| Date | May 2014 | Marks available | 5 | Reference code | 14M.2.hl.TZ2.2 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate and Find | Question number | 2 | Adapted from | N/A |
Question
The weights, in kg, of one-year-old bear cubs are modelled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\).
(a) Given that the upper quartile weight is 21.3 kg and the lower quartile weight is 17.1 kg, calculate the value of \(\mu \) and the value of \(\sigma \).
A random sample of 100 of these bear cubs is selected.
(b) Find the expected number of bear cubs weighing more than 22 kg.
Markscheme
(a) METHOD 1
\(\mu = \frac{1}{2} \times (17.1 + 21.3)\) (M1)
\(\mu = 19.2{\text{ (kg)}}\) A1
finding z value for the upper quartile \( = 0.674489{\text{K}}\)
\(0.674489{\text{K}} = \frac{{21.3 - 19.2}}{\sigma }\) or \( - 0.674489{\text{K}} = \frac{{17.1 - 19.2}}{\sigma }\) M1
\(\sigma = 3.11{\text{ (kg)}}\) A1
METHOD 2
finding z value for the upper quartile \( = 0.674489{\text{K}}\)
from symmetry the z value for a lower quartile is \( - 0.674489{\text{K}}\) M1
forming two simultaneous equations:
\( - 0.674489{\text{K}} = \frac{{17.1 - \mu }}{\sigma }\)
\(0.674489{\text{K}} = \frac{{21.3 - \mu }}{\sigma }\) M1
solving gives:
\(\mu = 19.2{\text{ (kg)}}\) A1
\(\sigma = 3.11{\text{ (kg)}}\) A1
[4 marks]
(b) using \(100 \times {\text{P}}(X > 22) = 100 \times 0.184241{\text{K}}\)
\( = 18\) A1
Note: Accept 18.4
[1 mark]
Total [5 marks]
