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

The interquartile range (IQR) is found by subtracting the first quartile (Q1) from the third quartile (Q3). Given that Q1 = 12 and Q3 = 25:

$IQR = Q3 - Q1 = 25 - 12 = 13$

To find the median angle, we need to determine the position of the median in the ordered data for the 70° angles. There are 60 students, so the median is at position:

$Median = \frac{60 + 1}{2} = 30.5$

The median value corresponds to the average of the 30th and 31st values in the ordered dataset, which can be estimated around the 70° values, leading to:

$Median \approx 68.5 \degree$

The lower quartile (Q1) represents the 25th percentile of the data. We can identify the position by calculating:

$Q1 = \frac{60 + 1}{4} = 15.25$

This indicates that Q1 is approximately the value at the 15th position, which falls in the range between 60 and 65 degrees, placing Q1 around 63°.

To check for outliers, we first need the upper quartile (Q3) which is given as 75° and the lower quartile (Q1) as 63°. The IQR is 13.

Outlier conditions:

• Upper outlier threshold: $Q3 + 1.5 * IQR = 75 + 1.5 * 13 = 81.5$
• Lower outlier threshold: $Q1 - 1.5 * IQR = 63 - 1.5 * 13 = 56.5$

The maximum is 84°, which is not greater than 81.5, and the minimum is 55°, which is just at the edge but not below 56.5. Thus, there are no outliers.

To create a box plot, plot the minimum (55), lower quartile (63), median (68.5), upper quartile (75), and maximum (84) on the grid provided, connecting these points with a box from Q1 to Q3.

The angles drawn were more accurate at 70° as shown by the median angle value being closer to the target of 70° than the median for the 20° angle. The analysis of the data further indicates tighter clustering of results around 70° compared to 20°.