9709 P6 - Nov 2008 - Q1
2439
Rachel measured the lengths in millimetres of some of the leaves on a tree. Her results are recorded below.
32, 35, 45, 37, 38, 44, 33, 39, 36, 45
Find the mean and standard deviation of the lengths of these leaves.
Solution
To find the mean, sum all the lengths and divide by the number of lengths:
\(\bar{x} = \frac{32 + 35 + 45 + 37 + 38 + 44 + 33 + 39 + 36 + 45}{10} = \frac{384}{10} = 38.4 \text{ mm}\)
\(To find the standard deviation, first calculate the variance:\)
\(s^2 = \frac{1}{10} \left( (32 - 38.4)^2 + (35 - 38.4)^2 + (45 - 38.4)^2 + (37 - 38.4)^2 + (38 - 38.4)^2 + (44 - 38.4)^2 + (33 - 38.4)^2 + (39 - 38.4)^2 + (36 - 38.4)^2 + (45 - 38.4)^2 \right)\)
\(s^2 = \frac{1}{10} (40.96 + 11.56 + 43.56 + 1.96 + 0.16 + 31.36 + 29.16 + 0.36 + 5.76 + 43.56)\)
\(s^2 = \frac{1}{10} (208.4) = 20.84\)
Then, the standard deviation is:
\(s = \sqrt{20.84} \approx 4.57 \text{ mm}\)
