‘World’s most premature baby' to survive celebrates 1st birthday

Baby was given 0% chance of survival
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Posted at 1:51 PM, Jun 21, 2021

MINNEAPOLIS, Minn. — The “world’s most premature baby" to survive recently celebrated his first birthday, according to Guinness World Records.

The organization says Richard Scott William Hutchinson was brought into the world early when his mother, Beth, experienced medical complications and unexpectedly went into labor on June 5, 2020. His due date wasn’t until Oct. 13.

At a mere 11.9 ounces, the baby was reportedly so small that he could fit in the palm of a hand.

Born about four months early, Guinness says the neonatology team at Children’s Minnesota hospital told Beth and Rick Hutchinson that their son had a 0% chance of survival.

Guinness says Richard was born at a gestational age of 21 weeks 2 days, making him 131 days premature. The standard gestational period for a baby is 40 weeks.

Guinness officially recognized Richard as the “world's most premature baby" to survive on his first birthday on June 5.

Richard celebrated his milestone birthday at home with his parents and other loved ones, according to Guinness.

Before Richard became the record holder, James Elgin Gill of Ottawa held the record for three-and-a-half decades. Guinness says he was born at 21 weeks 5 days, or 128 days premature, on May 20, 1987.