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Gennadij [26K]
3 years ago
11

Can u pls help me with this question ​

Mathematics
1 answer:
const2013 [10]3 years ago
5 0

Answer:

D.

Step-by-step explanation:

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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:

<u>The complete question is</u>: 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 = <u><em>birth weights of newborn babies</em></u>

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.

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