If the mean of a given dataset is85 and the standard deviation is12, where will a majority of thedata lie?A. 55 to 85B. 73 to 97C. 85 to 100

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:
[tex]\begin{gathered} z\text{ = }\frac{x-\varphi}{\sigma} \\ \text{where:} \\ \psi\text{ is the mean} \\ \text{and } \\ \sigma\text{ is the standard deviation} \end{gathered}[/tex]set z= 1:
[tex]\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}[/tex]set z = -1:
[tex]\begin{gathered} -1\text{ = }\frac{x_1-85}{12} \\ x_1-85\text{ = -12} \\ x_1=\text{ 85-12} \\ x_1=\text{ 73} \end{gathered}[/tex]Hence, the majority of the data would lie between 73 to 97
Answer:
73 to 97 (Option B)