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1 year ago

How to construct frequency table with sturge approximation

Mathematics

1 answer:

alexgriva [62]1 year ago

4 0

The stages to utilize to construct a frequency distribution table utilizing Sturge's approximation are as follows:

<h3>How to construct a frequency distribution table?</h3>

The steps to create a frequency distribution table utilizing Sturge's approximation exist as follows;

Step 1: Estimate the range of the data:

This exists simply by finding the difference between the biggest and the least values.

Step 2: Decide on the inaccurate number of classes in which the provided data exist to be grouped. The formula for this is;

K = 1 + 3.322logN

where; K= Number of classes

logN = Logarithm of the whole number of observations.

Step 3: Specify the approximate class interval size:

This exists acquired by splitting the range of data by the number of classes and exists symbolized by h class interval size.

Step 4: Find the starting point:

The lower class limit should take care of the least value in the raw data.

Step 5: Determine the remaining class boundaries:

When you have reached the lowest class boundary, then you can count the class interval size to the lower class boundary to obtain the upper-class boundary.

Step 6: Circulate the data into separate classes.

To learn more about the frequency distribution table refer to; brainly.com/question/27820465

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============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago

How to construct frequency table with sturge approximation​

How to construct frequency table with sturge approximation