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Volgvan
3 years ago
10

Need help quick im giving out brainliest.

Mathematics
2 answers:
Bond [772]3 years ago
5 0
False hope that helped
mojhsa [17]3 years ago
3 0
It’s False.
Hope that’s right :)
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The length of the human pregnancy is not fixed. It is known that it varies according to a distribution which is roughly normal,
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Answer:

a) Figure attached

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And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(234

And we can find this probability with this difference:

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An we know using the graph in part a that this area correspond to 0.95 or 95%

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And we can find this probability with this difference:

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An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%

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And using the z score we got:

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And that correspond to approximately 0.15%

Step-by-step explanation:

Part a

For this case we can see the figure attached.

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part b

Let X the random variable that represent the length of human pregnancy of a population, and for this case we know that:

Where \mu=266 and \sigma=16

We are interested on this probability

P(234

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(234

And we can find this probability with this difference:

P(-2

An we know using the graph in part a that this area correspond to 0.95 or 95%

Part c

P(250

And we can find this probability with this difference:

P(-1

An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%

Part d

We want this probability:

P(X

And using the z score we got:

P(X

And that correspond to approximately 0.15%

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