Normal probability problem (screenshot attached).

- anonymous

Normal probability problem (screenshot attached).

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- schrodinger

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- anonymous

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- anonymous

I'm a bit confused about how to solve c-e. It's probably something simple that I'm missing

- tkhunny

c. This is the definition of the 90th percentile.

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## More answers

- anonymous

@tkhunny Would I have to convert from X to Z, given that it's a normal probability as opposed to standard normal?

- tkhunny

If you were to calculate the sample mean and standard deviation, Z-Values could be calculated.

- tkhunny

...estimated Z-values

- 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

Okay, what is the numeric result? And what is the mean?

- anonymous

The mean is 1.34. I haven't calculated the SD yet.

- tkhunny

Good. Do you have a requirement for two decimal places?

- anonymous

No, but that was the value that I used for part a). X)
I got 0.01549 for the SD

- tkhunny

That's no good. Give it another go. Did you find a square root to get that?

- anonymous

Hmm, probably stupid calculation error. I used the formula that I posted above

- tkhunny

Try it again. You'll get it.

- anonymous

I used excel this time and got 0.3258

- tkhunny

You used stdevp() instead of stdev(). You need the sample SD, not the population SD.

- anonymous

I used STDEV(). Maybe I had input it wrong?
=STDEV(column array, mean)

- anonymous

which part would you like help with

- anonymous

Oh wait, I shouldn't have included the mean

- 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

Yeah, I realized that ahaha.
0.3365!

- tkhunny

There you go. Now for the Z-Score, using x = 1.3.

- 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

This is not a Mean estimate. Get rid of the \(\sqrt{n}\) and do the arithmetic, using the estimated values.

- tkhunny

It's not \(\overline{x}\), either. Just x. x = 1.3

- anonymous

Then what's \(\mu\)? You got x = 1.3 from part (a). right?

- tkhunny

You estimated \(\mu\) with \(\overline{x} = 1.34\) right up front, no?

- anonymous

Yeah!

- tkhunny

And the result is...?

- 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

I apologize! I can do physics and differential equations, but I have the hardest time grasping concepts of statistics X)

- 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

Sure, so then I got that
\[\large Z=\frac{ 1.34- \mu }{ 0.3365 }\]

- 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

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

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

It's in the problem statement.

- anonymous

For c?

- anonymous

@tkhunny

- tkhunny

D.
We finished C. a long time ago.

- 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

@tkhunny Did you get 0.4522?

- tkhunny

??
\(\dfrac{1.3−1.34}{0.3365} = \dfrac{-0.04}{0.3365} = -0.1189\)

- anonymous

Right, but it's asking for probability..?

- anonymous

So P(Z<-0.1189)

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