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Each of 60 students was asked to draw a 20° angle without using a protractor - Edexcel - A-Level Maths Statistics - Question 1 - 2015 - Paper 1

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Each of 60 students was asked to draw a 20° angle without using a protractor. The size of each angle drawn was measured. The results are summarised in the box plot b... show full transcript

Worked Solution & Example Answer:Each of 60 students was asked to draw a 20° angle without using a protractor - Edexcel - A-Level Maths Statistics - Question 1 - 2015 - Paper 1

Step 1

Find the range for these data.

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Answer

To find the range, subtract the minimum value from the maximum value. From the box plot, the minimum value is 9 and the maximum value is 48. Therefore, the range is calculated as:

Range=MaximumMinimum=489=39Range = Maximum - Minimum = 48 - 9 = 39

Step 2

Find the interquartile range for these data.

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Answer

The interquartile range (IQR) is found by subtracting the first quartile (Q1) from the third quartile (Q3). From the box plot, Q3 is 25 and Q1 is 12. Thus, the IQR is calculated as:

IQR=Q3Q1=2512=13IQR = Q3 - Q1 = 25 - 12 = 13

Step 3

Use linear interpolation to estimate the size of the median angle drawn. Give your answer to 1 decimal place.

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To find the median for the 70° angles, we need the combined cumulative frequency:

  • 55 ≤ a < 60: 6 students
  • 60 ≤ a < 65: 15 students (Cumulative: 21)
  • 65 ≤ a < 70: 11 students (Cumulative: 32)
  • 70 ≤ a < 75: 8 students (Cumulative: 40)
  • 75 ≤ a < 80: 7 students (Cumulative: 47)
  • 80 ≤ a < 85: 7 students (Cumulative: 54)

The median position is at (60/2) = 30. The median is found by interpolating between the 30th and 31st values: Since 32 students are less than 70°, we find that the median angle is approximately:

Median = 65 + rac{(30 - 21)}{(32 - 21)} imes (70 - 65) = 65 + 4.5 = 69.5°

Thus, the estimated median angle is approximately 69.5°.

Step 4

Show that the lower quartile is 63°.

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Answer

To verify the lower quartile (Q1), we look at the cumulative frequencies:

  • The first 15 (Q1 position, which corresponds to the 15th student) lie between 60-65°.
  • Continuing to 6 more (Cumulative: 21), we count further up to find that the angle at Q1 is 63° since the lower 25% of 60 is 15 students and their median is between 60 and 65 degrees.

Thus, it's shown that Q1 = 63°.

Step 5

Show that there are no outliers for these data.

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Answer

To determine outliers:

  • Calculate 1.5 * IQR = 1.5 * 13 = 19.5
  • Calculate the lower limit = Q1 - 19.5 = 63 - 19.5 = 43.5
  • Calculate the upper limit = Q3 + 19.5 = 75 + 19.5 = 94.5

We can see all angles (55 to 84) lie within [43.5, 94.5]. Therefore, there are no outliers in these data.

Step 6

Draw a box plot for these data on the grid on page 3.

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To draw the box plot:

  • The minimum value is 55°, the lower quartile (Q1) is 63°, the median is 69.5°, the upper quartile (Q3) is 75°, and the maximum is 84°. Plot these points on the grid and connect Q1 and Q3 with a box, marking the median within this box, and extending lines to the minimum and maximum to complete the plot.

Step 7

State which angle the students were more accurate at drawing. Give reasons for your answer.

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Answer

The median of the 70° angles (approximately 69.5°) is closer to 70° compared to the median of the 20° angles (which is 20°). Therefore, the students were more accurate at drawing the 70° angle as the discrepancies are smaller, and the IQR for 70° is also less than for 20°. This indicates fewer variations in their responses.

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