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.

(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]