2.5 – Modeling Real World Data:. Using Scatter Plots.
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Transcript of 2.5 – Modeling Real World Data:. Using Scatter Plots.
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 2000
Price ($1000)
122.9 121.5 130 140 152.5 169
Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.
a. Make a scatter plot of the data.
Year 1990 1992 1994 1996 1998 2000
Price ($1000)
122.9 121.5 130 140 152.5 169
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 10
Years Since 1990
Median House Prices
Price
($1000)
0 2 4 6 8 10
Years Since 1990
Median House Prices
Price
($1000)
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
b. Make a line of fit.
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
b. Make a line of fit.
Median House Prices
Price
($1000)
140
1200 2 4 6 8
10
Years Since 1990
b. Make a line of fit.
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.
