How risky is such a deficiency? Bayesian analysis is appropriate.
An excellent short pair of articles on this topic is here:
https://www.healthline.com/health-news/new-study-found-80-percent-of-covid-19-patients-were-vitamin-d-deficient
This is a good opportunity to use Bayesian statistics to go from the percentage of people who are vitamin D deficient, and have COVID, P(d'|C), to what we want to know, P(C|d'), the risk of having COVID, given that you are deficient in vitamin D.
Bayes theorem becomes
P(C|d') = P(C) [P(d'|C) / P(d')]
These articles note that of those with COVID, 80% were deficient in vitamin D,
P(d'|C) = 0.80
The term in brackets is the likelihood factor,
the multiplier that converts the general
prevalence probability, P(C), to the probability
for those who are deficient in vitamin D.
The fraction of the nursing home population deficient in
vitamin D, P(d'), is P(d')=0.60
https://www.medscape.com/answers/128762-54281/what-is-the-prevalence-of-vitamin-d-deficiency-in-the-us
This gives us a likelihood factor of 0.80/0.60= 1.33 elevation of their probability of having COVID from what they would have without a vitamin D deficiency. If P(C) for such a group is 0.05, they are 1.33 times this as likely to contract COVID due to their vitamin D deficiency, about 7%, elevated but far from 80%.
Vitamin D is a help, not a cure, not a certain preventative.
This review concluded it is not of demonstrated help:
https://www.msn.com/en-gb/health/medical/official-review-concludes-lack-of-evidence-linking-vitamin-d-and-lower-covid-risk/ar-BB1c0yW7?appid=hwbrowser&ctype=news
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