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nalin [4]
4 years ago
10

in constructing a box and whisker plot from the data set below at which value would you draw the left and right side of the box

Mathematics
1 answer:
KatRina [158]4 years ago
8 0
A box and whisker plot is a type of statistical graph that shows the median distributions of the data set. The box represents the data median and the medias of the quartiles. The whiskers represents the lines extended from the box to the lowest and the highest data.

Take this data as an example:
<span>3.9,  4.1,  4.2,  4.3,  4.3,  4.4,  4.4,  4.4,  4.4,  4.5,  4.5,  4.6,  4.7,  4.8,  4.9,  5.0,  5.1
</span>First, find the middle value. That would be the median. In this case, that would be 4.4. Let's denote this as Q₂ or Quartile 2. Now, take the group of data set before and after the median. These are:

<span>3.9,  4.1,  4.2,  4.3,  4.3,  4.4,  4.4,  4.4
</span>
and

<span>4.5,  4.5,  4.6,  4.7,  4.8,  4.9,  5.0,  5.1
</span>
The middle of the first data set is Q₁ = 4.3, while the middle of the second set is Q₃ = (4.7+4.8)/2 = 4.75

To construct the box and whisker plot, create line for the lowest value 3,9, Q₁, Q₂, Q₃ and the highest value, 4.4. The left side of the box would be Q₁, while the right side of the box is Q₃. The line in between is the media Q₂. The whiskers are the lines connecting the lowest to Q₁, and Q₃ to the highest on the other side. The resulting box and whisker plot is shown in the attached picture.

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notka56 [123]
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3 0
3 years ago
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were o
ad-work [718]

The question is incomplete. Here is the complete qeustion.

In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15

(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.

(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?

Answer: (a) 0.089 ≤ μ ≤ 0.171

(b) No

Step-by-step explanation:

(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample

X = \frac{0.1+0.13+0.16+0.15+0.14+0.08+0.15}{7}

X = 0.13

s = \sqrt{\frac{(0.1-0.13)^{2} + (0.13 - 0.13)^{2} + ... + (0.15 - 0.13)^{2}}{7-1} }

s = 0.029

The degrees of freedom is

N - 1 = 7 - 1 = 6

And since the confidence is of 99%:

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

The t-test statistics for t_{6,0.005} is 3.707

(Value found in the t-distribution table)

Now, calculate Error:

E = t_{6,0.005} . \frac{s}{\sqrt{N} }

E = 3.707. \frac{0.029}{\sqrt{7} }

E = 0.041

The interval will be:  

0.13 - 0.041 ≤ μ ≤ 0.13+0.041

0.089 ≤ μ ≤ 0.171

(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.

7 0
4 years ago
Surface area of a 4m and 4m and 4m cube
jeka94

Answer: 96 m²

Step-by-step explanation: Surface area is the sum of the areas of all of the faces of a 3-dimensional figure.

So to find the surface area of a cube, it's important to understand that each face of a cube has the same area so we can simply find the area of one face of a cube and multiply by the number of faces which is 6.

To find the area of one face of a cube, we can use the formula for the area of a square which is S².

Since the length of a side is 4 meters, we have (4 m)² or 4 meters x 4 meters which is 16 m².

Now, since there are six sides to a cube, we have 6 (16 m)² which is equal to 96 m². So the surface area of the cube is 96 m².

6 0
3 years ago
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Dafna11 [192]
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Zina [86]

Answer:

  84? Not sure but pretty sure

Step-by-step explanation:

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If 90° and reflex turns are allowed, then the number substantially increases.

Corner R: can only go to the adjacent diagonal O, and from there to one other O, then to any of the 3 M's, for a total of 3 paths.

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Middle R can go the three O's on the adjacent row, so can access the three paths available from each corner O along with the 10 paths available from the center O, for a total of 16 paths.

Then paths accessible from the top row of R's are 3 +10 +16 +10 +3 = 42 paths. There are two such rows of R's so a total of 84 paths.

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3 years ago
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