Abe plays for his school's cricket team. The number of runs scored by Abe in the eight games that he batted in, is shown below. (Abe was given out in all of the games.) 21 8 19 7 15 32 14 12 1.1.1 Determine the variance of the data. 1.1.2 Calculate the standard deviation 1.1.3 Interpret what the standard deviation tells us about the spread of the data. 1.2 The following table shows the number of hours studied and corresponding test scores of five students: Hours Studied (x) Test Scores (y) 2 50 4 65 6 70 8 85 10 90 1.2.1 Plot a scatter plot in ADDENDUM A using the given data 1.2.2 Describe the type of correlation shown in the scatter plot. 1.2.3 Draw an intuitive line of best fit on your graph and explain its significance. 1.2.4 Find the equation of the regression line in the form y = bx + a. Use one of the two tables on ADDENDUM B (attached) and then complete all the calculations. 1.2.5 Predict the test score for a student who studies for 7 hours using your equation. 1.2.6 Explain how regression analysis helps in making predictions.

Question 1

Answer · 2026-03-22T14:49:22.097Z

1.1 Determine the variance of the data (runs: 12, 14, 15, 18, 19, 20, 24, 32) Step 1: Calculate x_i and x_i^2. x_i = 12 + 14 + 15 + 18 + 19 + 20 + 24 + 32 = 154 x_i^2 = 12^2 + 14^2 + 15^2 + 18^2 + 19^2 + 20^2 + 24^2 + 32^2 = 144 + 196 + 225 + 324 + 361 + 400 + 576 + 1024 = 3250 Step 2: Calculate the mean x. x = ( x_i)/(n) = (154)/(8) = (77)/(4) Step 3: Calculate the population variance ^2. ^2 = ( x_i^2)/(n) - x^2 = (3250)/(8) - ( (77)/(4) )^2 = (6500)/(16) - (5929)/(16) = (571)/(16) (571)/(16) 1.2 Interpret the standard deviation Step 1: Calculate the standard deviation . = sqrt(^2) = sqrt((571)/(16)) = sqrt(571)4 ≈ 5.97 The standard deviation of approximately 5.97 runs shows the typical deviation of runs scored from the mean of 19.25 runs, measuring the spread or consistency of A's performance across the 8 games. Data for hours studied (x) and test scores (y): (0, 50), (2, 70), (5, 85) 1.3 Plot the scatter diagram The scatter plot has points at (0, 50), (2, 70), and (5, 85), showing an upward trend. 1.4 Describe the type of correlation Step 1: Observe the pattern. As hours studied increase from 0 to 5, test scores increase from 50 to 85. There is a strong positive linear correlation between hours studied and test scores. 1.5 Find the equation of the best fit line y = mx + c Step 1: Calculate the sums. n = 3, x = 0 + 2 + 5 = 7, y = 50 + 70 + 85 = 205 x^2 = 0^2 + 2^2 + 5^2 = 0 + 4 + 25 = 29 xy = 0 · 50 + 2 · 70 + 5 · 85 = 0 + 140 + 425 = 565 Step 2: Calculate slope m. m = (n xy - ( x)( y))/(n x^2 - ( x)^2) = (3 · 565 - 7 · 205)/(3 · 29 - 7^2) = (1695 - 1435)/(87 - 49) = (260)/(38) = (130)/(19) Step 3: Calculate intercept c. c = ( y - m x)/(n) = (205 - 130)/(19) · 73 = (205 - 910)/(19)3 = (3895 - 910)/(19)3 = (2985)/(19)3 = (2985)/(57) = (995)/(19) The equation is y = (130)/(19) x + (995)/(19). y = (130)/(19)x + (995)/(19) 1.6 Predict the test score for B who studied 3 hours Step 1: Substitute x = 3 into the regression equation. y = (130)/(19) · 3 + (995)/(19) = (390 + 995)/(19) = (1385)/(19) ≈ 72.89 Predicted test score is approximately 73 marks. 73 1.7 Explain how regression analysis makes predictions Regression analysis finds the best fit line y = mx + c by minimizing the sum of squared differences between observed y values and predicted y values on the line. Predictions are made by substituting a given x value into this equation to find y.

Question 2

1.1 Determine the variance of the data (runs: 12, 14, 15, 18, 19, 20, 24, 32) Step 1: Calculate x_i and x_i^2. x_i = 12 + 14 + 15 + 18 + 19 + 20 + 24 + 32 = 154 x_i^2 = 12^2 + 14^2 + 15^2 + 18^2 + 19^2 + 20^2 + 24^2 + 32^2 = 144 + 196 + 225 + 324 + 361 + 400 + 576 + 1024 = 3250 Step 2: Calculate the mean x. x = ( x_i)/(n) = (154)/(8) = (77)/(4) Step 3: Calculate the population variance ^2. ^2 = ( x_i^2)/(n) - x^2 = (3250)/(8) - ( (77)/(4) )^2 = (6500)/(16) - (5929)/(16) = (571)/(16) (571)/(16) 1.2 Interpret the standard deviation Step 1: Calculate the standard deviation . = sqrt(^2) = sqrt((571)/(16)) = sqrt(571)4 ≈ 5.97 The standard deviation of approximately 5.97 runs shows the typical deviation of runs scored from the mean of 19.25 runs, measuring the spread or consistency of A's performance across the 8 games. Data for hours studied (x) and test scores (y): (0, 50), (2, 70), (5, 85) 1.3 Plot the scatter diagram The scatter plot has points at (0, 50), (2, 70), and (5, 85), showing an upward trend. 1.4 Describe the type of correlation Step 1: Observe the pattern. As hours studied increase from 0 to 5, test scores increase from 50 to 85. There is a strong positive linear correlation between hours studied and test scores. 1.5 Find the equation of the best fit line y = mx + c Step 1: Calculate the sums. n = 3, x = 0 + 2 + 5 = 7, y = 50 + 70 + 85 = 205 x^2 = 0^2 + 2^2 + 5^2 = 0 + 4 + 25 = 29 xy = 0 · 50 + 2 · 70 + 5 · 85 = 0 + 140 + 425 = 565 Step 2: Calculate slope m. m = (n xy - ( x)( y))/(n x^2 - ( x)^2) = (3 · 565 - 7 · 205)/(3 · 29 - 7^2) = (1695 - 1435)/(87 - 49) = (260)/(38) = (130)/(19) Step 3: Calculate intercept c. c = ( y - m x)/(n) = (205 - 130)/(19) · 73 = (205 - 910)/(19)3 = (3895 - 910)/(19)3 = (2985)/(19)3 = (2985)/(57) = (995)/(19) The equation is y = (130)/(19) x + (995)/(19). y = (130)/(19)x + (995)/(19) 1.6 Predict the test score for B who studied 3 hours Step 1: Substitute x = 3 into the regression equation. y = (130)/(19) · 3 + (995)/(19) = (390 + 995)/(19) = (1385)/(19) ≈ 72.89 Predicted test score is approximately 73 marks. 73 1.7 Explain how regression analysis makes predictions Regression analysis finds the best fit line y = mx + c by minimizing the sum of squared differences between observed y values and predicted y values on the line. Predictions are made by substituting a given x value into this equation to find y.

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