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2 years ago

How to construct frequency table with sturge approximation

Mathematics

1 answer:

alexgriva [62]2 years 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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Elan Coil [88]

Answer:

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3 years ago

Read 2 more answers

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bulgar [2K]

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Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
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$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
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(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

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3 years ago

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beks73 [17]

SOLUTION

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