Correct Answer  Option 4 : more than 0% but less than 50%
Explanation:
The ratio of Y's bid price to X's cost 
Markup 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% markup
= (1  0.74104)
= 0.25896
= 25.896%