anonymous
  • anonymous
Normal probability problem (screenshot attached).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I'm a bit confused about how to solve c-e. It's probably something simple that I'm missing
tkhunny
  • tkhunny
c. This is the definition of the 90th percentile.

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anonymous
  • anonymous
@tkhunny Would I have to convert from X to Z, given that it's a normal probability as opposed to standard normal?
tkhunny
  • tkhunny
If you were to calculate the sample mean and standard deviation, Z-Values could be calculated.
tkhunny
  • tkhunny
...estimated Z-values
anonymous
  • anonymous
Well, I already have an estimated mean. The SD would just be\[\huge \sigma = \sqrt{\frac{ \sum(x-\bar{x})^2 }{ n-1 }}\] right?
tkhunny
  • tkhunny
Okay, what is the numeric result? And what is the mean?
anonymous
  • anonymous
The mean is 1.34. I haven't calculated the SD yet.
tkhunny
  • tkhunny
Good. Do you have a requirement for two decimal places?
anonymous
  • anonymous
No, but that was the value that I used for part a). X) I got 0.01549 for the SD
tkhunny
  • tkhunny
That's no good. Give it another go. Did you find a square root to get that?
anonymous
  • anonymous
Hmm, probably stupid calculation error. I used the formula that I posted above
tkhunny
  • tkhunny
Try it again. You'll get it.
anonymous
  • anonymous
I used excel this time and got 0.3258
tkhunny
  • tkhunny
You used stdevp() instead of stdev(). You need the sample SD, not the population SD.
anonymous
  • anonymous
I used STDEV(). Maybe I had input it wrong? =STDEV(column array, mean)
anonymous
  • anonymous
which part would you like help with
anonymous
  • anonymous
Oh wait, I shouldn't have included the mean
tkhunny
  • tkhunny
Okay, I don't know why you included the mean. That will artificially reduce the standard deviation since the SD already includes all the data. You just added one centrally-located value.
anonymous
  • anonymous
Yeah, I realized that ahaha. 0.3365!
tkhunny
  • tkhunny
There you go. Now for the Z-Score, using x = 1.3.
anonymous
  • anonymous
The conversion that I have is \[\text{Standardization}: \large \frac{ \bar{x}-\mu }{ \sigma / \sqrt{n} }\] But isn't \(\bar{x}\) already \(\mu\) in this case?
tkhunny
  • tkhunny
This is not a Mean estimate. Get rid of the \(\sqrt{n}\) and do the arithmetic, using the estimated values.
tkhunny
  • tkhunny
It's not \(\overline{x}\), either. Just x. x = 1.3
anonymous
  • anonymous
Then what's \(\mu\)? You got x = 1.3 from part (a). right?
tkhunny
  • tkhunny
You estimated \(\mu\) with \(\overline{x} = 1.34\) right up front, no?
anonymous
  • anonymous
Yeah!
tkhunny
  • tkhunny
And the result is...?
anonymous
  • anonymous
Oh, I thought I meant that I estimated \(\mu\) *as* 1.34. I have not made a separate calculation for \(\mu\). But wouldn't they be the same process and give me the same number?
anonymous
  • anonymous
I apologize! I can do physics and differential equations, but I have the hardest time grasping concepts of statistics X)
tkhunny
  • tkhunny
You can't make a calculation for \(\mu\). It is what it is. Use \(\overline{x} = 1.34\) Statistics can be a bear. Just go slowly and deliberately.
anonymous
  • anonymous
Sure, so then I got that \[\large Z=\frac{ 1.34- \mu }{ 0.3365 }\]
anonymous
  • anonymous
Just curious -- why did you get rid of the \(\sqrt{n}\)? In my notes I have that the standardization of \(\bar{x}\) includes it.
tkhunny
  • tkhunny
No, \(Z = \dfrac{x - \overline{x}}{0.3365} = \dfrac{1.3 - 1.34}{0.3365}\) \(\sqrt{n}\) applies to estimates of the MEAN, not estimates of a single value, X.
anonymous
  • anonymous
Ohh, I think I understand what we're doing here. We don't have \(\mu\), but we can estimate it using the estimator of \(\hat{\mu}=\bar{x}\). So where did the x = 1.3 come from?
tkhunny
  • tkhunny
It's in the problem statement.
anonymous
  • anonymous
For c?
anonymous
  • anonymous
@tkhunny
tkhunny
  • tkhunny
D. We finished C. a long time ago.
anonymous
  • anonymous
I see. When I asked my question after your initial response I was referring to c), but I think I understand where we're at now. So the value that I found for Z is -0.1189
anonymous
  • anonymous
@tkhunny Did you get 0.4522?
tkhunny
  • tkhunny
?? \(\dfrac{1.3−1.34}{0.3365} = \dfrac{-0.04}{0.3365} = -0.1189\)
anonymous
  • anonymous
Right, but it's asking for probability..?
anonymous
  • anonymous
So P(Z<-0.1189)

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