Question

A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 623 babies born in New York. The mean weight was 3448 grams with a standard deviation of 862 grams. Assume that birth weight data are approximately bell-shaped. Estimate the number of newborns who weighed between 1724 grams and 5172 grams. Round to the nearest whole number.

Answer 1

Answer:

The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3448, \sigma = 862$

Proportion of newborns who weighed between 1724 grams and 5172 grams.

This is the pvalue of Z when X = 5172 subtracted by the pvalue of Z when X = 1724. So

X = 5172

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{5172 - 3448}{862}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 1724

$Z = \frac{X - \mu}{s}$

$Z = \frac{1724 - 3448}{862}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

Estimate the number of newborns who weighed between 1724 grams and 5172 grams.

0.9544 out of 623 babies. SO

0.9544*623 = 595

The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.