Question

The weights of newborn babies are distributed normally, with a mean of approximately 105 Oz and a standard deviation of 15 Oz. If a newborn baby is selected at random. What if the probably that the baby weighs less than 150 Oz? help please!! Also I need to round answer to nearest thousandth

Answer 1

Answer:

0.999 = 99.9% probability that the baby weighs less than 150 Oz.

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.

The weights of newborn babies are distributed normally, with a mean of approximately 105 Oz and a standard deviation of 15 Oz.

This means that $\mu = 105, \sigma = 15$

What if the probabability that the baby weighs less than 150 Oz?

This is the pvalue of Z when X = 150. SO

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

$Z = \frac{150 - 105}{15}$

$Z = 3$

$Z = 3$ has a pvalue of 0.999

0.999 = 99.9% probability that the baby weighs less than 150 Oz.