Why is it best to compare medians and interquartile ranges for these data, rather than comparing means and standard deviations
? A. Some of these data have vastly different centers, and the median and interquartile range provide more reliable information in these circumstances. B. These data are fairly symmetric and do not have outliers, and the median and interquartile range provide more reliable information in these circumstances. C. Some of these data have outliers and/or are skewed, and the median and interquartile range are resistant to outliers. D. Some of these data have outliers and/or are skewed, and the median and interquartile range amplify the effects of outliers and skewness.
Answer:Some of these data have outliers and/or are skewed, and the median and interquartile range are resistant to outliers.
Step-by-step explanation:
Outliers often characterize a lot of data. Outliers are extreme values which have obvious impact on the value of the standard deviation of a given distribution.
However, the median and interquartile range are resistant to outliers hence they can be used even in the presence of outliers in the distribution.
I believe that she had $852.50. When you multiply 550 by .55(55%) you get 302.50. This is how much she has left. Now you add that to the 550 to get $852.50
Let the no. of girls be x and boys be y...... x/y=1/3 =>3x=y ...............(I) now 4 boys leave the class is then x/y-4=2/5 =>5x=2y-8 (I)=>5x=2(3x)-8 x=8 now 3x=y =>y=24 therefore no. of girls is 8 and the no. of boy is 24.............. I have tried my best.....hope i helped............
