Which measure of central tendency is most appropriate for skewed data with an outlier?

Fresh day Khathutshelo, let's solve. Question 1: Step 1: Analyze the given data for monthly salaries. The salaries are: 85 000, 92 000, 75 000, 88 000, 72 000, 242 000, 74 000, 85 000, 70 000, 86 000, 82 000, 101 000. Step 2: Identify any outliers in the data. Most salaries are in the range of R70 000 to R101 000. However, R242 000 is significantly higher than the other salaries, making it an outlier. Step 3: Evaluate the appropriateness of each statistic. • The arithmetic mean is sensitive to outliers, meaning an outlier can significantly skew its value, making it less representative of the typical salary. • The geometric mean is typically used for data that grows exponentially or for ratios, not usually for salary data with outliers. • The median is the middle value of an ordered dataset. It is robust to outliers, meaning extreme values have little effect on it, making it a good measure of central tendency when outliers are present. • The range shows the spread of the data but does not represent a central tendency and is heavily influenced by outliers. Step 4: Conclude the most appropriate statistic. Due to the presence of the outlier (R242 000), the median is the most appropriate statistic as it provides a better representation of the typical salary without being distorted by the extreme value. The correct option is C. C. Median Question 2: Step 1: List all the monthly salaries. The salaries are: 85 000, 92 000, 75 000, 88 000, 72 000, 242 000, 74 000, 85 000, 70 000, 86 000, 82 000, 101 000. Step 2: Count the number of employees (n). There are 12 employees, so n = 12. Step 3: Calculate the sum of all salaries ( x). x = 85000 + 92000 + 75000 + 88000 + 72000 + 242000 + 74000 + 85000 + 70000 + 86000 + 82000 + 101000 x = 1152000 Step 4: Calculate the mean monthly salary using the formula x = ( x)/(n). x = (1152000)/(12) x = 96000 The mean monthly salary is R96 000. The correct option is A. A. 96 000 Send me the next one 📸