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Sloan [31]
1 year ago
15

68. Traveling to work How long do people travel each day to get to work ? The following table gives the average travel times to

work (in minutes ) for workers in each state and the District of Columbia who are at least 16 years old and don't work at home .
a) Make a histogram to display the time travel time data using intervals of within 2 minutes, starting at 14 minutes.

b) Describe the shape of the distribution. What is the most common interval of travel times?

Mathematics
1 answer:
vitfil [10]1 year ago
5 0

A histogram is an image of data that resembles a bar graph and groups several categories into columns along the horizontal x-axis. The numerical count or percentage of occurrences for each column in the data are shown on the vertical y-axis. To see how data distribution patterns look, utilize columns.

(a) Use classes with a 2 minute width and a 14 minute starting point to create a histogram of the journey times. Thus, the first lesson lasts between 14 and 16 minutes.

Describe the distribution's shape. What is the average journey time interval?

If a histogram has a bell shape, its center and spread can be used to parsimoniously characterize it. The axis of symmetry is located at the middle. The spread is the separation between the center and a particular point of inflection. The inflection points of the bell-shaped histogram are indicated here.

To learn more about  Histogram refer to:

brainly.com/question/25983327

#SPJ13

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Answer:

a) A sample size of 5615 is needed.

b) 0.012

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}}

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z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that \pi = 0.2

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

0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}

0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}

\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}

(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}

n = 5615

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now \pi = 0.12, n = 5615.

We have to find M.

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

M = 2.81\sqrt{\frac{0.12*0.88}{5615}}

M = 0.012

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3 years ago
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