Answer: 0.052
Explanation:
Given : 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 9.2 gigagrams per year.
i.e.
and
![\sigma=9.2](https://img.qammunity.org/2020/formulas/mathematics/college/snyk5xauu6xi2gzf4hjphend4siy2zu7fq.png)
Let x denotes the yearly amounts of carbon emissions from cars in Belgium.
Then, the probability that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 12.8 gigagrams and 14.0 gigagrams per year will be :-
![P(12.8<x<14.0)=P((12.8-13.9)/(9.2)<(x-\mu)/(\sigma)<(14.0-13.9)/(9.2))\\\\=P(-0.12<z<0.01)\ \ [\because\ z=(x-\mu)/(\sigma)]\\\\=P(z<0.01)-P(z<-0.12)\\\\=P(z<0.01)-(1-P(z<0.12))\ \ [\because P(Z<-z)=1-P(z<z)]\\\\=0.5040-(1-0.5478)\ \ [\text{By using z-table}]\\\\\=0.5040-0.4522=0.0518\approx0.052](https://img.qammunity.org/2020/formulas/mathematics/college/6sdx7icbtin5dbkqx3gm23qj65wmh5qc64.png)
Hence, the required probability = 0.052