In a group of 200 students, the mean and the standard deviation of scores were found to be 40 and 15, respectively. Later on it was found that the two scores 43 and 35 were misread as 34 and 53, respectively. The corrected mean of scores is:
Step-by-step Solution:
1. Given Information: - Original average (\( \text{Average}_{\text{old}} \)) = 40 - Number of students = 200 - Change in values: \[ (43 - 34) + (35 - 53) = 9 - 18 = -11 \] 2. Calculate the Average of the Change: The average of the change is calculated by dividing the total change by the number of students: \[ \text{Average of change} = \frac{-11}{200} = -0.055 \] 3. Calculate the New Average: The new average is obtained by adding the average of the change to the original average: \[ \text{Average}_{\text{new}} = \text{Average}_{\text{old}} + \text{Average of change} \] \[ \text{Average}_{\text{new}} = 40 + (-0.055) = 40 - 0.055 = 39.945 \] Therefore, the new average is: \[ \boxed{39.945} \]