Given:
mean = 85
Standard deviation = 12
Assuming that the data is normally distributed, about 68% of the data would lie one standard deviation from the mean.
Using the z-score formula:
![\begin{gathered} z\text{ = }(x-\varphi)/(\sigma) \\ \text{where:} \\ \psi\text{ is the mean} \\ \text{and } \\ \sigma\text{ is the standard deviation} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pyzlwnupe7z0xzjhi8fjjpxnvu33zz8rx2.png)
set z= 1:
![\begin{gathered} 1\text{ = }\frac{x_2-\text{ 85}}{12} \\ x_2-\text{ 85 = 12} \\ x_2=\text{ 12 + 85} \\ x_2=\text{ 97} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1o4xf2wjyqn0f7tw7vku8maakeqidm5mfa.png)
set z = -1:
![\begin{gathered} -1\text{ = }(x_1-85)/(12) \\ x_1-85\text{ = -12} \\ x_1=\text{ 85-12} \\ x_1=\text{ 73} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sq7jojuouln25icwjl3li5n9mpxtj65xo9.png)
Hence, the majority of the data would lie between 73 to 97
Answer:
73 to 97 (Option B)