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### Course: Statistics and probability>Unit 4

Lesson 3: Effects of linear transformations

# Transforming data problem

It is very common to take data and apply the same transformation to every data point in the set. For example, we may take a set of temperatures taken in degrees fahrenheit and convert them all to degrees celsius. How would this conversion impact the measures of center of spread in the data set? Let's look at a simpler example to think about this situation.

## Part 1: Adding a constant

Five friends took a $10$ question multiple choice quiz in class. Their raw scores on the quiz are shown in the dotplot below along with summary statistics.
$\overline{x}$${s}_{x}$$\text{M}$$\text{IQR}$range
Scores$8$$1.41$$8$$3$$4$
The teacher told everyone that she would add $1$ to every student's score as extra credit. Their new scores are shown below.
Problem A (part 1)
Find the summary statistics for the new scores.
Hint: You don't need to calculate these using their formulas. Look at the dotplots to visually assess the center and spread of each distribution.
$\overline{x}$${s}_{x}$$\text{M}$$\text{IQR}$range
Raw scores$8$$1.41$$8$$3$$4$
New scores

problem b (part 1)
Describe the effects adding the constant had on the measures of center and spread.

## Part 2: Multiplying a constant

The teacher always scores her quizzes out of $100$ points. For this $10$-question quiz, she multiplies the new scores by $10$ to get the students' final grades which are shown in the dotplot below.
$\overline{x}$${s}_{x}$$\text{M}$$\text{IQR}$range
Scores$8$$1.41$$8$$3$$4$
New scores$9$$1.41$$9$$3$$4$
Final grades$?$$?$$?$$?$$?$
problem a (part 2)
Find the mean and median of the final grades.
Hint: You don't need to calculate these using their formulas. Think about how the center has or hasn't changed.
mean =
median =

problem b (part 2)
Find the range of the final grades.
range =

problem c (part 2)
Considering what happened to the range, guess what the IQR and standard deviation are for the final grades.
Hint: You don't need to calculate these using their formulas. Think about how the spread has or hasn't changed.
$\text{IQR}=$
${s}_{x}=$

problem d (part 2)
Describe the effects multiplying these data by a constant had on the measures of center and spread.

## Part 3: Putting it all together

Let's look at a temperature conversion example. Suppose a set of temperature measurements has a mean of ${104}^{\circ }\text{F}$ and a standard deviation of ${9}^{\circ }\text{F}$, and we convert all of the temperatures to degrees celsius.
Here's the conversion formula: ${}^{\circ }\text{C}=\left({}^{\circ }\text{F}-32\right)×\frac{5}{9}$
problem A (part 3)
What is the mean of the temperatures in degrees celsius?
mean =
${}^{\circ }\text{C}$

problem B (part 3)
What is the standard deviation of the temperatures in degrees celsius?
standard deviation =
${}^{\circ }\text{C}$

## Want to join the conversation?

• Why subtracting does not impact the standard deviation?

Could anyone break it down?
• Shifting all of the data points left or right on a number line doesn't impact the standard deviation of a data set. This is because standard deviation is a measure of how spread out data points are. Because adding and subtracting don't change the spread of the data, the standard deviation doesn't change. This also holds true for the range and IQR.
• I'm confused about why this is the answer. What is the math/logic behind leaving one part of the conversion problem out of the scenario when dealing with SD? I see that it works, and get that we're measuring distances from the mean in degrees, not actual degrees, so subtracting 32 wouldn't make much sense and it would make the answer negative, which doesn't work for SD. But why does the multiplying by 5/9 part work? I guess what I'm confused about is what is the 5/9ths part to the temperature conversion formula?
• One way to think about this is to compare to distance measurements. If you look at a ruler, you will see that inches are wider than centimeters. If you look at a thermometer, you will see that degrees Celsius are bigger (sort of "wider") than degrees Fahrenheit.
A ruler starts counting both inches and centimeters in the same place, so the conversion formula only involves multiplication (0 inches = 0 cm; 1 inch = 2.5 cm, 2 inches = 5 cm; etc). Nothing would be added or subtracted from the mean or the SD.

But a thermometer starts counting in different places: 0 C = 32 F. That's why we add or subtract 32 when converting temperatures (how much warmer is it than the freezing point of water). Mean is a temperature (how warm it is on an average day), so we have to add or subtract 32 when converting between C and F.
Standard deviation does not measure how hot it is, but rather how different a set of days are from the average day. SD is a measure of spread. (In different places, the same "width" of spread could be around a mean temperature 0 C, 20 C, or 1000 C.) Because of that, the only thing that matters is that a degree Celsius is 9/5 "wider" than a degree Fahrenheit, aka, a degree Fahrenheit is 5/9 "narrower" than a degree Celsius. That's why we only multiply the units, we don't add or subtract.

Another comparison to reinforce the point: The Kelvin and Celsius scales use the same size ("width") units, but zero Kelvin is "absolute zero", the coldest possible temperature. Zero Celsius is the freezing point of water. Same units, just shifted by hundreds of degrees. What happens to the mean and standard deviation when you convert between Kelvin and Celsius? [For an explanation of absolute zero, take a physics course; here just trust me it's waaaay colder than 0 C. Right now, focus on the math, which is all you need to answer my questions.]
[pause to think, then scroll down for answers and a new question]

Answers: No multiplication involved. The mean shifts but the SD stays the same. Why?
• How do you arrive at 3 for the IQR? Quatile one equals 7 and quartile three equals 9, so wouldn't that make the IQR equal to 2?
• This problem is just asking for the S.D. in degrees Celsius. If we know the S.D. is in degrees Fahrenheit, all we need to do is convert it to celsius. Why are you removing the 32. This is not about scaling this is just about converting. If the S.D. was 9F the SD is not actually 5C because 9F (the original S.D.) is not equal to 5C
• The measures of spread are affected by multiplying/deviding but not by adding/subtracting
• why do i not subtract the 32
• You don't subtract the 32 when solving for the standard deviation because the standard deviation in statistics is affected by multiplication/divison data tranformations
• I don't understand how to convert a standard deviation of 9F to degrees celsius.

C = (F-32) * 5/9

Why we can replace (F-32) with a standard deviation of 9F?
Why we don't calculate using the formula? : C = (9F-32)*5/9
• Converting a standard deviation from Fahrenheit to Celsius involves applying the same transformation as for individual data points. In this case, since the conversion from Fahrenheit to Celsius is linear, you can directly substitute the standard deviation in Fahrenheit into the conversion formula without needing to recalculate it. The formula C = (F - 32) * 5/9 represents the conversion of a single temperature value, where (F - 32) is the difference in Fahrenheit and then multiplied by the scale factor 5/9 to convert it to Celsius. Therefore, substituting the standard deviation of 9°F directly into (F - 32) in the conversion formula gives you the standard deviation in Celsius.
(1 vote)
• ts was tricky ngl
• Hi!

Is the range affected when a set is shifted?

Also, for the last problem concerning standard deviation, why don't we subtract 32? Doesn't that result in an incorrect conversion?

Thanks!
• The dataset is {6,7,8,9,10}.
So IQR=IQ3-IQ1~9.5-6.5=3