
\[\Large \sigma = \sqrt{\frac{p*(1-p)}{n}}\] \[\Large \sigma = \sqrt{\frac{p*(1-p)}{100}}\] \[\Large \sigma = \sqrt{\frac{p*(1-p)}{2*50}}\] \[\Large \sigma = \sqrt{\frac{1}{2}}*\sqrt{\frac{p*(1-p)}{50}}\] \[\Large \sigma = \sqrt{\frac{1}{2}}*0.07\] \[\Large \sigma \approx 0.0495 \approx 4.95\%\]
jim_thompson5910
  • jim_thompson5910
The quantity \[\Large \sqrt{\frac{p*(1-p)}{50}}\] is the given standard deviation of 7% = 0.07

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