The given table shows the data of the marks for 6 students in a test.
a) What is the mean of the marks shown in the table?
b) What is the mean absolute deviation of the marks?
Answer
\boxed{\text{Final Answer:}} \newline a) \text{The mean of the marks is approximately } \boxed{84.83}. \newline b) \text{The mean absolute deviation of the marks is approximately } \boxed{5.22}.
Step 1 ：Calculate the mean of the marks: \(\frac{85+90+78+92+88+76}{6} = \frac{509}{6} \approx 84.83\)
Step 2 ：Calculate the mean absolute deviation: \(\frac{|85-84.83|+|90-84.83|+|78-84.83|+|92-84.83|+|88-84.83|+|76-84.83|}{6} \approx 5.22\)
Step 3 ：\boxed{\text{Final Answer:}} \newline a) \text{The mean of the marks is approximately } \boxed{84.83}. \newline b) \text{The mean absolute deviation of the marks is approximately } \boxed{5.22}.
