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Anestetic [448]
3 years ago
6

Which statement correctly compares the mean number of students for the data in the plots?

Mathematics
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

<em><u>"There are about 2 more students in each class at Oak Middle School than at Poplar Middle School."</u></em>

Step-by-step explanation:

We need to compute the mean for each.

<u>Poplar Middle School:</u>

Lets list the numbers

20,20,20,21,21,21,21,21,22,22,22,22,22,22,22,23,23,23,23,24

The average is sum of all divided by number of numbers, that will be:

435/20 = 21.75

<u>Oak Middle School:</u>

The numbers are:

20,21,21,22,22,23,23,23,23,24,24,24,25,25,26,26,27,27,28,29

The average is sum divided by number of numbers, that will be:

483/20 = 24.15

The difference in mean is:

24.15 - 21.75 = 2.4

We can say, that is about 2 more students in Oak Middle School, correct answer choice is "2nd answer choice" - <em><u>"There are about 2 more students in each class at Oak Middle School than at Poplar Middle School."</u></em>

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2 years ago
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2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for
kirza4 [7]

a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

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

The margin of error is given by:

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

The variables are listed as follows:

  • \pi is the sample proportion, which is also the estimate of the parameter.
  • z is the critical value.
  • n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of \frac{1+0.90}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

n = 150, \pi = \frac{15}{150} = 0.1

The margin of error is of:

M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%

The interval is given by the estimate plus/minus the margin of error, hence:

  • The lower bound is: 10 - 2.45 = 7.55%.
  • The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

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

0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}

\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}

(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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Answer:

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Step-by-step explanation:

The LCM is 18.

9x2=18

2x9=18

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