Question

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

Answer 1

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;

$z = \dfrac{x - \mu }{\sigma }$

Where:

x = Observed score = 11.5 Gigagrams/year

We have;

$z = \dfrac{11.5 - 13.9 }{5.8 } = \dfrac{-2.4 }{5.8 } = 0.4138$

From the z-score table relations/computation, the probability (p-value) = 0.6605

Where:

x = Observed score = 14 Gigagrams/year

We have;

$z = \dfrac{14- 13.9 }{5.8 } = \dfrac{0.1}{5.8 } = 0.01724$

From the z-score table relations/computation, the probability (p-value) = 0.4931

Therefore, the probability, $p_{ca}$, 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;

$p_{ca}$ = 0.6605 - 0.4931 = 0.1674 ≈ 0.167

Therefore, the correct option is c. 0.167.