Question

1. the following data shows the world record times for the 200 meter race for flying start machines (a kind of bicycle). what would you predict the record time to be in 1988?

Explanation:

Step1: Assign variables

Let \(x\) be the number of years since 1974. So for 1974, \(x = 0\); for 1975, \(x=1\); for 1976, \(x = 2\); for 1977, \(x=3\); for 1979, \(x = 5\); for 1985, \(x=11\); for 1986, \(x = 12\); for 1992, \(x=18\). Let \(y\) be the record time.

Step2: Assume a linear regression model \(y=ax + b\)

We use the formula for the slope \(a=\frac{n\sum(xy)-\sum x\sum y}{n\sum(x^{2})-(\sum x)^{2}}\) and \(b=\frac{\sum y - a\sum x}{n}\), where \(n = 9\) (the number of data - points).
First, calculate \(\sum x\):
\(\sum x=0 + 1+2 + 3+5+11+12+18\)
\(\sum x=52\)
Calculate \(\sum y\):
\(\sum y=10.4+10.0 + 9.4+9.1+8.8+7.2+6.8+6.5\)
\(\sum y=68.2\)
Calculate \(\sum(xy)\):
\(\sum(xy)=0\times10.4+1\times10.0+2\times9.4+3\times9.1+5\times8.8+11\times7.2+12\times6.8+18\times6.5\)
\(\sum(xy)=0 + 10.0+18.8+27.3+44.0+79.2+81.6+117.0\)
\(\sum(xy)=377.9\)
Calculate \(\sum(x^{2})\):
\(\sum(x^{2})=0^{2}+1^{2}+2^{2}+3^{2}+5^{2}+11^{2}+12^{2}+18^{2}\)
\(\sum(x^{2})=0 + 1+4+9+25+121+144+324\)
\(\sum(x^{2})=638\)

Then, calculate the slope \(a\):

$$
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$$

Calculate the intercept \(b\):

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$$

The regression equation is \(y=-0.0478x + 7.854\)

For 1988, \(x = 1988 - 1974=14\)

Substitute \(x = 14\) into the equation:
\(y=-0.0478\times14 + 7.854\)
\(y=-0.6692+7.854\)
\(y = 7.1848\approx7.2\)

Answer:

The predicted record time in 1988 is approximately \(7.2\) seconds.