anonymous
  • anonymous
Problem regarding confidence intervals and probability (screenshot attached)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I know for the expected value, I need n and p. I think I can define n as the observed population, 1000. Is 95% p in this case?
ganeshie8
  • ganeshie8
Yes, p = 0.95
anonymous
  • anonymous
Okay, so then the expected value should be 950, right?

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ganeshie8
  • ganeshie8
Yes, this is a sampling distribution confidence level of 95% means that 95% of the samples will capture the population mean
anonymous
  • anonymous
Okay, so how might I proceed to solve the second question?
ganeshie8
  • ganeshie8
The statement, " 95% of the samples will capture the population mean" is same as saying "each sample has 95% probability to containt he population mean" yes ?
ganeshie8
  • ganeshie8
the samples were selected independently and we have binomial distrabution : \(n=1000\) and \(p=0.95\)
ganeshie8
  • ganeshie8
use the normal approximation formulas to find the probability for 950 to 970 samples to contain the population mean
anonymous
  • anonymous
Normal approximation as in standardizing to Z?
ganeshie8
  • ganeshie8
since the number of samples are large enough (1000), the sampling distribution would be approxmately normal. so we can use the approximation formulas
ganeshie8
  • ganeshie8
basically we're converting binomial distribution to the normal distribution so that we can use zscores to compute the probability easily
ganeshie8
  • ganeshie8
you could also work the probability in binomial distribution, but computing 20 binomial coefficients and adding them is a pain
anonymous
  • anonymous
That makes sense So \(P(950
anonymous
  • anonymous
right?
anonymous
  • anonymous
Or is the \(\sqrt{n}\) not needed here.. I'm not certain when it is needed and when it is not
ganeshie8
  • ganeshie8
nope, you need to use the formulas for approximating "binomial distribution" as normal distribution
ganeshie8
  • ganeshie8
|dw:1447475619376:dw|
anonymous
  • anonymous
Okay but isn't it the same formula though (but without the \(\sqrt{n}\)) ?
ganeshie8
  • ganeshie8
zscore formula remains same, you need to find standard deviation using that approximation formula
anonymous
  • anonymous
Oh, okay! I wasn't going to do that but I wanted to make sure that the formula I was using was correct. I thought you were stating that the formula in general was completely wrong, with or without the sqrt(n). I got that \(\sigma=6.89\) So then\[P (\frac{ 950-950 }{ 6.89 }<\text{Z}<\frac{970-950}{6.89})\]\[=P(0<\text{Z}<2.9)\]\[=\Phi(2.9)-\Phi(0)\]
ganeshie8
  • ganeshie8
That looks good, but there is a slight mistake
ganeshie8
  • ganeshie8
binomial distribution is discrete normal distribution is continuous when converting form binomial to normal, we need to account for that
ganeshie8
  • ganeshie8
|dw:1447476412535:dw|
ganeshie8
  • ganeshie8
simply widen the interval : 949.5 < X < 970.5
anonymous
  • anonymous
Darnit X) So \[P (\frac{ (950-0.5)-950 }{ 6.89 }<\text{Z}<\frac{(970+0.5)-950}{6.89})\]
ganeshie8
  • ganeshie8
Looks good !
anonymous
  • anonymous
Okay perfect, thank you! X)
ganeshie8
  • ganeshie8
np :)

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