Solution:
Given that;
The amount of 5th and 6th graders is normally distributed with a mean of 4 inches and standard deviation of 1 inch.
To find the percentage of 5th and 6th graders that grows between 1 and 3 inches, we will apply the z score formula
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ Where \\ x\text{ is the observed value} \\ \mu\text{ is the mean} \\ \sigma\text{ is the standard deviation} \end{gathered}[/tex]Where, x = 1 inch
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{1-4}{1}=\frac{-3}{1}=-3 \\ z=-3 \end{gathered}[/tex]Where, x = 3 inches
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{3-4}{1}=\frac{-1}{1}=-1 \\ z=-1 \end{gathered}[/tex]Using the z score table,
[tex]P(-3For the 5th graders, the percentage is[tex]=0.15731\times100\%=15.731\%[/tex]For the 6th graders, the percentage is the same as the percentage of the 5th graders.
The percentage of the 6th graders is
[tex]15.731\%[/tex]The percentage of 5th and 6th graders that grows between 1 and 3 inches will be
Since, they have different peaks, it is not appropriate to use normal distribution.
Hence, the answer is A