Mathematics, 16.03.2020 18:49 reecedstceklein

According to a recent survey of households, the poll found that 41 % of the adults watching a special event on TV enjoyed the commercials more than the event itself. Consider a random sample of nine adults who watched the special event. Complete parts a through d below.

a. For this experiment, define the event that represents a "success".
i. A success is an adult who did not enjoy the commercials more than the event on TV itself.
ii. A success is an adult who enjoyed the commercials more than the event on TV itself.

b. What is the probability that less than four people enjoyed the commercials more than the event?
c. What is the probability that exactly six or seven people enjoyed the commercials more than the event?
d. What is the mean for this binomial distribution?

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Answer from: jamarstand

a) ii. A success is an adult who enjoyed the commercials more than the event on TV itself.

b) P(X < 4) = 0.4576

c) P(X=6) + P(X=7) = 0.1064

d) Mean = 3.69

Step-by-step explanation:

a) The population parameter stated in the survey is the event that represents a success. Hence, A success is an adult who enjoyed the commercials more than the event on TV itself.

b) The probability that less than four people enjoyed the commercials more than the event

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of adults sampled = 9

x = Number of successes required = number of people that enjoyed the commercial more than the event on TV = less than 4

p = probability of success = probability of enjoying the commercial more than the event = 0.41

q = probability of failure = probability of NOT enjoying the commercial more than the event = 1 - 0.41 = 0.59

P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0.00866299582 + 0.05418043148 + 0.15060323326 + 0.24419846296

P(X < 4) = 0.45764512351 = 0.4576

c) The probability that exactly six or seven people enjoyed the commercials more than the event = P(X=6) + P(X=7)

Using the Binomial distribution formula

P(X=6) + P(X=7) = 0.08194801935 + 0.02440582659 = 0.106353846 = 0.1064

d) Mean = expected value of people that would enjoy the commercial more than the main event = np = 9×0.41 = 3.69

Hope this Helps!!!

Answer from: aaliyahnv07

a) ii. A success is an adult who enjoyed the commercials more than the event on TV itself.

b) $P(X=0)=(9C0)(0.41)^0 (1-0.41)^{9-0}=0.0087$

$P(X=1)=(9C1)(0.41)^1 (1-0.41)^{9-1}=0.0.54$

$P(X=2)=(9C2)(0.41)^2 (1-0.41)^{9-2}=0.151$

$P(X=3)=(9C3)(0.41)^3 (1-0.41)^{9-3}=0.244$

And adding we got:

P(X

c) $P(6 \leq X \leq 7)$

$P(X=6)=(9C6)(0.41)^6 (1-0.41)^{9-6}=0.082$

$P(X=7)=(9C7)(0.41)^7 (1-0.41)^{9-7}=0.0244$

And adding we got:

$P(6 \leq X \leq 7)=0.082+0.0244=0.106$

d) $\mu = np =9*0.41=3.69$

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:

$X \sim Binom(n=9, p=0.41)$

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

The success on this case is:

ii. A success is an adult who enjoyed the commercials more than the event on TV itself.

Part b

For this case we want this probability:

P(X

And we can find the individual probabilities and we got:

$P(X=0)=(9C0)(0.41)^0 (1-0.41)^{9-0}=0.0087$

$P(X=1)=(9C1)(0.41)^1 (1-0.41)^{9-1}=0.0.54$

$P(X=2)=(9C2)(0.41)^2 (1-0.41)^{9-2}=0.151$

$P(X=3)=(9C3)(0.41)^3 (1-0.41)^{9-3}=0.244$

And adding we got:

P(X

Part c

For this case we want this probability:

$P(6 \leq X \leq 7)$

$P(X=6)=(9C6)(0.41)^6 (1-0.41)^{9-6}=0.082$

$P(X=7)=(9C7)(0.41)^7 (1-0.41)^{9-7}=0.0244$

And adding we got:

$P(6 \leq X \leq 7)=0.082+0.0244=0.106$

Part d

For this case the mean is given by:

$\mu = np =9*0.41=3.69$

Answer from: Quest

Bi’m pretty sure hope it works

Answer from: Quest

5días gucgivuzu 32 horas y horas 6y t7

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