VCAA exam 2010, standard deviations confusion

[Further Maths 101]

Hey all, just a tad confused how to answer question 14 (finding the standard deviation).

Could someone please explain how the answer is B and if so what is the actual standard deviation value.

Thanks in advance.

[AngelWings]

Here's a very rough way to do this and note that I run through generally how to do it rather than actually doing it for you.
From the box plot, you know where the median of the data is, which is usually the same as the mean in normally distributed data. Even though the data's a little skewed and not exactly normally distributed, you can use the median, note where the mean should've been and work out roughly where the standard deviations are (there's three on either side of the mean usually, but we'll go by median here). Since there are 14 total bars and 6 sections from the SDs, you can tell that the SDs are 2.33 bars from the median (14/6 = 2.33). If you work out where this roughly is and round to closest integer, you'll get the right answer. Hope that helps.

[S_R_K]

Here's a very rough way to do this and note that I run through generally how to do it rather than actually doing it for you.
From the box plot, you know where the median of the data is, which is usually the same as the mean in normally distributed data. Even though the data's a little skewed and not exactly normally distributed, you can use the median, note where the mean should've been and work out roughly where the standard deviations are (there's three on either side of the mean usually, but we'll go by median here). Since there are 14 total bars and 6 sections from the SDs, you can tell that the SDs are 2.33 bars from the median (14/6 = 2.33). If you work out where this roughly is and round to closest integer, you'll get the right answer. Hope that helps.

Can you explain how this method works in general? Won't it depend on the scale of the histogram? For instance, if in this case the histogram was divided up into larger intervals (say, or width 4), then wouldn't your method give the result 7/6 ~ 1.17, which doesn't give B as the answer. And surely standard deviation doesn't change just because you change the scale on the graph... Or am I just completely misunderstanding your approach?

[passbleh24]

In this question you add up the frequencies using 180 as your mean in this case. So 68% of the data usually lies approximately 1 standard deviation from the mean. So if you count and add up the frequencies starting from 180 until you reach 34% on both forward and backward direction you get the interval. 178 - 182.
Standard deviation is the measure of spread from the mean.
182 - 2 = 2
180 - 178 = 2
So the standard deviation is 2 degrees.

If the intervals change as you said, so would the amount that represents 1 standard from the mean. In this case 1 standard deviation is represented by a 2 degree change from the mean of 180.

Although the number representing the 1 standard deviation is different for different intervals it is still referring to the same amount in frequencies.
 

[AngelWings]

Can you explain how this method works in general? Won't it depend on the scale of the histogram? For instance, if in this case the histogram was divided up into larger intervals (say, or width 4), then wouldn't your method give the result 7/6 ~ 1.17, which doesn't give B as the answer. And surely standard deviation doesn't change just because you change the scale on the graph... Or am I just completely misunderstanding your approach?

As I said, it's a very rough approach used to eyeball the question and I wouldn't recommend doing this often. Basically, the intention is to mentally divide the data up from the centre outwards so that the SDs can be approximately calculated. I'm afraid I'm not exactly sure what you meant.
If you meant what happens if each bar represented a difference of 2^oC rather than one, then the data would consist of 7 bars, where each bar would be the average of every two bars that currently exist (e.g. between 178^oC and 180^oC, it currently has two bars that are ~13% and ~21%, so this would be ~17% as one bar). Using the method I provided, there will be approximately 6 sections from the SDs, so each SD would be 7/6 of these new 2^oC difference bars away from the mean (or where you'd assume the mean to be according to the skew of the data and the median), which is ~180^oC and wouldn't have changed. Since each new bar represents 2^oC, you'd have to multiply the 7/6 by 2 (which I omitted in my previous post since anything multiplied by 1 is exactly the same), which should give ~2.33 again and rounds down to 2 as an integer. Does that make sense?

Note: I did some things whilst posting and realised Passbleh24 posted in between me typing this up and posting. Leaving this here anyway.

[S_R_K]

Since each new bar represents 2^oC, you'd have to multiply the 7/6 by 2

This was the only bit that I was a bit troubled by, but let me see if I've got this right:

If you've got a roughly normal distribution shown in a histogram, then the standard deviation can be read off in the following way. Since ~100% of the data is located within 3 sd from the mean, we divide up the data into 6 equally sized intervals. If the range of the data is the interval [a,b], then we approximate a standard deviation by (b–a)/6. If this is what's going on, then I can see why this works as a useful rough method.

I think what was confusing me is that's not really the number of columns in the histogram that matters, it's just the distance between the endpoints (ie. the range of the data). As you say, if the range of 14 were broken up into 7 bars of width 2, we'd still (effectively) do 14/6.

I hadn't thought of or seen this method before, so thanks.

[AngelWings]

This was the only bit that I was a bit troubled by, but let me see if I've got this right:

If you've got a roughly normal distribution shown in a histogram, then the standard deviation can be read off in the following way. Since ~100% of the data is located within 3 sd from the mean, we divide up the data into 6 equally sized intervals. If the range of the data is the interval [a,b], then we approximate a standard deviation by (b–a)/6. If this is what's going on, then I can see why this works as a useful rough method.

I think what was confusing me is that's not really the number of columns in the histogram that matters, it's just the distance between the endpoints (ie. the range of the data). As you say, if the range of 14 were broken up into 7 bars of width 2, we'd still (effectively) do 14/6.

I hadn't thought of or seen this method before, so thanks.

Pretty much. We already know that 99.7% of the data will be within 3 SDs either side of the mean, which is close to 100% of the data. Note that this'd only work in MC questions or as a quick check and you'd have to show more working out if you were doing this as a short answer question. Sorry I missed that step in the first post. 

[lucas2057]

Standard deviation is approximately a quarter of the range. idk what it's on about using the 68-95-99.7 rule.

[Lear]

Standard deviation is approximately a quarter of the range. idk what it's on about using the 68-95-99.7 rule.

Uh no

[studyingg]

Standard deviation is approximately a quarter of the range. idk what it's on about using the 68-95-99.7 rule.

If you have the Nelson book, there is a part in the book that says that st.dev is range/4, this is a very rough estimate, and it's not very reliable, and I feel like this should have been noted in the text-book as this shortcut does not apply to symetrical distributions.

 If you understood it like that then you can interpret the st.dev for bell shaped distributions as range/6 instead  (this can be inferred from the 68-95-99.7 rule)  You can try applying this to the 2002 VCAA exam 1 and 2010 exam 1 which test this theory.

[Lear]

If you have the Nelson book, there is a part in the book that says that st.dev is range/4, this is a very rough estimate, and it's not very reliable, and I feel like this should have been noted in the text-book as this shortcut does not apply to symetrical distributions.

It's quite bizarre that they would put that in there considering it isn't even remotely accurate for typical single variable data.