hazardous waste: following is a list of the number of hazardous - waste sites in a sample of states of the…

hazardous waste: following is a list of the number of hazardous - waste sites in a sample of states of the united states in a recent year. the list has been sorted into numerical order.\n2 3 5 9 9 11 12 12 13 13\n14 15 15 16 19 19 20 21 25 26\n32 32 38 40 48 49 49 52 86 97\nsend data to excel\npart 1 of 4\nfind the first and third quartiles of these data.\nthe first quartile is 12.\nthe third quartile is 38.\npart 2 of 4\nfind the median of these data.\nmedian = 19\npart 3 of 4\nfind the lower and upper outlier boundaries.\nlower outlier boundary is - 27.\nupper outlier boundary is 77.\npart 4 of 4\nare there any outliers? if so, list them. otherwise, select \none.\

hazardous waste: following is a list of the number of hazardous - waste sites in a sample of states of the united states in a recent year. the list has been sorted into numerical order.\n2 3 5 9 9 11 12 12 13 13\n14 15 15 16 19 19 20 21 25 26\n32 32 38 40 48 49 49 52 86 97\nsend data to excel\npart 1 of 4\nfind the first and third quartiles of these data.\nthe first quartile is 12.\nthe third quartile is 38.\npart 2 of 4\nfind the median of these data.\nmedian = 19\npart 3 of 4\nfind the lower and upper outlier boundaries.\nlower outlier boundary is - 27.\nupper outlier boundary is 77.\npart 4 of 4\nare there any outliers? if so, list them. otherwise, select \none.\

Answer

Explanation:

Step1: Count data points

There are 30 data - points.

Step2: Find the median position

The median position for (n = 30) data - points is (\frac{n+1}{2}=\frac{30 + 1}{2}=15.5). The median is the average of the 15th and 16th ordered data - points. The 15th value is 19 and the 16th value is 19, so the median (M = 19).

Step3: Find the first - quartile position

The position of the first quartile (Q_1) is (\frac{n + 1}{4}=\frac{30+1}{4}=7.75). The first quartile is the value at the 8th position, so (Q_1 = 12).

Step4: Find the third - quartile position

The position of the third quartile (Q_3) is (\frac{3(n + 1)}{4}=\frac{3\times(30 + 1)}{4}=23.25). The third quartile is the value at the 24th position, so (Q_3=38).

Step5: Calculate the inter - quartile range (IQR)

(IQR=Q_3 - Q_1=38 - 12 = 26).

Step6: Calculate the lower outlier boundary

Lower outlier boundary (=Q_1-1.5\times IQR=12-1.5\times26=12 - 39=-27).

Step7: Calculate the upper outlier boundary

Upper outlier boundary (=Q_3 + 1.5\times IQR=38+1.5\times26=38 + 39 = 77).

Answer:

Median: 19 First quartile ((Q_1)): 12 Third quartile ((Q_3)): 38 Lower outlier boundary: - 27 Upper outlier boundary: 77