Question

For questions, 1-2, determine whether experiment is a binomial experiment. If it is, identify a success, specify the values of n, p, and q, and list the possible values of the random variable x. If it is not a binomial experiment, explain why. 1. A survey found that 49% of U.S. households own a dedicated game console. Eight U.S. households are randomly selected. The random variable represents the number of U.S. households that own a dedicated game console. You draw five cards, one at a time, from a standard deck. You do not replace a card once it is drawn. The random variable represents the number of cards that are hearts. 2. 39% of US adults think that the government should help fight childhood obesity. You randomly select six U.S. adults. Find the probability that the number of them who think the government should help fight childhood obesity is (a) exactly two, and (b) at least five. 3. 34% of voters think that Congress should help write standards for school food. You randomly select six voters and ask them whether Congress should help write standards for school food. The random variable represents the number of voters who think that Congress should help write standards for school food. Construct the binomial distribution. 4, 5-69% of adults think that musicians should be allowed to sing potentially offensive songs. You randomly select four adults and ask them whether they think musicians should be allowed to sing potentially offensive songs. The random variable represents the number of adults who think that musicians should be allowed to sing potentially offensive songs. Find the mean, variance, and standard deviation of the binomial distribution for the random variable. Interpret the results by describing any unusual outcomes.

Answer 1

Answer:

Part 1

X="number of U.S. households that own a dedicated game console"

Is a binomial experiment we an event defined with the associated probability and we have specific trials.

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

$X \sim Binom(n=8, p=0.49)$

Y= "number of cards that are hearts."

Thats not a binomial experiment since the probability for each trial changes since we are doing the experiment without replacment

Part 2

a) $P(X=2)=(6C2)(0.39)^2 (1-0.39)^{6-2}=0.316$

b) $P(X \geq 5) = P(X=5)+P(X=6)$

$P(X=5)=(6C5)(0.39)^5 (1-0.39)^{6-5}=0.033$

$P(X=6)=(6C6)(0.39)^6 (1-0.39)^{6-6}=0.00352$

$P(X \geq 5) = P(X=5)+P(X=6)=0.033+0.00352=0.0365$

Part 3

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

$X \sim Binom(n=6, p=0.34)$

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

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

Part 4

$E(X)=np=4*0.69=2.76$

The variance $\sigma^2 = np(1-p) = 4*0.69*(1-0.69) =0.8556$

$\sigma=\sqrt{np(1-p)}=\sqrt{4*0.69(1-0.69)}=0.925$

Unusual outcomes would be considered values above or below 2 deviations from the mean for example 2.76-(2*0.925) =0.91 or 2.76+2(0.925)=4.61[/tex]. X=0 and X=4 would be considered as unusual values.

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".

Part 1

X="number of U.S. households that own a dedicated game console"

Is a binomial experiment we an event defined with the associated probability and we have specific trials.

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

$X \sim Binom(n=8, p=0.49)$

Y= "number of cards that are hearts."

Thats not a binomial experiment since the probability for each trial changes since we are doing the experiment without replacment

Part 2

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

$X \sim Binom(n=6, p=0.39)$

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:

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

And we want to find this probability:

a) $P(X=2)=(6C2)(0.39)^2 (1-0.39)^{6-2}=0.316$

b) $P(X \geq 5) = P(X=5)+P(X=6)$

$P(X=5)=(6C5)(0.39)^5 (1-0.39)^{6-5}=0.033$

$P(X=6)=(6C6)(0.39)^6 (1-0.39)^{6-6}=0.00352$

$P(X \geq 5) = P(X=5)+P(X=6)=0.033+0.00352=0.0365$

Part 3

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

$X \sim Binom(n=6, p=0.34)$

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

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

Part 4

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

$X \sim Binom(n=4, p=0.69)$

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

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

The expected value is given by:

$E(X)=np=4*0.69=2.76$

The variance $\sigma^2 = np(1-p) = 4*0.69*(1-0.69) =0.8556$

$\sigma=\sqrt{np(1-p)}=\sqrt{4*0.69(1-0.69)}=0.925$

Unusual outcomes would be considered values above or below 2 deviations from the mean for example 2.76-(2*0.925) =0.91 or 2.76+2(0.925)=4.61[/tex]. X=0 and X=4 would be considered as unusual values.