Question

750,000

85,000

29,000

12,000

10,900

6,700

12,000

26,000

78,000

485,000

11,800

14,000

30,000

115,000

1,250,000

2,600,000

375,000

40,000

19,500

12,000

1. Generate the mean, median, mode, standard deviation, range, 30th percentile, rank for $40,000, percentile rank of $40,000 and the trimmean cutting off 10%. Write a statement that interprets each one. (Points-)

2. Compare the mean, median and mode. How can you explain the differences among them? Which one do you think is most appropriate to present central tendency for this data and why? (points-)

Answer 1

Data	(x-xbar)^2	Sorted
750000	2.04218E+11	6700
85000	45409479025	10900
29000	72412119025	11800
12000	81850349025	12000
10900	82480968025	12000
6700	84911046025	12000
12000	81850349025	14000
26000	74035689025	19500
78000	48441809025	26000
485000	34933479025	29000
11800	81964827025	30000
14000	80709969025	40000
30000	71874929025	78000
115000	33523779025	85000
1250000	9.06123E+11	115000
2600000	5.29877E+12	375000
375000	5914379025	485000
40000	66613029025	750000
19500	77615174025	1250000
12000	81850349025	2600000
5961900	7.5155E+12	Total

Mean,

= 5961900/20 = 298095

Median = average of (n/2)th and (n/2+1)st data.

= average of 10th and 11th data

= (29000+30000)/2 = 29500

Mode = 12000

Standard deviation

10% trimmed mean,

It exclude 10% data 2smallest and 2 largrest data

range = 2600000-6700 = 2593300

30th percentile.

= (30*20)/100 = 6th data = 12000

rank for $40,000 = 12

Percentile rank for $40000 = (12*100)/20 = 60th percentile.

The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers. The "median" is the "middle" value in the list of numbers. To find the median, your numbers have to be listed in numerical order from smallest to largest, so you may have to rewrite your list before you can find the median. The "mode" is the value that occurs most often. If no number in the list is repeated, then there is no mode for the list.

As here, Mean > Median > Mode, the data is positively skewed.

Here, Mean is the most appropriate to present central tendency for this data as the range of the data is high.

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