The Solution:
Given:
[tex]\begin{gathered} \bar{x}=mean=40g \\ \sigma=\text{ standard deviation}=2g \end{gathered}[/tex]We are required to find the percentage that is between 34 grams and 42 grams.
By z-score statistic,
The lower limit is:
[tex]Z_1=\frac{x-\mu}{\sigma}=\frac{34-40}{2}=\frac{-6}{2}=-3[/tex]The upper limit is:
[tex]Z_2=\frac{x-\mu}{\sigma}=\frac{42-40}{2}=\frac{2}{2}=1[/tex]The probability is:
[tex]P(Z_1Converting to percent, we multiply by 100:[tex]0.840\times100=84\text{\%}[/tex]Therefore, the correct answer is 84%
