2.5 – Modeling Real World Data:
2.5 – Modeling Real World Data:
Using Scatter Plots
Ex.1 The table below shows the median
selling price of new, privately-owned,
one-family houses for some recent years.
Ex.1 The table below shows the median
selling price of new, privately-owned,
one-family houses for some recent years.
Year
1990 1992
1994 1996 1998
Price
($1000)
122.9 121.5 130
140
152.5
2000
169
Ex.1 The table below shows the median
selling price of new, privately-owned,
one-family houses for some recent years.
Year
1990 1992
1994 1996 1998
Price
($1000)
122.9 121.5 130
140
a. Make a scatter plot of the data.
152.5
2000
169
Years Since 1990
Price
Years Since 1990
Price
($1000)
Years Since 1990
Median House Prices
Price
($1000)
Years Since 1990
Median House Prices
Price
($1000)
0
Years Since 1990
Median House Prices
Price
($1000)
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
b. Make a line of fit.
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
b. Make a line of fit.
10
Median House Prices
Price
($1000)
140
120
0
2
4
6
8
Years Since 1990
b. Make a line of fit.
10
c. Find a prediction equation for line of fit.
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1
x2 - x1
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130
x2 - x1
8–4
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5
x2 - x1
8–4
4
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63
x2 - x1
8–4
4
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63
x2 - x1
8–4
4
*So use x1 = 4
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63
x2 - x1
8–4
4
*So use x1 = 4, y1 = 130
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63
x2 - x1
8–4
4
*So use x1 = 4, y1 = 130, and m ≈ 5.63
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63
x2 - x1
8–4
4
*So use x1 = 4, y1 = 130, and m ≈ 5.63
y – y1 = m(x – x1)
c. Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63
x2 - x1
8–4
4
*So use x1 = 4, y1 = 130, and m ≈ 5.63
y – y1 = m(x – x1)
c.
Find a prediction equation for line of fit.
*Use the best two ordered pairs from b. to
find the slope for the line!
(4, 130) and (8, 152.5)
m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63
x2 - x1
8–4
4
*So use x1 = 4, y1 = 130, and m ≈ 5.63
y – y1 = m(x – x1)
y – 130 = 5.63(x – 4)
y – 130 = 5.63(x) – 5.63(4)
y – 130 = 5.63x – 22.52
y = 5.63x + 107.48
d. Predict the price in 2020.
d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
y = 5.63x + 107.48
d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
y = 5.63x + 107.48
y = 5.63(30) + 107.48
d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
y = 5.63x + 107.48
y = 5.63(30) + 107.48
y = 168.9 + 107.48
d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
y = 5.63x + 107.48
y = 5.63(30) + 107.48
y = 168.9 + 107.48
y = 276.38
d. Predict the price in 2020.
2020 means when x=30 (yrs after 1990)
*Plug 30 in for x!
y = 5.63x + 107.48
y = 5.63(30) + 107.48
y = 168.9 + 107.48
y = 276.38
So, in 2020 the price will be $276,380.