Question

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

Answer 1

The z-score for 2,500 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 2,500 g.

Explanation:

did the assignment on edge:)

Answer 2

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

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 =
`(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(
`(X-\mu)/(\sigma)` <
`(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.