USA and WWE(R) Unveil 'The Sexiest Woman on Television'

2006 Diva Search Winner to Be Chosen During Live TV Special

NEW YORK, Aug. 14 -- USA Network and WWE will unveil "The Sexiest Woman on Television" when the $250,000 WWE Diva Search winner is announced from the Hard Rock Cafe during their live 1-hour television event, "The Sexiest Woman on Television: The WWE Diva Search Finals," this Wednesday, August 16, 2006 (10-11p.m. ET/ 9-10p.m. CT).

Thousands of hopefuls from across North America answered an open casting call and WWE fans have been voting via online and SMS for their favorite Diva on Monday Night RAW(R) (USA 9 p.m. ET/PT) and on "Friday Night SmackDown(R)" (UPN 8 p.m. ET / 7 p.m. CT). After five weeks of elimination rounds, USA will televise the event as the winner will be crowned before a live audience at the Hard Rock Cafe, becoming the newest WWE Diva and earning $250,000 for being named "The Sexiest Woman on Television."

Fans who watched the WWE's broadcasts were asked to consider each finalists performance each week to determine whether each contestant possessed the following qualities of a WWE Diva (1) Glamour; (2) Beauty; (3) Physical Fitness; (4) Poise; (5) Presence; (6) Charisma; (7) Personality; and (8) Talent. The finalist with the fewest fan votes has been eliminated each week. As of today, the final four contestants competing for the title of 2006 Diva Search Winner are Jen England, Layla El, Milena Roucka and J.T. Tinney.

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SOURCE USA Network