Q on IQR and SD

S

Sonal7

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I am not sure how to do do Q 11. Its easy to find 11 numbers that add to 0 but I cant see how the IQR and SD are connected.
 
I don't see a Q11. Do you mean #3 at the bottom?

I don't think there's any direct connection between IQR and SD, though you may have been taught some rule of thumb that I don't know.

I would start by just picking an easy set of numbers, finding their IQR and SD, and then trying to adjust them to reach the goal. It may be useful to think about what you can change that will increase or decrease the SD without changing the IQR.
 
Not sure of a direct connection between IQR and SD though. However, as IQR covers only the middle 50% of the data ( between Q1 and Q3), any change below 25th percentile and above 75th percentile of the values will not alter the IQR but will affect the SD alone.

Tried altering the values beyond 2.5 and -2.5 to make my SD twice the IQR. I guess the ulterior concept here is that outliers in data never affects the Q1, Q2, Q3 values. I tried explaining in the way I understood. Hope this helps :)

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Yes, those are the ideas I was hinting at.

Your work reminds me, though, that there are different ways to define/calculate quartiles, so we really need to see Sonal7's own work, which may use a different method than yours. (And I think you mislabeled your Q1 and Q3 anyway!)
 
Totally agree!
Oops yes typing mistake for Q1 and Q2. Lables should be interchanged.
 
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I have used the numbers that they gave in the ans book which I dont get.
They made a to be square root of 42.4. Where did i go wrong?
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Please show how you got [MATH]\sqrt{\frac{6^2+4a^2}{11}}[/MATH] from. It's wrong. I think it's just a silly mistake.
 
Dr.Peterson said:
Please show how you got [MATH]\sqrt{\frac{6^2+4a^2}{11}}[/MATH] from. It's wrong. I think it's just a silly mistake.
Click to expand...
its from the formula of standard deviation. the 6^2 and 4a^2 is the sqaure of the x values. I just realised my mistake it should be 1^2 multiplied by 6 to give the sum of the square of the x, so it should be 6 not 6^2. I will redo this. We want double of the IQR, so I got this forumula. I have another question on SD which I am confused about. I will post as a separate thread. Please ignore the pencilled bit at the top.

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Sonal7 said:
its from the formula of standard deviation. the 6^2 and 4a^2 is the sqaure of the x values. I just realised my mistake it should be 1^2 multiplied by 6 to give the sum of the square of the x, so it should be 6 not 6^2. I will redo this. We want double of the IQR, so I got this forumula. I have another question on SD which I am confused about. I will post as a separate thread. Please ignore the pencilled bit at the top.
Click to expand...
Yes, that's exactly the silly mistake I expected, perhaps from thinking 6*1^2 and unconsciously forgetting the order of operations. And I thought you'd see it when you tried to explain it. You didn't disappoint.
 
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