Unit 2
5 marks · HardThe temperature of a chemical experiment is monitored every 2 minutes, giving the readings in °C: 20.7, 21.5, 21.9, 22.6, 23.1, 22.3. The experimenter models the temperature change with the linear equation T = a + bt, where t is the time in minutes.
Calculate the mean and median temperature from the data.
Given that at t = 0, T = 20.7, and at t = 10, T = 23.1, find the values of a and b.
According to the model, what will the temperature be after 12 minutes? Give your answer to 1 decimal place.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) Mean ≈ 21.87 °C, Median = 21.9 °C · must show Mean: correctly calculated, Median: data sorted, value read off
(b) a = 20.7; b = (23.1-20.7)/10 = 0.24 · must show Correct values for a and b found using linear equation
(c) 23.6 °C · must show Substitutes t = 12 into model, Rounds answer to 1 decimal place
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Attempts to find mean or median correctly from listed data.
A1 (a) · 1 mark · Both mean and median correct to 2 decimal places (mean) and 1 dec. place (median, if not whole).
M2 (b) · 1 mark · Uses two points to form equation and finds value for b using (T2-T1)/(t2-t1).
A2 (b) · 1 mark · Correct values for a and b both found and stated. If wrong a, b but correct method, award method mark only.
A3 (c) · 1 mark · Substitutes into model for t=12 and gives answer to 1 decimal place. (follow-through)
Worked solution
a) Ordered temperatures: 20.7, 21.5, 21.9, 22.3, 22.6, 23.1
Mean: (20.7 + 21.5 + 21.9 + 22.6 + 23.1 + 22.3)/6 = 131.1 / 6 ≈ 21.85
Median: Middle two values are 21.9 and 22.3, so median = (21.9 + 22.3)/2 = 22.1
b) T = a + bt. At t=0, T=20.7 ⇒ a=20.7. At t=10, T=23.1 ⇒ 23.1 = 20.7 + 10b ⇒ 10b = 2.4 ⇒ b = 0.24
c) For t = 12, T = 20.7 + 0.24 × 12 = 20.7 + 2.88 = 23.58 = 23.6°C