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What percentage of babies born in the United States are classified as having a low birthweight (<2500g)? explain how you got

lawyer [7]

Answer:

2.28% of babies born in the United States having a low birth weight.

Step-by-step explanation:

The complete question is: In the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g. What percent of babies born in the United States are classified as having a low birth weight (< 2,500 g)? Explain how you got your answer.

We are given that in the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g.

Let X = birth weights of newborn babies

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 3,500 g

$\sigma$ = standard deviation = 500 g

So, X ~ N( $\mu=3500, \sigma^{2} = 500$ )

Now, the percent of babies born in the United States having a low birth weight is given by = P(X < 2500 mg)

P(X < 2500 mg) = P( $\frac{X-\mu}{\sigma}$ < $\frac{2500-3500}{500}$ ) = P(Z < -2) = 1 - P(Z $\leq$ 2)

= 1 - 0.97725 = 0.02275 or 2.28%

The above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.97725.

4 0

4 years ago

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .