Question

A study of U.S. births published on the website Medscape from WebMD reported that the average birth weight of babies was 7.5 pounds and the standard deviation was about 1.2 pounds. Assume the distribution is approximately normal. Find the proportion of babies with birth weights of 6 pounds or less. Round your answer to 2 decimal places.

Answer 1

Answer:

The proportion of babies with birth weights of 6 pounds or less is 0.1056 = 10.56%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Average birth weight of babies was 7.5 pounds and the standard deviation was about 1.2 pounds.

This means that $\mu = 7.5, \sigma = 1.2$

Find the proportion of babies with birth weights of 6 pounds or less.

This is the pvalue of Z when X = 6. So

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

$Z = \frac{6 - 7.5}{1.2}$

$Z = -1.25$

$Z = -1.25$ has a pvalue of 0.1056

The proportion of babies with birth weights of 6 pounds or less is 0.1056 = 10.56%.