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RUDIKE [14]
3 years ago
14

A study was being conducted about birth weights of babies at a local hospital and found the average to be 7.6 pounds with a stan

dard deviation of 1.3 pounds (the distribution was approximately normal). Many pre-mature births weights are in the lowest 1% of births. What would be the birth weight associated with the lowest 1%?

Mathematics
1 answer:
dangina [55]3 years ago
7 0

Answer:

The birth weight associated with the lowest 1% is 4.6 pounds.

Step-by-step explanation:

Let <em>X</em> represent the birth weights of babies.

It is provided that X\sim N(7.6,1.3^{2})

It is also provided that many pre-mature births weights are in the lowest 1% of births.

Let <em>x</em> represent the births weights that are in the lowest 1% of births.

That is, P (X < x) = 0.01.

⇒ P (Z < z) = 0.01

The corresponding <em>z</em>-score is, <em>z</em> = -2.33.

Compute the value of <em>x</em> as follows:

z=\frac{s-\mu}{\sigma}\\\\-2.33=\frac{x-7.6}{1.3}\\\\x=7.6-(1.3\times 2.33)\\\\x=4.571\\\\x\approx 4.6

Thus, the birth weight associated with the lowest 1% is 4.6 pounds.

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Step-by-step explanation:

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so,

b = 40°

due to the parallel nature of the 2 lines there is a symmetry effect for such shapes inscribed a circle. the upper and the lower triangle must be similar. and when applying a vertical line through the central crossing point, everything to the left is mirrored by everything on the right.

so, angle c must be equal to angle b.

c = 40°

and as the sum of all angles in a triangle is always 180°, d is then

d = 180 - 40 - 40 = 100°

the interior angle of the arc angle a is the supplementary angle of d (together they are 180°), because together with d they cover the full down side of the top-left to bottom-right line.

interior angle to a = 180 - 100 = 80°

due to the symmetry again, the arc angle opposite to a is the same as a.

as we know, the interior angle to a pair of opposing arc angles is the mean value of the 2 angles.

so, we have

(a + a)/2 = 80

2a/2 = 80

a = 80°

there might (and actually should) be some more direct approaches for "a" out of the other pieces of information, but that was the most straight one right out of my mind, and I don't spend time on finding additional shortcuts, when I have already a working approach.

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I need three dollars seventy-eight cents more to buy a pair of tennis shoes. The tennis shoes cost eight dollars less the one hu
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3 years ago
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4x-2y=14
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Get just one variable on one side for both equations.

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The General Social Survey asked the question: "For how many days during the past30 days was your mental health, which includes s
serg [7]

Answer:

a) For this case we can conclude that we are 95% confident that the true mean for the variable of interest is between 3.4 and 4.24 days in the US population.

b) The 95% confident means that if we select 100 different samples and calculate the 95% confidence interval for each sample selected, then we will have approximately 95 out of 100 confidence intervals will contain the true mean for the parameter of interest.

c) If we have the same info but we want more confidence that implies that the confidence interval would be wider, since the margin of error increase with more confidence.

d) Assuming that we have the same confidence level and the value for the deviation not changes. If we see if we decrease the sample size, then the margin of error would be lower since the original sample size was 1151. So then if we use 500 Americans we would have a lower value for the margin of error for this new interval. And then our confidence interval would smaller than the original.  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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 confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The confidence interval after apply the formulas was (3.40 ,4.24)

Part a

For this case we can conclude that we are 95% confident that the true mean for the variable of interest is between 3.4 and 4.24 days in the US population.

Part b

The 95% confident means that if we select 100 different samples and calculate the 95% confidence interval for each sample selected, then we will have approximately 95 out of 100 confidence intervals will contain the true mean for the parameter of interest.

Part c

If we have the same info but we want more confidence that implies that the confidence interval would be wider, since the margin of erroe increase with more confidence, because the critical value increase.

Part d

We need to take in count that the margin of error is given by:

ME=t_{\alpha/2}\frac{s}{\sqrt{n}}

Assuming that we have the same confidence level and the value for the deviation s not changes. If we see if we decrease the sample size, then the margin of error would be lower since the original sample size was 1151. So then if we use 500 Americans we would have a lower value for the margin of error for this new interval. And then our confidence interval would smaller than the original.  

4 0
3 years ago
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