Need help with a statistics problem

Factual Questions
1

So I’m trying to solve this statistics problem, for no other reason than someone showed it to me and I didn’t know how to do it (I’ve never actually taken statistics, but I was bored and curious so I took a stab at it). So I’ve been watching videos all morning on how to do this, but I think I’m missing something.

We have a normal distribution with a standard deviation of 10 and a median of -12.
They want me to find the area between the values of -9 and 44.

So I got as far as getting a Z value of each of these ----I get .3 and 5.6 (which, according to the answer key, is correct) (I get 5.6 because 44 - (-12) / 10 =5.6)

But I can’t follow from there…everything I watch tells me to go look up these values on the Z score table. But 5.6 is too big, it’s not on the Z score table, so I’m clearly not understanding the steps involved here. All of the videos I find online about how to do this use examples with smaller numbers. (FYI the answer key says the answer is .3821, I just can’t make heads or tails out of how to get there)

So does anyone here know how to solve a problem like this?

2

I haven’t taken stats in a while, but since 44 is so far from the median of -12, more than 5 standard deviations, you’re effectively just looking for the probability that something is greater than -9, or one minus the probability that something is less than 9.

I don’t remember the next step, going from Z scores to probabilities, and I’m sure someone will come along and tell me why I’m wrong about all of this.

So, my FQ answer is that 44 is so far from the median, given the SD, that the area under the curve there is essentially zero, so you should be able to fine the area using an open ended function that calculates one of the above – greater than or less than, and adjusting appropriately, with 1- if necessary.

3

In the end you want to compute this:

\int _{-9}^{44}\frac{e^{-\frac{1}{200} (x+12)^2}}{10 \sqrt{2 \pi }}dx

which is:

\frac{1}{2} \left(\text{erf}\left(\frac{14 \sqrt{2}}{5}\right)-\text{erf}\left(\frac{3}{10 \sqrt{2}}\right)\right)

or approx:

0.382089

But it sounds like you want (or are required) to use a look up table?

4

Yes…he’s not allowed a calculator, but he is allowed a table.
As for RitterSport’s answer…I thought so too…but I can’t make sense of his answer. It says “A=B-C”, and B appears the 5.6 value (which he’s giving a value of .5, which is the whole right side of the graph, right?), and C has a value of .1179, which I can’t quite see where’s he’s getting. Z table for .03 is .6179 , but then he subtracts .1179 from .5 to arrive at his answer of .3821 and I can’t figure out why he’s using .1179.

5

0.5 is the area from negative infinity to -12. So, my guess is that 0.1179 is from -12 to -9, sum those up and you get the probability that it’s less than 0.9, and one minus that is what you’re looking for.

1-(n+0.5) = 0.5 - n, where n is the difference area between -12 and -9.

ETA: I think. I’m sorry I’m filling FQ with my guesses. I’ll stop now.

6

You can just use 1 for the value of z=5.6 (look at the trend of the table to convince yourself this is fine). Then you just look up the value of z=0.3 (which is 0.6179) and subtract.

1 - 0.6179 = 0.3821

7

That sounds plausible to me, and it works…hopefully someone who is confident in this sort of thing can come along and confirm.

8

Thanks Chingon—trying to make sense of his drawing, this now seems to be exactly what he’s doing.

9

Chingon has it. It’s similar (in this case equivalent) he method I taught my students when I taught stats.

10

For the record the look up table (or R function equivalent) of Z=5.8 would be about 0.9999999893. Since you are subtracting a middle range number with limited precision you may as well call it one as Chingon did. But if you were interested in something like what is the area above 44 (which actually comes up more often in statistics than computing the probability between two values) then this extra precision would be crucial. This is why modern statisticians view look up tables the same way that modern engineers view slide rules.

11

I got up to the problem definition and stopped reading, thinking I could do it in my head.

Normal distribution, so median=mean. Therefore we can adjust everything upward to center it at 0, asking for the area between 3 and 56. That 56 is so many standard distributions out that it might as well be at infinity, and we’re totally on the right half of the distribution, so really we’re asking for 0.5 minus the area between 0 and 3.

I remember that the two-sided area for 1-sd out is about 2/3. So 1/3 for one side. The curve isn’t flat there, but I’m going to pretend that it is, and scale it by 3/10, which is the proportion 0-3 is relative to 0-10. 1/3*3/10=1/10.

Therefore the answer is roughly 0.5 - 0.1 = 0.4. And reading on to your answer of 0.3821, really isn’t too far off.

I could have gotten closer by fudging the 0.1 upward a bit, since I knew the peak was higher in the center. I’m not quite sure what I would have bumped it to, but thinking that it was probably between 0% and 50%, I might have chosen 25%. So the final answer would have been 0.375 instead, which is not bad at all.

12

Note that not all Z tables are set up the same way. Most commonly they give the area to the left of a particular z (so, P(z < #)), in which case @Chingon’s calculation is correct; but I’ve seen tables that instead give the area between a particular z and z=0.

13

Yeah, I found that out the hard way too after trying to follow what these tutorial were doing, and then looking at the tables I found on line.
Anyway, I’m not statistician and was more just curious what I was misunderstanding about the solution these various tutorials were showing me, because I couldn’t replicate it.