Question #147177
(5.2.26) A survey showed that 83​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 21 adults are randomly​ selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight​ correction?
The probability that no more than 1 of the 21 adults require eyesight correction is?
1
Expert's answer
2020-12-02T10:39:21-0500

p = 83% = 0.83

n = 21

Binomial probability:

P(X=x)=C(n,x)px(1p)nxP(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{n-x}

P(X1)=P(X=0)+P(X=1)P(X\le1)=P(X=0)+P(X=1)

P(X1)=21!0!21!0.8300.1721+21!1!20!0.8310.1720=7.153047281015P(X\le1)=\frac{21!}{0!21!}\cdot 0.83^0\cdot 0.17^{21}+\frac{21!}{1!20!}\cdot 0.83^1\cdot 0.17^{20}=7.15304728\cdot10^{−15}

The probability that no more than 1 of the 21 adults require eyesight correction is close to zero. Yes, 1 is a significantly low number of adults requiring eyesight​ correction.


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