Question

1. 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 below 10 20 30 40 50 size of angle (a) Find the range for these data. (b) Find the interquartile range for these data. The students were then asked to draw a 70° angle. The results are summarised in the table below Angle, a, (degrees) Number of students 60#0<65 65 a < 70 70 a < 75 75 #0 < 80 80#0<85 15 13 7 (c) Use linear interpolation to estimate the size of the median angle drawn. Give your answer to I decimal place (d) Show that the lower quartile is 63° For these data, the upper quartile is 75, the minimum is 55° and the maximum is 84 An outlier is an observation that falls either more than 1.5 x (interquartile range) above the upper quartile or more than 1.5% (interquartile range) below the lower quartile. (e) (i) Show that there are no outliers for these data. (ii) Draw a box plot for these data on the grid on page 3 (D) State which angle the students were more accurate at drawing. Give reasons for your answer

Answer 1

Question 1:

a) The range here is computed as:

= Max - Min

These values are end points of the box plot.

Therefore, Range = 39 - 9 = 30

Therefore 30 is the required range here.

b) The interquartile range is computed as

IQR = Q₃ - Q₁ = 25 - 12 = 13

( These are obtained from the end points of the box )

Therefore 13 is the required inter quartile range here.

c) Total students = 60

The median is thus computed using linear interpolation as:

$= 65 + \frac{(70-65)}{(34-21)}*(30-21) = 68.46$

d) Now the lower quartile is computed here as:

60*0.25 = 15

The first quartile is now computed as:

$= 60 + \frac{(65-60)}{(21-6)}*(15-6) = 63$

Therefore proved that 63 is the required value here.