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The data in the chart shows the maximum and corresponding resting heart rate of a healthy 20-year old man. Fit aregression line to the data. Then find the coefficient of correlation. Finally, use the regression line to predict themaximum heart rate for a resting heart rate of 40, 55 and 76.60 70 8050resting90maximum 162 164 166 168 170Choose the correct regression line.O A. y 0.20x 152.0O B. y 152.0x 0.20O C. y 0.20x 152.0O D. y 152.0

Question
The data in the chart shows the maximum and corresponding resting heart rate of a healthy 20-year old man. Fit a
regression line to the data. Then find the coefficient of correlation. Finally, use the regression line to predict the
maximum heart rate for a resting heart rate of 40, 55 and 76.
60 70 80
50
resting
90
maximum 162 164 166 168 170
Choose the correct regression line.
O A. y 0.20x 152.0
O B. y 152.0x 0.20
O C. y 0.20x 152.0
O D. y 152.0
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The data in the chart shows the maximum and corresponding resting heart rate of a healthy 20-year old man. Fit a regression line to the data. Then find the coefficient of correlation. Finally, use the regression line to predict the maximum heart rate for a resting heart rate of 40, 55 and 76. 60 70 80 50 resting 90 maximum 162 164 166 168 170 Choose the correct regression line. O A. y 0.20x 152.0 O B. y 152.0x 0.20 O C. y 0.20x 152.0 O D. y 152.0

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Step 1

The given data is

Resting
Maximum
80
50
60
70
90
162
164
166
168
170
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Resting Maximum 80 50 60 70 90 162 164 166 168 170

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Step 2

We know that,

For a given set of data the equation of the regression line is given by,
у %3 ах + b
ху — ху
(x)2 x2
where,
and
b yax
a=
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For a given set of data the equation of the regression line is given by, у %3 ах + b ху — ху (x)2 x2 where, and b yax a=

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Step 3

Let’s calculate a and b to...

ΣΥ-1 Χ.
X
50 6070 80 + 90
= 70
5
n
ΣΕ1
У
162 164166 + 168 + 170
i=1
166
5
n
Σ-1xν.
ху
50 x 162 60 x 164 +70 x 166 +80 x 168 + 90 x 170
i=1
5
n
= 11660
Σ-
x2
2
502 602702 802 902
i=1
5100
5
n
(70)(166) (11660)
(70)2 (5100)
ху — ху
(x)2 -x2
0.2
a=
b%3D ў— ах
= (166) - (0.2)(70)
152
Непсе,
the regression line will be y = 0.2x152
Hence,
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ΣΥ-1 Χ. X 50 6070 80 + 90 = 70 5 n ΣΕ1 У 162 164166 + 168 + 170 i=1 166 5 n Σ-1xν. ху 50 x 162 60 x 164 +70 x 166 +80 x 168 + 90 x 170 i=1 5 n = 11660 Σ- x2 2 502 602702 802 902 i=1 5100 5 n (70)(166) (11660) (70)2 (5100) ху — ху (x)2 -x2 0.2 a= b%3D ў— ах = (166) - (0.2)(70) 152 Непсе, the regression line will be y = 0.2x152 Hence,

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