Question

1. marsha recorded the time it took six children of different ages to run one lap around the track. using the line of best fit, which is closest to the length of time it should take marsha’s 8 year old son to run one lap?

age of child	time (seconds)
6	205
10	132
13	119
14	108
17	62

a. 168 seconds
b. 176 seconds
c. 182 seconds
d. 190 seconds

Explanation:

Step1: Calculate mean of x and y

Let age = \(x\), time = \(y\).
\(\bar{x} = \frac{4+6+10+13+14+17}{6} = \frac{64}{6} \approx 10.67\)
\(\bar{y} = \frac{230+205+132+119+108+62}{6} = \frac{856}{6} \approx 142.67\)

Step2: Compute slope (m)

\(m = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2}\)
Numerator:
\((4-10.67)(230-142.67)+(6-10.67)(205-142.67)+(10-10.67)(132-142.67)+(13-10.67)(119-142.67)+(14-10.67)(108-142.67)+(17-10.67)(62-142.67)\)
\(\approx (-6.67)(87.33)+(-4.67)(62.33)+(-0.67)(-10.67)+(2.33)(-23.67)+(3.33)(-34.67)+(6.33)(-80.67)\)
\(\approx -582.5-291.1+7.17-55.25-115.5-510.7 \approx -1547.9\)

Denominator:
\((4-10.67)^2+(6-10.67)^2+(10-10.67)^2+(13-10.67)^2+(14-10.67)^2+(17-10.67)^2\)
\(\approx 44.49+21.81+0.45+5.43+11.09+40.07 \approx 123.34\)

\(m \approx \frac{-1547.9}{123.34} \approx -12.55\)

Step3: Find y-intercept (b)

\(b = \bar{y} - m\bar{x} \approx 142.67 - (-12.55)(10.67) \approx 142.67 + 133.9 \approx 276.57\)

Step4: Predict time for age=8

Line equation: \(y = -12.55x + 276.57\)
For \(x=8\): \(y \approx -12.55(8) + 276.57 = -100.4 + 276.57 \approx 176.17\)

Answer:

B. 176 seconds