1.

(a) We assume that the probability is binomially distribution, i.e. with n = 12 and p = 0.2.

The probability of half your team being off sick is given by:

12C6 × 0.26 × (1 − 0.2)6 = 0.016

(b) You are taking on an additional 8 members of staff who each earn £1,800 × 3 = £5,400 for three months. Therefore to cover 8 eight staff you are required to pay an additional 8

× 5,400 = £43,200

(c) We assume the staff earn on average £1,800 with a standard deviation of £200. To calculate the probability that a member of staff earns between £1,600 and £1,900 we must calculate the z-score associated with each value. For £1,600 we have a z-score of:

z = 1,600 − 1,800 = −1 200 and for £1,900 we have a z-score of z = 1,900 − 1,800 = 1 200 2 Using the z table we find P (z < 12 ) = 0.6915 and P (z < −1) = P (z > 1) = 1 − P (z < 1) = 1 − 0.8413 = 0.1587. We therefore find:
P (−1 < z < 12 ) = 0.6915 − 0.1587 = 0.5328

(d) We calculate the expected return on each of the three possibilities.
Lay off staff = 0.1 × 10 + 0.2 × 5 + 0.7 × (−20) = −12 Retain original staff levels = 0.1 × 20 + 0.2 × (−5) + 0.7 × (−30) = −20 Retain original and temporary staff = 0.1 × 40 + 0.2 × (−20) + 0.7 × (−35) = −24.5 Therefore the optimum outcome, in financial terms, would be to lay off staff.