Searcy is located at 35°14'49" North, 91°44'1" West (35.247043, -91.733706)1.
As of the census2 of 2000, there are 18,928 people, 6,822 households, and 4,495 families residing in the city. The population density is 497.2/km² (1,287.4/mi²). There are 7,405 housing units at an average density of 194.5/km² (503.6/mi²). The racial makeup of the city is 90.24% White, 6.60% Black or African American, 0.31% Native American, 0.50% Asian, 0.02% Pacific Islander, 1.09% from other races, and 1.25% from two or more races. 2.06% of the population are Hispanic or Latino of any race.
There are 6,822 households out of which 28.5% have children under the age of 18 living with them, 51.8% are married couples living together, 10.8% have a female householder with no husband present, and 34.1% are non-families. 29.5% of all households are made up of individuals and 13.1% have someone living alone who is 65 years of age or older. The average household size is 2.32 and the average family size is 2.86.
In the city the population is spread out with 19.7% under the age of 18, 23.4% from 18 to 24, 23.3% from 25 to 44, 17.8% from 45 to 64, and 15.9% who are 65 years of age or older. The median age is 30 years. For every 100 females there are 89.1 males. For every 100 females age 18 and over, there are 87.4 males.
The median income for a household in the city is $32,321, and the median income for a family is $41,334. Males have a median income of $32,445 versus $21,142 for females. The per capita income for the city is $16,553. 15.0% of the population and 11.7% of families are below the poverty line. Out of the total population, 18.1% of those under the age of 18 and 8.0% of those 65 and older are living below the poverty line.