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Vika [28.1K]
3 years ago
13

Pls help fast 98 points again point spree)) What is the mean absolute deviation for 9, 7, 6, 8, 7, and 5?

Mathematics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

6

Step-by-step explanation:

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It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir
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Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

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Now, we find the margin of error M as such

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In which \sigma is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

M = z*\frac{\sigma}{\sqrt{n}}

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

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Step-by-step explanation:

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