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3 years ago

Help me ASAP it’s timed

Mathematics

1 answer:

Rama09 [41]3 years ago

3 0

G and h
No explanation
Sorry
Tried my best

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The following six independent length measurements were made (in feet) for a line: 736.352, 736.363, 736.375, 736.324, 736.358, a

horrorfan [7]

Answer:

Assuming population data

$\sigma = \sqrt{0.000354}=0.0188$

Assuming sample data

$s = \sqrt{0.000425}=0.0206$

Step-by-step explanation:

For this case we have the following data given:

736.352, 736.363, 736.375, 736.324, 736.358, and 736.383.

The first step in order to calculate the standard deviation is calculate the mean.

Assuming population data

$\mu = \frac{\sum_{i=1}^6 X_i}{6}$

The value for the mean would be:

$\mu = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359$

And the population variance would be given by:

$\sigma^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6}$

And we got $\sigma^2 =0.000354$

And the deviation would be just the square root of the variance:

$\sigma = \sqrt{0.000354}=0.0188$

Assuming sample data

$\bar X = \frac{\sum_{i=1}^6 X_i}{6}$

The value for the mean would be:

$\bar X = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359$

And the population variance would be given by:

$s^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6-1}$

And we got $s^2 =0.000425$

And the deviation would be just the square root of the variance:

$s = \sqrt{0.000425}=0.0206$

6 0

3 years ago