Algorithm End_to_Beginning Step 1. Set all the values in the array H to 0. Step 2. Starting with H[n-1] and going down to H[0] try to increase the value of H[i] as follows: Step 3. Starting with index i+1 and going up to n-1 (the last index in the array A) repeat Steps 4 and 5: Step 4. See if any element is smaller or equal to A[i] and has its H value also bigger to H[i]. Step 5. If yes, then A[i] can be followed by that element in a sorted subsequence, thus set H[i] to be 1 plus the H value of that element. Step 6. Calculate the largest (maximum) value in array H. By adding 1 to that value we have the length of a longest reversely sorted subsequence. Step 7. Identify a longest subsequence by identifying elements in array A that have decreasing H values, starting with the largest (maximum) value in array H.

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Explain the circumstances for which the interquartile range is the preferred measure of dispersion. what is an advantage that the standard deviation has over the interquartile range? choose the correct answer below. a. the interquartile range is preferred when the distribution is symmetric. an advantage of the standard deviation is that it increases as the dispersion of the data increases. b. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation. c. the interquartile range is preferred when the distribution is symmetric. an advantage of the standard deviation is that it is resistant to extreme values. d. the interquartile range is preferred when the data are bell shaped. an advantage of the standard deviation is that it is resistant to extreme values. e. the interquartile range is preferred when the data are skewed or have outliers. an advantage of the standard deviation is that it uses all the observations in its computation. f. the interquartile range is preferred when the data are bell shaped. an advantage of the standard deviation is that it increases as the dispersion of the data increases.