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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:

To rewrite these equations from point-slope form to standard form, you can do the following:

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1. y -6 = -6(x +2)

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<em>Additional comment</em>

The "standard form" of a linear equation is ax+by=c for integers a, b, c. The leading coefficient (generally, 'a') should be positive, and all coefficients should be mutually prime (have no common factors). That is why we multiply by -1 in problem 2.

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