Correlation and regression Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Correlation and regression questions. See exactly how to solve problems on regression line, prediction, substitution, inverse prediction.

regression linepredictionsubstitutioninverse predictionrearrangegradient
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A cafe owner records data over several days. They record the daily maximum temperature (°C) as xx and the number of iced drinks sold as yy. The regression line of yy on xx is y=10+2xy = 10 + 2x, valid for 0≤x≤200 \le x \le 20. Use the model to predict yy when x=4x = 4.

Worked solution

  1. Write down the regression line of y on x.

    y=10+2xy = 10 + 2x

    Begin from the given equation of the line of y on x.

  2. Substitute x = 4 into the equation.

    y=10+2×4y = 10 + 2 \times 4

    Replace x with the required value.

  3. Evaluate to obtain the predicted value.

    y=18y = 18

    This is the value of y predicted by the model.

Answer
y=18y = 18
Question 2
2 markseasy
A teacher compares revision data across a class. They record the number of hours revised as xx and the test score (%) as yy. The regression line of yy on xx is y=20+3xy = 20 + 3x, valid for 2≤x≤322 \le x \le 32. Use the model to predict yy when x=9x = 9.

Worked solution

  1. Write down the regression line of y on x.

    y=20+3xy = 20 + 3x

    Begin from the given equation of the line of y on x.

  2. Substitute x = 9 into the equation.

    y=20+3×9y = 20 + 3 \times 9

    Replace x with the required value.

  3. Evaluate to obtain the predicted value.

    y=47y = 47

    This is the value of y predicted by the model.

Answer
y=47y = 47
Question 3
2 markseasy
A garage studies used cars of one model. They record the age of the car (years) as xx and the resale value (£1000s) as yy. The regression line of yy on xx is y=5−1.5xy = 5 - 1.5x, valid for 5≤x≤205 \le x \le 20. Use the model to predict yy when x=8x = 8.

Worked solution

  1. Write down the regression line of y on x.

    y=5−1.5xy = 5 - 1.5x

    Begin from the given equation of the line of y on x.

  2. Substitute x = 8 into the equation.

    y=5−1.5×8y = 5 - 1.5 \times 8

    Replace x with the required value.

  3. Evaluate to obtain the predicted value.

    y=−7y = -7

    This is the value of y predicted by the model.

Answer
y=−7y = -7
Question 4
2 markseasy
A biologist studies plants in a greenhouse. They record the weekly water given (litres) as xx and the plant height (cm) as yy. The regression line of yy on xx is y=50+4xy = 50 + 4x, valid for 1≤x≤261 \le x \le 26. Use the model to predict yy when x=11x = 11.

Worked solution

  1. Write down the regression line of y on x.

    y=50+4xy = 50 + 4x

    Begin from the given equation of the line of y on x.

  2. Substitute x = 11 into the equation.

    y=50+4×11y = 50 + 4 \times 11

    Replace x with the required value.

  3. Evaluate to obtain the predicted value.

    y=94y = 94

    This is the value of y predicted by the model.

Answer
y=94y = 94
Question 5
2 markseasy
A marketing team reviews weekly figures. With the advertising spend (£100s) as xx and the units sold as yy, the regression line is y=30+2.5xy = 30 + 2.5x. Estimate the value of xx for which y=70y = 70.

Worked solution

  1. Set y equal to the given value in the equation.

    70=30+2.5x70 = 30 + 2.5x

    Substitute the known y-value into the line.

  2. Rearrange to make x the subject.

    x=70−302.5x = \frac{70 - 30}{2.5}

    Subtract the intercept and divide by the gradient.

  3. Evaluate x.

    x=16x = 16

    This is the required value of x.

Answer
x=16x = 16

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