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the table shows the femur lengths (in centimeters) and heights (in cent…

Question

the table shows the femur lengths (in centimeters) and heights (in centimeters) of several people. do the data show a linear relationship? if so, write an equation of a line of fit and use it to (a) estimate the height of a person whose femur is 35 centimeters long? (b) estimate the femur length if a persons height is 190 centimeters.

femur length (cm) xheight (cm) y
45183
32151
50195
37162
41174
30141
34151
47185
45182

equations:

show the work for (a)

show the work for (b)

Explanation:

Step1: Calculate slope and y - intercept

Let \(x\) be femur length and \(y\) be height. First, find the mean of \(x\) values \(\bar{x}=\frac{40 + 45+32+50+37+41+30+34+47+45}{10}=39.1\) and the mean of \(y\) values \(\bar{y}=\frac{170 + 183+151+195+162+174+141+151+185+182}{10}=169.4\).
The slope \(m=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}\).
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})(y_{i}-\bar{y})=(40 - 39.1)(170 - 169.4)+(45 - 39.1)(183 - 169.4)+\cdots+(45 - 39.1)(182 - 169.4)=574.9\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=(40 - 39.1)^{2}+(45 - 39.1)^{2}+\cdots+(45 - 39.1)^{2}=288.9\)
\(m=\frac{574.9}{288.9}\approx2\)
The y - intercept \(b=\bar{y}-m\bar{x}=169.4-2\times39.1 = 169.4 - 78.2=91.2\).
The equation of the line of fit is \(y = 2x+91.2\).

Step2: Solve part (a)

Substitute \(x = 35\) into \(y = 2x+91.2\).
\(y=2\times35 + 91.2=70+91.2 = 161.2\) cm.

Step3: Solve part (b)

Set \(y = 190\) in \(y = 2x+91.2\).
\(190=2x+91.2\)
\(2x=190 - 91.2\)
\(2x=98.8\)
\(x = 49.4\) cm.

Answer:

Equations: \(y = 2x+91.2\)
(a) 161.2 cm
(b) 49.4 cm