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balandron [24]
3 years ago
11

Refer to diagram. The percent of all voters are registered Democrats is part of which of the following distributions?

Mathematics
2 answers:
djyliett [7]3 years ago
8 0
A, C and D are all irrelevant to the question.

The answer is B) The marginal distribution of political party.

Hope this helps!

kozerog [31]3 years ago
5 0

Answer:

The percent of all voters that are registered Democrats is part of the marginal distribution of political party - Answer B.

Step-by-step explanation:

Marginal distribution is the probability function or description of a subset of variables from a  particular sample data. It is created or described when, in a sample we have a specific categorization odf our data. When we take into account both subsets simultaneously we talk of a joint distribution.

In your table there are two subsets or categories. Gender and political party. The question only talks about the democrats, one value of the sepcific category of political parties., so its not a joint distribution. As we are talking only of political parties , withput considering gender of democrats, the marginal distribution to consider is the one of political parties

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Answer:

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Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....

where, n = number of samples taken = 150

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           p = probability of success which in our question is % of bulbs that

                  are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

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<u>Mean of X</u>, \mu = n \times p = 150 \times 0.10 = 15

<u>Standard deviation of X</u>, \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.10 \times (1-0.10)} = 3.7

So, X ~ N(\mu = 15, \sigma^{2} = 3.7^{2})

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    P(X < 20.5) = P( \frac{X-\mu}{\sigma} < \frac{20.5-15}{3.7} ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = n \times p = 150 \times 0.10 = 15.

                                           

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3 years ago
198 -176 - {48 -(24 + 8 + 4)}}​
Igoryamba

Answer:

the answer is 10

Step-by-step explanation:

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