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2 years ago

Identify the domain and range for the table from problem #5 of your assignment. Also, determine if it represents a function.

Mathematics

1 answer:

Lana71 [14]2 years ago

6 0

Answer:

Domain: {4,5,7,10}

Range: {2,4,8}

Not a function

Step-by-step explanation:

The domain is all the x-values.

The range is all the y-values.

However, this isn't a function because two different inputs have the same output. If it were a function, there would be one unique output for every input.

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$P(8 < x \leq 12)=P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(15C9)(0.8)^{9} (1-0.8)^{15-9}=0.0429$

$P(X=10)=(15C10)(0.8)^{10} (1-0.8)^{15-10}=0.1032$

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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:

$X \sim Binom(n=15, p=1-0.2=0.8)$

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:

$P(8 < x \leq 12)=P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(15C9)(0.8)^{9} (1-0.8)^{15-9}=0.0429$

$P(X=10)=(15C10)(0.8)^{10} (1-0.8)^{15-10}=0.1032$

$P(X=11)=(15C11)(0.8)^{11} (1-0.8)^{15-11}=0.1878$

$P(X=12)=(15C12)(0.8)^{12} (1-0.8)^{15-12}=0.2501$

$P(8 < x \leq 12)=0.0429+0.1032+0.1878 +0.2501=0.5839$

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