The yearly amounts of carbon emissions from cars in Belgium are normally distributed with a mean of 13.9 gigagrams per year and a standard deviation of 5.8 gigagrams per year. Find the probability that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 11.5 gigagrams and 14.0 gigagrams per year. a. 0.340 b. 0.660 c. 0.167 d. 0.397

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.

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)