Question

Normal probability problem (screenshot attached).

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

...estimated Z-values

Answer 6

Well, I already have an estimated mean. The SD would just be\[\huge \sigma = \sqrt{\frac{ \sum(x-\bar{x})^2 }{ n-1 }}\]
right?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

I got 0.01549 for the SD

Answer 11

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

Answer 12

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

Answer 13

Try it again. You'll get it.

Answer 14

I used excel this time and got 0.3258

Answer 15

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

Answer 16

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

Answer 17

which part would you like help with

Answer 18

Oh wait, I shouldn't have included the mean

Answer 19

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.

Answer 20

Yeah, I realized that ahaha.
0.3365!

Answer 21

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

Answer 22

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?

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

Yeah!

Answer 28

And the result is...?

Answer 29

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?

Answer 30

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

Answer 31

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.

Answer 32

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

Answer 33

Just curious -- why did you get rid of the \(\sqrt{n}\)? In my notes I have that the standardization of \(\bar{x}\) includes it.

Answer 34

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.

Answer 35

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?

Answer 36

It's in the problem statement.

Answer 37

For c?

Answer 38

@tkhunny

Answer 39

D.

We finished C. a long time ago.

Answer 40

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

Answer 41

@tkhunny Did you get 0.4522?

Answer 42

??

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

Answer 43

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

Answer 44

So P(Z<-0.1189)