A recent survey found that 71% of senior adults wear glasses for driving. In a group of 20 senior adults, how like that no more than 10 wear glasses for driving? Group of answer choices 95.2% 4.8% 3.1% 1.7%

Respuesta :

Answer:

[tex] P(X\leq 10) = 1- 0.961525= 0.0385[/tex]  

The nearest answer for this case would be 3.1%

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=20, p=0.71)[/tex]

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

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

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

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

For this case we can begin finding the probability P(X>10). If we find the individual probabilities we got:

[tex]P(X=11)=(20C11)(0.71)^{11} (1-0.71)^{20-11}=0.0563[/tex]

[tex]P(X=12)=(20C12)(0.71)^{12} (1-0.71)^{20-12}=0.1034[/tex]

[tex]P(X=13)=(20C13)(0.71)^{13} (1-0.71)^{20-13}=0.1558[/tex]

[tex]P(X=14)=(20C14)(0.71)^{14} (1-0.71)^{20-14}=0.1907[/tex]

[tex]P(X=15)=(20C15)(0.71)^{15} (1-0.71)^{20-15}=0.1867[/tex]

[tex]P(X=16)=(20C16)(0.71)^{16} (1-0.71)^{20-16}=0.1429[/tex]

[tex]P(X=17)=(20C17)(0.71)^{17} (1-0.71)^{20-17}=0.082[/tex]

[tex]P(X=18)=(20C18)(0.71)^{18} (1-0.71)^{20-18}=0.036[/tex]

[tex]P(X=19)=(20C19)(0.71)^{19} (1-0.71)^{20-19}=0.0087[/tex]

[tex]P(X=20)=(20C20)(0.71)^{20} (1-0.71)^{20-20}=0.00106[/tex]

And if we add the values we got:

[tex] P(X>10)= P(X=11) +.... +P(X=20) = 0.961525[/tex]

And if we use the complement rule the probability that "no more than 10 wear glasses for driving" we can do this:

[tex] P(X\leq 10) = 1- 0.961525= 0.0385[/tex]  

The nearest answer for this case would be 3.1%

Answer:

A,C,E

Explantion:

A,C,E

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