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Statistics and Probability - Lines of best fit

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Scatter Diagram is a graphical representation of the relationship between two variables, where each point represents a pair of data values.

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Correlation describes the nature of the relationship: Positive correlation means as xx increases, yy increases; Negative correlation means as xx increases, yy decreases.

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The Strength of correlation (Strong, Moderate, or Weak) indicates how closely the data points cluster around a straight line.

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The Line of Best Fit (or Trend Line) is a straight line that best represents the data on a scatter plot. It should pass through the mean point (xˉ,yˉ)(\bar{x}, \bar{y}). Natural variation means roughly half the points should be above the line and half below.

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Interpolation is the process of estimating a value within the range of the given data set. This is generally considered reliable.

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Extrapolation is the process of predicting a value outside the range of the given data. This is less reliable as the trend may not continue.

📐Formulae

Mean of x:xˉ=∑xn\text{Mean of } x: \bar{x} = \frac{\sum x}{n}

Mean of y:yˉ=∑yn\text{Mean of } y: \bar{y} = \frac{\sum y}{n}

Mean Point=(xˉ,yˉ)\text{Mean Point} = (\bar{x}, \bar{y})

Equation of the Line: y=mx+c\text{Equation of the Line: } y = mx + c

Gradient (Slope): m=y2−y1x2−x1\text{Gradient (Slope): } m = \frac{y_2 - y_1}{x_2 - x_1}

💡Examples

Problem 1:

A student records the number of hours spent studying (xx) and the test scores (yy) for five students: (2,40),(4,60),(6,70),(8,80),(10,90)(2, 40), (4, 60), (6, 70), (8, 80), (10, 90). Calculate the mean point (xˉ,yˉ)(\bar{x}, \bar{y}).

Solution:

First, calculate xˉ\bar{x}: xˉ=2+4+6+8+105=305=6\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6. Then, calculate yˉ\bar{y}: yˉ=40+60+70+80+905=3405=68\bar{y} = \frac{40 + 60 + 70 + 80 + 90}{5} = \frac{340}{5} = 68. The mean point is (6,68)(6, 68).

Explanation:

The mean point is the average of all xx-coordinates and all yy-coordinates. Every line of best fit must pass through this point.

Problem 2:

A line of best fit passes through the mean point (6,68)(6, 68) and another point (10,92)(10, 92). Find the equation of the line in the form y=mx+cy = mx + c.

Solution:

Find the gradient mm: m=92−6810−6=244=6m = \frac{92 - 68}{10 - 6} = \frac{24}{4} = 6 Substitute m=6m = 6 and point (6,68)(6, 68) into y=mx+cy = mx + c: 68=6(6)+c68 = 6(6) + c 68=36+c68 = 36 + c c=68−36=32c = 68 - 36 = 32 The equation is y=6x+32y = 6x + 32.

Explanation:

Using the gradient formula and one known point (the mean point), we can determine the specific linear relationship between the variables.

Problem 3:

Using the equation y=6x+32y = 6x + 32, predict the test score for a student who studies for 77 hours and 1515 hours. Identify which prediction is interpolation.

Solution:

For x=7x = 7: y=6(7)+32=42+32=74y = 6(7) + 32 = 42 + 32 = 74. For x=15x = 15: y=6(15)+32=90+32=122y = 6(15) + 32 = 90 + 32 = 122. Since the original data range for study hours was between 22 and 1010, the prediction for 77 hours is interpolation.

Explanation:

Predictions within the observed xx range (2≤x≤102 \le x \le 10) are interpolations. Predictions outside this range are extrapolations and may result in impossible values (like a score of 122122 if the test is out of 100100).

Lines of best fit Grade 9 Notes & Examples | IB Maths