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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.

THE ANSWER IS:
The z-score for 2,500 g is –2.According to the empirical rule, 95% of babies have a birth weight of between 2,500 g and 4,500 g.5% of babies have a birth weight of less than 2,500 g or greater than 4,500 g.Normal distributions are symmetric, so 2.5% of babies weigh less than 2,500g.

Respuesta :

merlynthewhizz
First, we sketch this information on a normal distribution graph as shown below

We are looking to find the probability of baby weight <2500gram

We need to standardize the value 2500

[tex]z= \frac{2500-3500}{500} [/tex]
[tex]z=-2[/tex]

Next, we need to look at the z-table. If your table only shows value for Z<z, then use the property of symmetry. It's shown in the last diagram. We read the probability of  P(z<2)=0.9772, then we do [tex]1-0.9772[/tex] to obtain the area to the right of [tex]z=2[/tex], which is the same size with the area of [tex]z\ \textless \ -2[/tex]

[tex]P(Z,-2)=0.0228[/tex]

Which means the percentage of baby with weight <2500 is 0.0228×100=2.28%
Ver imagen merlynthewhizz
ogorwyne

Based on the empirical rule, 95% of babies have a birth weight of between 2,500 g and 4,500g, therefore, 2.5% of babies weigh less than 2,500g.

What does the empirical rule of the question state?

It states that 68 percent of the values are in 1 standard deviation from the mean average. 0.95 are from 2 SD of the mean and 99.7 from the 3 SD.

(μ) = 3,500 g and the Standard deviation is 500 grams

This shows that confirmation that 95% of babies have a birth weight of between two standard deviation = 2,500 g and 4,500 g.

Hence we concludce that 2.5% of babies weigh less than 2,500g.

Read more on probability here: https://brainly.com/question/24756209

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