Correct Answer - Option 4 : more than 0% but less than 50%
Explanation:
| The ratio of Y's bid price to X's cost |
Mark-up of Y's bid |
Number of bids |
| 1.02 |
2% |
6 |
| 1.04 |
4% |
12 |
| 1.06 |
6% |
3 |
| 1.10 |
10% |
6 |
| 1.12 |
12% |
3 |
| |
|
Total (Σn) =30 |
Mean of the bid to total cost ratio;
\(\mu = \left[ {\frac{{1.02 \times 6 + 1.04 \times 12 + 1.06 \times 3 + 1.10 \times 6 + 1.12 \times 3}}{{30}}} \right] = 1.058\)
Standard deviation,
\(\sigma = \sqrt {\frac{{{\text{Σ }}{{\left( {{x_i} - \mu } \right)}^2}}}{N}} \)
\( = \sqrt {\frac{{6{{\left( {1.02 - 1.058} \right)}^2} + 12{{\left( {1.04 - 1.058} \right)}^2} + 3{{\left( {1.06 - 1.058} \right)}^2} + 6{{\left( {1.10 - 1.058} \right)}^2} + 3{{\left( {1.12 - 1.058} \right)}^2}}}{{30}}} \)
= 0.034
Thus, Z at 8% markup level = \(\left( {\frac{{x - \mu }}{\sigma }} \right)\)
= \(\left( {\frac{{1.08 - 1.058}}{{0.034}}} \right)\) = 0.0647
From normal deviate table;
| Z |
Probability (%) |
| 0.60 |
72.6 |
| 0.70 |
75.8 |
By interpolation, probability for Z = 0.647
\( = 72.6 + \left( {\frac{{75.8 - 72.6}}{{0.70 - 0.60}}} \right)\left( {0.647 - 0.60} \right)\)
= 74.104%
Hence, the probability of winning against the competitor Y at 8% mark-up
= (1 - 0.74104)
= 0.25896
= 25.896%
