Question

(CO 3) Fifty-four percent of US teens have heard of a fax machine. You randomly select 12 US teens. Find the probability that the number of these selected teens that have heard of a fax machine is exactly six (first answer listed below). Find the probability that the number is more than 8 (second answer listed below).

Answer 1

Answer:

a) $P(X=6)=(12C6)(0.54)^6 (1-0.54)^{12-6}=0.217$  

b) $P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(12C9)(0.54)^9 (1-0.54)^{12-9}=0.0836$  

$P(X=10)=(12C10)(0.54)^{10} (1-0.54)^{12-10}=0.0294$  

$P(X=11)=(12C11)(0.54)^{11} (1-0.54)^{12-11}=0.00628$  

$P(X=12)=(12C12)(0.54)^{12} (1-0.54)^{12-12}=0.000615$  

And adding the values we got:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)= 0.0836+0.0294+0.00628+0.000615 = 0.120$

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 "number of teens that have heard of a fax machine", on this case we now that:  

$X \sim Binom(n=12, p=0.54)$  

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

Part a

For this case we want this probability:

$P(X=6)=(12C6)(0.54)^6 (1-0.54)^{12-6}=0.217$  

Part b

For this case we want this probability:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(12C9)(0.54)^9 (1-0.54)^{12-9}=0.0836$  

$P(X=10)=(12C10)(0.54)^{10} (1-0.54)^{12-10}=0.0294$  

$P(X=11)=(12C11)(0.54)^{11} (1-0.54)^{12-11}=0.00628$  

$P(X=12)=(12C12)(0.54)^{12} (1-0.54)^{12-12}=0.000615$  

And adding the values we got:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)= 0.0836+0.0294+0.00628+0.000615 = 0.120$

Answer 2

Answer:

21.71% probability that the number of these selected teens that have heard of a fax machine is exactly six

11.98% probability that the number is more than 8

Step-by-step explanation:

For each teen, there are only two possible outcomes. Either they have heard of a fax machine, or they have not. The probability of a teen having heard of a fax machine is independent of other teens. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Fifty-four percent of US teens have heard of a fax machine.

This means that $p = 0.54$

You randomly select 12 US teens.

This means that $n = 12$

Find the probability that the number of these selected teens that have heard of a fax machine is exactly six

This is P(X = 6).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{12,6}.(0.54)^{6}.(0.46)^{6} = 0.2171$

21.71% probability that the number of these selected teens that have heard of a fax machine is exactly six

Find the probability that the number is more than 8

$P(X > 8) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.54)^{9}.(0.46)^{3} = 0.0836$

$P(X = 10) = C_{12,10}.(0.54)^{10}.(0.46)^{2} = 0.0294$

$P(X = 11) = C_{12,11}.(0.54)^{11}.(0.46)^{1} = 0.0062$

$P(X = 12) = C_{12,12}.(0.54)^{12}.(0.46)^{0} = 0.0006$

$P(X > 8) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.0836 + 0.0294 + 0.0062 + 0.0006 = 0.1198$

11.98% probability that the number is more than 8