Answer:
The correct option is;
c. 0.167
Explanation:
The parameters given are;
The mean, μ = 13.9 Gigagrams/year
The standard deviation, σ = 5.8 Gigagrams/year
The z-score formula is given as follows;
Where:
x = Observed score = 11.5 Gigagrams/year
We have;
From the z-score table relations/computation, the probability (p-value) = 0.6605
Where:
x = Observed score = 14 Gigagrams/year
We have;
From the z-score table relations/computation, the probability (p-value) = 0.4931
Therefore, the probability, 
, that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 15.5 Gigagrams/year and 14 Gigagrams/year = The area under the normal curve bounded by the p-values for the two amounts of carbon emission
Which gives;
= 0.6605 - 0.4931 = 0.1674 ≈ 0.167
Therefore, the correct option is c. 0.167.
, then the bond is non-polar.
, then the bond will be covalent.
, then the bond will be ionic.