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frosja888 [35]
2 years ago
9

Using a proportion, what can you infer about the number of households in her town that have more than three children?

Mathematics
1 answer:
eduard2 years ago
4 0

Answer:

About 296 households have more than three children.

Step-by-step explanation:

The complete question is:

Carmen used a random number generator to simulate a survey of how many children live in the households in her town. There are 1,346 unique addresses in her town with numbers ranging from zero to five children. The results of 50 randomly generated households are shown below. Children in 50 Households Number of Children Number of Households 0 5 1 11 2 13 3 10 4 9 5 2 Using a proportion, what can you infer about the number of households in her town that have more than three children? About 242 households have more than three children. About 269 households have more than three children. About 296 households have more than three children. About 565 households have more than three children.

Solution:

The data provided for the number of children and number of households is:

Number of Children (<em>X</em>)          Number of Households (<em>f</em> (X))

               0                                                   5

               1                                                    11

               2                                                   13

               3                                                   10

               4                                                    9

               5                                                    2

           TOTAL                                             50

Compute the probability of households having more than three children as follows:

P (More than 3 children) = P (4 children) + P (5 children)

                                        =\frac{9}{50}+\frac{2}{50}

                                        =\frac{11}{50}

                                        =0.22

Let the random variable <em>Y</em> be defined as the number of households having more than three children.

The probability of the random variable <em>Y </em>is, <em>p</em> = 0.22.

The entire population, i.e. total number of addresses in Carmen's town with numbers ranging from zero to five children, consists of <em>N</em> = 1,346 households.

Compute the expected number of households having more than three children as follows:

E(Y) =N\times p

         =1346\times 0.22\\=296.12\\\approx 296

Thus, about 296 households have more than three children.

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Step-by-step explanation:

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Answer:

a) n = 1037.

b) n = 1026.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

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So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

​(a) Assume that nothing is known about the percentage to be estimated.

We need to find n when M = 0.04.

We dont know the percentage to be estimated, so we use \pi = 0.5, which is when we are going to need the largest sample size.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 2.575*0.5

(\sqrt{n}) = \frac{2.575*0.5}{0.04}

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n = 1037.

(b) Assume prior studies have shown that about 55​% of​ full-time students earn​ bachelor's degrees in four years or less.

\pi = 0.55

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0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}

(\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}

(\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}

n = 1025.7

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n = 1026.

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