chapter 1

Chapter 1: Overview and Descriptive Statistics54.a. The lower half of the data set: 4.4 16.4 22.2 30.0 33.1 36.6, whose median, and22.2 + 30.0( ) therefore, the lower quartile, is + 26.1.2The top half of the data set: 36.6 40.4 66.7 73.7 81.5 109.9, whose median, and66.7 + 73.7( ) therefore, the upper quartile, is = 70. 2So, the IQR = (70.2 – 26.1) = 44.12.b.A boxplot (created in Minitab) of this data appears below:050sheer strength100There is a slight positive skew to the data. The variation seems quite large. There are nooutliers.c. An observation would need to be further than 1.5(44.1) = 66.15 units below the lowerquartile [( 26.1− 66.15)= − 40.05 units]or above the upper quartile[( 70 .2+ 66.15)= 136.35 units]to be classified as a mild outlier. Notice that, in thiscase, an outlier on the lower side would not be possible since the sheer strength variablecannot have a negative value.An extreme outlier would fall (3)44.1) = 132.3 or more units below the lower, or abovethe upper quartile. Since the minimum and maximum observations in the data are 4.4and 109.9 respectively, we conclude that there are no outliers, of either type, in this dataset.d. Not until the value x = 109.9 is lowered below 73.7 would there be any change in thevalue of the upper quartile. That is, the value x = 109.9 could not be decreased by morethan (109.9 – 73.7) = 36.2 units.26