From the given information, we know that the mean and standard deviation are, respectively,
[tex]\begin{gathered} \mu=19 \\ \sigma=4.2 \end{gathered}[/tex]From the 68-95-99 rule, we know that approximately 95% falls between 2 standard deviation of the mean, that is,
[tex]-2=\frac{x-19}{4.2}...(A)[/tex]and
[tex]2=\frac{x-19}{4.2}...(B)[/tex]From equation (A), we have
[tex]x-19=-8.4[/tex]then
[tex]x=10.6[/tex]Now, from equation (B), we get
[tex]\begin{gathered} x-19=8.4 \\ then \\ x=27.4 \end{gathered}[/tex]Therefore, the answer for part a is: Approximately 95% of the students have commute between 10.6 and 27.4 miles
Part b.In this case, we need to find the z score value for 6.4 miles and 31.6 miles and then obtain the corresponding probabilty from the z-table.
For 6.4 miles, the z score is
[tex]z=\frac{6.4-19}{4.2}=-3[/tex]and for 31.6 miles, the z score is
[tex]z=\frac{31.6-19}{4.2}=3[/tex]Now, we need to find the corresponding probability between z=-3 and z=3, which is 0.9973
Therefore, the answer for part b is: Approximately. 0.9973 of the students have commute distances between 6.4 miles and 31.6 miles
