Question

Birth weights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz. The proportion of infants with birth weights between 125 oz and 140 oz is:

Answer 1

Answer:

The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 110, \sigma = 0.15$

The proportion of infants with birth weights between 125 oz and 140 oz is

This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So

X = 140

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

$Z = \frac{140 - 110}{15}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 125

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

$Z = \frac{125 - 110}{15}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.9772 - 0.8413 = 0.1359

The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.