3.6 The following tables list the median family incomes for the 13 Canadian prov
ID: 3043886 • Letter: 3
Question
3.6 The following tables list the median family incomes for the 13 Canadian provinces and territories in 2000 and 2011 and for 13 states of the United States in 1999 and 2012. For the provinces and the states, compute the mean and median family income for each year and compare the two measures of central tendency. Which measure of central tendency is greater for each year? Are the distributions skewed? In which direction?
Median Income for Thirteen States, 1999 and 2012 (U.S. dollars)
State 1999 2012
Alabama 36,213 43,464
Alaska 51,509 63,348
Arkansas 29,762 39,018
California 43,744 57,020
Connecticut 50,798 64,247
Illinois 46,392 51,738
Kansas 37,476 50,003
Maryland 52,310 71,836
Michigan 46,238 50,015
New York 40,058 47,680
Ohio 39,617 44,375
South Dakota 35,962 49,415
Texas 38,978 51,926
Explanation / Answer
for year 1999
mean=sum/total
=36213+51509+29762+....+35962+38978/12
=549057/13
MEAN =42235.15
sor the numbers
29762 35962 36213 37476 38978 39617 40058 43744 46238 46392 50798
51509 52310
median=middlemost value
MEDIAN=40058
mean> median
income varaible is positively skewed in 1999.
Skewed to right
For Year 2012
mean=sum/total
=684085/13
=52621.92
mean =52621.92
median=50015
mean is greater for both the years
MEAN IS HIGHER
POSITIVELY SKEWED
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.