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

`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

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=(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