Final answer:
We’ve verified that the range of the data set is 9, so statement 'b' is correct. The IQR is 6.5, not 5, making statement 'a' incorrect. There are no outliers based on the IQR criteria, and thus, statement 'c' is incorrect. The distribution is skewed right, making 'd' correct, and the median is less than the mean, confirming statement 'e' as correct.
Step-by-step explanation:
To address the student's question regarding the data set (18, 14, 12, 14, 11, 11, 19, 20, 16, 11), we first need to compute various statistical measures. First, let's arrange the data in ascending order to make the computations easier: (11, 11, 11, 12, 14, 14, 16, 18, 19, 20).
Interquartile Range (IQR)
The IQR is the difference between the third quartile (Q3) and the first quartile (Q1). To find Q1, we take the median of the lower half of the data, and to find Q3, we take the median of the upper half of the data, excluding the median of the entire data set if the number of data points is odd. Q1 is 11.5 (the average of the 3rd and 4th values), and Q3 is 18 (the average of the 7th and 8th values), so the IQR = Q3 - Q1 = 18 - 11.5 = 6.5. Therefore, statement 'a' is not correct.
Range
The range is calculated by subtracting the minimum value from the maximum value in the data set. So, range = 20 - 11 = 9. Therefore, statement 'b' is correct.
Outliers
To check for outliers, one common method is to consider any data point that falls more than 1.5*IQR below Q1 or above Q3 as an outlier. However, in this data set, the smallest value is 11 and the largest is 20, which are both within the acceptable range (Q1 - 1.5*IQR to Q3 + 1.5*IQR), so statement 'c' is not correct.
Skewness
The skewness of the distribution indicates the direction and degree of asymmetry. If the mean is greater than the median, the distribution is skewed right. We find the mean by adding all the numbers and dividing by the count: (18 + 14 + 12 + 14 + 11 + 11 + 19 + 20 + 16 + 11) / 10 = 14.6. The median is 14 (the average of the 5th and 6th values). Since the mean is greater than the median, statement 'd' is correct.
Median vs. Mean
As calculated above, the median is 14 and the mean is 14.6. Therefore, the median is indeed less than the mean, making statement 'e' correct.