Question

A survey showed that 82​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 15 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?

Answer 1

Answer:

$P(X \leq 1)= P(X=0) +P(X=1)$

And using the probability mass function we can find the individual probabiities

$P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}$

$P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}$

And replacing we got:

$P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}$

And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight​ correction in a sample of 15 since the probability obtained is very near to 0

Step-by-step explanation:

Let X the random variable of interest "number of adults who need correction", on this case we now that:

$X \sim Binom(n=15, p=0.82)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We want to find this probability:

$P(X \leq 1)= P(X=0) +P(X=1)$

And using the probability mass function we can find the individual probabiities

$P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}$

$P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}$

And replacing we got:

$P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}$

And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight​ correction in a sample of 15 since the probability obtained is very near to 0