# Linear Transformations 2.1b

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Linear Transformations 2.1b
Target Goal: I can transform linear data and determine its effect on center and spread. HW: pg 107: 19-21, 23

Changing the Unit of Measurement
Linear Transformations Changes the original variable x into the new variable xnew given by an equation of the form: xnew = a + bx

Linear Transformations xnew = a + bx
Adding the constant a shifts all values of x upward or downward by the same amount. Multiplying by the positive constant b changes the size of the unit of measurement. Adding a constant to each observation does not change the spread. Linear transformations do not change the shape (symmetric, skewed right or left) of a distribution.

Effects of Linear Transformations
Linear transformations: miles/kilometers Fahrenheit/Celsius Multiplying each observation by a positive number b multiplies both measures of center (mean and median) and measures of spread (standard deviation and IQR) by b. Adding the same number a (either positive or negative) to each observation: adds a to measures of center and to quartiles but does not change measures of spread.

Ex: Los Angeles Laker’s Salaries (2000)
= \$4.14 mil, M = \$2.6 mil, s = 4.76 mil

Ex. Los Angeles Laker’s Salaries
Plot (a) stems = millions, Describe distribution: Right skewed with two outliers.

Five-number summary is: \$0.3 mil, \$1.0 mil, \$2.6 mil, \$4.6 mil, \$17.1
(a) Suppose that each member of the team receives a \$100,000 bonus for winning the NBA championship. How will this affect the shape, center and spread? \$100,000 = \$0.1 mill

Five-number summary was: \$0.3 mil, \$1.0 mil, \$2.6 mil, \$4.6 mil, \$17.1
The linear transformation is xnew = a + bx ; xnew = [each value in (a) and the mean and median will all increase by 0.1: plot (b)] xnew = x

Five-number summary was: \$0.3 mil, \$1.0 mil, \$2.6 mil, \$4.6 mil, \$17.1
We added 0.1; Old: mean = \$4.14 mil, M = \$2.6 mil, s = 4.76 mil meannew = \$4.24 mil mediannew = \$2.7 mil snew = no change

Five-number summary was: \$0.3 mil, \$1.0 mil, \$2.6 mil, \$4.6 mil, \$17.1
The new five-number summary is: \$0.4 mil, \$1.1 mil, \$2.7 mil, \$4.7mil, \$17.2 Does the spread change? No

Ex. L.A. Lakers cont. (b) Each player is offered a 10% increase base salary. The linear transformation is xnew = a + bx xnew = x

Calculate the following:
mean = \$4.14(1.10) = median = \$2.6(1.10) = Spread and IQR also increased by 10%. \$4.55 \$2.86

Five-number summary was: \$0.3 mil, \$1.0 mil, \$2.6 mil, \$4.6 mil, \$17.1
The new five-number summary is: [see graph (c)] xnew = x \$0.33mil, \$1.10 mil, \$2.86 mil, \$5.06 mil, \$18.81mil

Example: Cockroaches 1.4 2.2 1.1 1.6 1.2 Mean = 7.5/5 = 1.5 s = .436
Maria measures the lengths of 5 cockroaches that she finds at school. Here are her results (in inches): Find the mean and standard deviation of Maria’s measurements. (use calc) Enter data into L1, STAT:CALC:1-VAR Stats: enter: L1 or, 2nd STAT(LIST):MATH:stdDev(7): L1 Mean = 7.5/5 = 1.5 s = .436

Example: Cockroaches xnew = a + bx
Maria’s science teacher is furious to discover that she has measured the cockroach lengths in inches rather than centimeters (There are 2.54 cm in 1 inch). She gives Maria two minutes to report the mean and standard deviation of the 5 cockroaches in centimeters. Maria succeeded. Will you? xnew = a + bx

To obtain mean and s in centimeters, multiply the results in inches by 2.54:
New Mean = 1.5 x2.54 = 3.81cm New s = .436 x 2.54 = 1.107 cm

c. Considering the 5 cockroaches that Maria found as a small sample from the population of all cockroaches at her school, what would you estimate as the average length of the population of cockroaches? How sure of the estimate are you? The average cockroach length can be est. by the mean length of the sample ; 1.5 inches. This is a questionable estimate because our sample is so small.

Example: mean and s are not enough
The mean x bar and standard deviation s measure center and spread but are not a complete description of a distribution. Data sets with different shapes can have the same mean and standard deviation.

To demonstrate this fact, use your calculator to find the mean and s for the following two small data sets: Set A: mean = 7.501, s = 2.032 Set B: mean = 7.501, s = 2.031

Then make a stem-plot of each and comment on the shape of each distribution.
Set A is skewed to the left. Set B has a high outlier.