Answer: 0.052
Step-by-step 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. [tex]\mu=13.9[/tex] and [tex]\sigma=9.2[/tex]
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 :-
[tex]P(12.8<x<14.0)=P(\dfrac{12.8-13.9}{9.2}<\dfrac{x-\mu}{\sigma}<\dfrac{14.0-13.9}{9.2})\\\\=P(-0.12<z<0.01)\ \ [\because\ z=\dfrac{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[/tex]
Hence, the required probability = 0.052
