| Overview for Searcy County,
AR |
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Searcy County is one
of about 3,141 counties and county
equivalents in the United States. It has
667.1 sq. miles in land area and a
population density of 11.9 per square
mile. In the last three decades of the
1900s its population grew by 6.9%. On
the 2000 census form, 98.7% of the
population reported only one race, with
<0.1% of these reporting
African-American. The population of this
county is 1.0% Hispanic (of any race).
The average household size is 2.33
persons compared to an average family
size of 2.83 persons.
In 2008 health care and social
assistance was the largest of 20 major
sectors. It had an average wage per job
of $20,073. Per capita income grew by
22.4% between 1998 and 2008 (adjusted
for inflation). |
People
& Income Overview
(By Place of Residence) |
Value |
Industry Overview (2008)
(By Place of Work) |
Value |
| Population
(2009) |
7,944 |
Covered
Employment |
1,801 |
| Growth
(%) since 1990 |
1.3% |
Average wage
per job |
$21,316 |
| Households
(2000) |
3,523 |
Manufacturing - % all jobs in County |
13.0% |
| Labor Force
(persons) (2009) |
3,617 |
Average wage
per job |
$18,110 |
|
Unemployment Rate (2009) |
7.2 |
Transportation & Warehousing - % all
jobs in County |
1.0% |
| Per Capita
Personal Income (2008) |
$23,976 |
Average wage
per job |
$34,727 |
| Median
Household Income (2008) |
$25,547 |
Health
Care, Social Assist. - % all jobs in
County |
18.5% |
| Poverty
Rate (2008) |
24.2 |
Average wage
per job |
$20,073 |
| H.S.
Diploma or More - % of Adults 25+ (2000) |
68.0 |
Finance and
Insurance - % all jobs in County |
2.1% |
| Bachelor's
Deg. or More - % of Adults 25+ (2000) |
8.4 |
Average wage
per job |
$25,977 |
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