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katen-ka-za [31]
3 years ago
8

Pls help a girl out .......

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

Yes

Step-by-step explanation:

The sentence does correctly state the equation written.

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How many women must be randomly selected to estimate the mean weight of women in one age group? We want 90% confidence that the
ki77a [65]

Answer:

155 women must be randomly selected.

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 - 0.9}{2} = 0.05

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

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

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

The population standard deviation is known to be 28 lbs.

This means that \sigma = 28

We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean. How many women must be sampled?

This is n for which M = 3.7. So

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

3.7 = 1.645\frac{28}{\sqrt{n}}

3.7\sqrt{n} = 1.645*28

\sqrt{n} = \frac{1.645*28}{3.7}

(\sqrt{n})^2 = (\frac{1.645*28}{3.7})^2

n = 154.97

Rounding up:

155 women must be randomly selected.

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3 years ago
2h + (6 + h) + 6 ∙ 2
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