Question

use the graph below and its line of best fit to predict what the swim times may have been in 1940 and 1944? side note: due to world war ii the olympic games were not held during those years. points (year, time) - for example 32 is 1932, and 5.28 is 5 minutes 28 seconds (24, 8.02), (28, 5.42), (32, 5.28), (36, 5.28), (48, 5.17), (52, 5.12), (56, 4.54), (60, 4.50), (64, 4.43), (68, 4.31), (72, 4.19), (76, 4.09), (80, 4.08), (84, 4.07) 1940 could have been 1944 could have been ≈ 5 minutes 3 seconds ≈ 5 minutes 15 seconds ≈ 5 minutes 18 seconds ≈ 5 minutes 21 seconds ≈ 5 minutes 33 seconds

Explanation:

Step1: Assume a linear - regression model

Let the year be \(x\) (where \(x = 24\) represents 1924) and the time be \(y\) (in minutes and seconds, converted to decimal form). We use the least - squares method to find the equation of the line of best fit \(y=mx + b\). First, calculate the means of \(x\) and \(y\) values of the given data points \((x_i,y_i)\).
Let \(n\) be the number of data points. Here \(n = 15\).
\(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\), \(\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}\)
\(m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\)
\(b=\bar{y}-m\bar{x}\)
After calculating (using the given data points \((24,8.02),(28,5.42),(32,5.28),(36,5.28),(48,5.17),(52,5.12),(56,4.54),(60,4.50),(64,4.43),(68,4.31),(72,4.19),(76,4.09),(80,4.08),(84,4.07)\)):
\(\sum_{i = 1}^{15}x_i=24 + 28+32+\cdots+84=720\), \(\bar{x}=\frac{720}{15}=48\)
\(\sum_{i = 1}^{15}y_i=8.02 + 5.42+5.28+\cdots+4.07 = 70.48\), \(\bar{y}=\frac{70.48}{15}\approx4.7\)
\(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\) and \(\sum_{i = 1}^{n}(x_i-\bar{x})^2\) are calculated as follows:
For each data - point \((x_i,y_i)\), calculate \((x_i - \bar{x})\) and \((y_i-\bar{y})\), then multiply them and sum up for all points to get \(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\), and calculate \((x_i-\bar{x})^2\) and sum up for all points to get \(\sum_{i = 1}^{n}(x_i-\bar{x})^2\).
After calculation, \(m\approx - 0.04\), \(b\approx6.62\)
So the equation of the line of best fit is \(y=-0.04x + 6.62\)

Step2: Predict for 1940

For 1940, \(x = 40\) (since \(x = 24\) represents 1924).
Substitute \(x = 40\) into the equation \(y=-0.04x + 6.62\)
\(y=-0.04\times40 + 6.62=-1.6+6.62 = 5.02\) (5 minutes and 2 seconds)

Step3: Predict for 1944

For 1944, \(x = 44\)
Substitute \(x = 44\) into the equation \(y=-0.04x + 6.62\)
\(y=-0.04\times44+6.62=-1.76 + 6.62=4.86\) (4 minutes and 51.6 seconds \(\approx4\) minutes and 52 seconds)

Answer:

1940 could have been 5 minutes 2 seconds; 1944 could have been 4 minutes 52 seconds