Question

According to the National Vital Statistics, full-term babies' birth weights are Normally distributed with a mean of 7.5 pounds and a standard deviation of 1.1 pounds. What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

Answer 1

Answer:

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

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:

$\mu = 7.5, \sigma = 1.1$

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So

X = 8.6

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

$Z = \frac{8.6 - 7.5}{1.1}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 6.4

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

$Z = \frac{6.4 - 7.5}{1.1}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds