Question 1197079: Here is the statistics for the high temperatures in the city during October:
*Mean of 65.3°F
*Median of 63.5°F
*Standard deviation of 9.3°F
* IQR are of a 7.1°F
Recall that the temperature C measured in degrees Celsius is related to the temperature F measured in degrees Fahrenheit by C = 5/9 ( F -32)
A. Describe how the value of each statistic changes when 32 is subtracted from the temperature in degrees Fahrenheit.
B. Describe how the value of each statistic further changes when the new values are multiplied by 5/9.
C. describe how to find the value of each statistic when the temperature is measured in degrees Celsius.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part A
Answers:
The mean and median decrease by 32
The standard deviation and IQR remain the same.
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Explanation:
Consider the set {1,2,3,4,10}
The mean is 4 since we add up the values to get 20, and divide by 5 (the number of items in the set).
(1+2+3+4+10)/5 = 20/5 = 4
Now let's shift everything up by 3 units to get {4,5,6,7,13}
The mean of this new set is 7 which is an increase of 3
It's not a coincidence that the mean has gone up by the same exact value as each item in the data set.
Here's a proof of why this is the case
https://www.algebra.com/algebra/homework/Probability-and-statistics/statistics-transformations1.lesson
This idea also applies to subtraction as well.
If we subtract 32 from each data value, then the mean goes down by 32.
Let's go back to the set {1,2,3,4,10}
subtract 5 from each item to get {-4,-3,-2,-1,5}
The median of the original set was 3; the median of the new set is -2
The jump from 3 to -2 is "minus 5".
Therefore, if you subtract 5 from each item, then the median also goes down by the same amount.
Using the idea of the previous paragraph, we can see that subtracting 32 from each temperature will have the median go down by 32.
The standard deviation will remain the same no matter what you add or subtract to each data value. This is because the standard deviation measures how spread out a data set is. Shifting a data set up or down will preserve the spread.
Use your calculator to find the standard deviation of these two sets
{1,2,3,4,10}
{4,5,6,7,13}
and you'll find the population standard deviation of each set is roughly 3.16227766016838
Recall that
IQR = Q3 - Q1
Q1 = first quartile
Q3 = third quartile
Q1 and Q3 decrease by 32 when you subtract 32 from each item.
So,
new IQR = (new Q3) - (new Q1)
new IQR = (Q3 - 32) - (Q1 - 32)
new IQR = Q3 - 32 - Q1 + 32
new IQR = (Q3 - Q1) + (-32+32)
new IQR = (old IQR) + 0
new IQR = old IQR
The IQR has not changed
We should expect it to remain the same since it's a measure of spread like the standard deviation is.
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Part B
Answers:
Mean, Median, Standard Deviation and IQR are multiplied by 5/9
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Explanation:
I'll go back to {1,2,3,4,10}
Recall the mean of this is 4
Multiply each item by 2 to get {2,4,6,8,20}
The mean is now 8
This is a jump of "times 2" since 8 = 2*4
So,
new mean = 2*(old mean)
The median of the old set was 3
The median of the new set is 6
This is also a jump of "times 2"
6 = 2*3
These ideas help show why the mean and median get scaled by 5/9 when we multiply each temperature by 5/9. The same applies for the IQR as well.
new IQR = (new Q3) - (new Q1)
new IQR = (5/9)*Q3 - (5/9)*(Q1)
new IQR = (5/9)*(Q3 - Q1)
new IQR = (5/9)*(old IQR)
The standard deviation proof is a bit more complicated. But imagine that the scaling operation basically either spreads the data out further (if multiplying by something larger than 1) or shrinks the spread down (if multiplying by some value k such that 0 < k < 1).
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Part C
Answers:
The mean and median decrease by 32 then multiply by 5/9
The standard deviation and IQR are multiplied by 5/9
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Reason:
We combine the ideas of the two previous parts A and B to arrive at the answer shown above.
We can rewrite C = (5/9)(F - 32) as C = (5/9)*F - (5/9)*32 or C = (5/9)*F - 160/9
Think of that as y = mx+b
y = temperature in degrees Celsius
m = 5/9
x = temperature in degrees Fahrenheit
b = -160/9
Then refer to this article
https://stattrek.com/random-variable/transformation
which talks about how linear transformations affect various statistics such as sample mean and standard deviation.
That article reiterates the ideas I mentioned earlier.
Transforms like this are useful if all we care about are the summary statistics. It allows us to skip the tedious busywork of having to recompute each item for each transformation applied. It's also useful for times when we need to convert from one measure system to another.
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