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Statistics and Probability - Scatter graphs and bivariate data

Grade 10IB

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

🔑Concepts

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Bivariate data involves the relationship between two variables, typically plotted as (x,y)(x, y) coordinates on a scatter graph to identify patterns or trends.

A scatter plot showing points distributed with a general upward trend.
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Correlation describes the strength and direction of the relationship: Positive correlation (both variables increase together), Negative correlation (one increases as the other decreases), or No correlation (no discernible pattern).

Diagram showing negative correlation with points falling along a downward sloping line.
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The Line of Best Fit (Trend Line) is a straight line that passes through the 'middle' of the points. It must pass through the Mean Point (xˉ,yˉ)(\bar{x}, \bar{y}). It is used for Interpolation (predicting within the data range) and Extrapolation (predicting outside the data range).

A line of best fit passing through the mean point of a data set.
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Outliers are data points that lie significantly far away from the general trend of the other points. They can affect the position of the line of best fit.

📐Formulae

xˉ=∑xn\bar{x} = \frac{\sum x}{n}

yˉ=∑yn\bar{y} = \frac{\sum y}{n}

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

y=mx+cy = mx + c

💡Examples

Problem 1:

A student records the number of hours spent studying (xx) and the test scores (yy) for 5 students: (2,40),(4,60),(6,70),(8,90),(10,90)(2, 40), (4, 60), (6, 70), (8, 90), (10, 90). Calculate the mean point (xˉ,yˉ)(\bar{x}, \bar{y}) for this data.

Solution:

xˉ=2+4+6+8+105=305=6\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6 yˉ=40+60+70+90+905=3505=70\bar{y} = \frac{40 + 60 + 70 + 90 + 90}{5} = \frac{350}{5} = 70 The mean point is (6,70)(6, 70).

Explanation:

To find the mean point, calculate the arithmetic mean of all xx-coordinates and all yy-coordinates separately.

Problem 2:

Given a line of best fit equation y=5x+20y = 5x + 20 where xx is the number of hours worked and yy is the total earnings. Predict the earnings for someone working 77 hours.

Solution:

Substitute x=7x = 7 into the equation: y=5(7)+20y = 5(7) + 20 y=35+20y = 35 + 20 y=55y = 55 The predicted earnings are 5555.

Explanation:

Interpolation involves substituting a known independent variable value into the linear equation derived from the scatter graph.

Problem 3:

Determine the type of correlation for the following data points: (1,10),(2,8),(3,5),(4,3),(5,1)(1, 10), (2, 8), (3, 5), (4, 3), (5, 1).

Solution:

As xx increases (1→51 \to 5), the values of yy decrease (10→110 \to 1). Therefore, the data shows a strong negative correlation.

Explanation:

Correlation is identified by observing the direction in which yy moves as xx increases. Since yy decreases as xx increases, the gradient of the relationship is negative.

Problem 4:

A researcher investigates the relationship between the outside temperature xx (∘C^\circ\text{C}) and the number of hot drinks sold yy. The data collected is: (5,50),(10,40),(15,30),(20,20),(25,10)(5, 50), (10, 40), (15, 30), (20, 20), (25, 10). Plot the data and determine the equation of the line of best fit if it passes through (0,60)(0, 60) and (30,0)(30, 0).

Scatter plot with a downward sloping line of best fit for temperature vs drinks sold.

Solution:

m=0−6030−0=−6030=−2m = \frac{0 - 60}{30 - 0} = \frac{-60}{30} = -2 c=60c = 60 y=−2x+60y = -2x + 60

Explanation:

The points show a perfect negative linear correlation. Using the coordinates provided for the line of best fit, we calculate the gradient mm and the y-intercept cc to form the linear equation.

Problem 5:

Calculate the mean point (xˉ,yˉ)(\bar{x}, \bar{y}) for the following heights (xx in cm) and weights (yy in kg) of four athletes: (160,60),(170,70),(180,80),(190,90)(160, 60), (170, 70), (180, 80), (190, 90). Use this to verify if the line y=x−100y = x - 100 is a valid line of best fit.

Scatter plot showing height and weight with a mean point and a line of best fit.

Solution:

xˉ=160+170+180+1904=175\bar{x} = \frac{160+170+180+190}{4} = 175 yˉ=60+70+80+904=75\bar{y} = \frac{60+70+80+90}{4} = 75 Mean Point=(175,75)\text{Mean Point} = (175, 75) Substitute into line: 75=175−100  ⟹  75=75\text{Substitute into line: } 75 = 175 - 100 \implies 75 = 75

Explanation:

The mean point is calculated by averaging the xx and yy values separately. Since the mean point satisfies the equation y=x−100y = x - 100, it is a valid candidate for the line of best fit.