WHO says coronavirus is deadlier than seasonal flu, doesn't spread as easily

Organization puts mortality rate at 3.4%
Posted at 8:24 AM, Mar 04, 2020

The head of the World Health Organization said Tuesday that COVID-19, better known as coronavirus, has a higher mortality rate than that of the seasonal flu, according to The New York Times and ABC News.

In a press conference in Geneva on Tuesday, WHO director general Dr. Tedros Adhanom Ghebreyesus said that the mortality rate of reported COVID-19 cases is "about 3.4 percent." Ghebreyesus added that seasonal flu kills less than 1 percent of people infected.

The New York Times notes that the 3.4 percent mortality rate does not include mild cases where patients are not hospitalized, and is also skewed by cases in Wuhan, China, where cases have bee more severe and the death rate is higher.

According to ABC News, Ghebreyesus also said during Tuesday's press conference that COVID-19 does not spread as easily as the typical seasonal flu, but because the viral strain is so new, it still poses an issue.

"While many people globally have built up immunity to seasonal flu strains, COVID-19 is a new virus to which no one has immunity," he said, according to ABC News. "That means more people are susceptible to infection, and some will suffer severe disease."

The New York Times reports that Ghebreyesus' statement was the first time the WHO has confirmed a mortality rate for COVID-19. The announcement also comes as nine people have died of the disease in the United States in the past week, all of them in Washington state. So far, 128 cases of COVID-19 have been confirmed in the United States.

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