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I could get the mean using the formulae but not the value of SD and IQR from using the formulae. What is the problem?

  • $\begingroup$ Yes, what is the problem? This question asks you to plug three numbers into the formula. If you succeeded with one calculation, what prevented you from succeeding with the other two calculations, which are carried out in the same way? $\endgroup$ – whuber Nov 5 '14 at 15:59
  • $\begingroup$ I have no idea. Can you give me some hint??? $\endgroup$ – Lim Zhi Jian Nov 5 '14 at 16:01
  • $\begingroup$ To obtain the answer for the SD, let $F=3$ and compute $C = \frac{5}{9}(F-32) = \frac{5}{9}(3-32)$. The other two calculations are carried out the same way. $\endgroup$ – whuber Nov 5 '14 at 16:05
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    $\begingroup$ @whuber: That would give a negative standard deviation. $\endgroup$ – Scortchi - Reinstate Monica Nov 5 '14 at 16:15
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    $\begingroup$ @Scortchi You're right--I was not paying any attention to what the numbers mean! Way too hurried this morning... . Lim Zhi, because both the SD and the IQR compare temperatures, the "$-32$" in the conversion formula is inapplicable: all that matters is that each degree Fahrenheit is $5/9$ degrees Celsius, whence an SD of $3$ degrees F is an SD of $3\times 5/9$ degrees C, and similarly for the IQR. $\endgroup$ – whuber Nov 5 '14 at 16:19

To convert the standard deviation & inter-quartile range—measuring spread—, it's only the relative size of a degree Fahrenheit & a degree Celsius that's relevant, not the arbitrary zero points of each scale. To convert means, of course the location matters.

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