Question

The mean birth weight for children born in a certain country at full term (after 40 weeks) is 3500grams; the standard deviation for birth weight is 500 grams. The shape of the distribution is symmetric and unimodal. About 34% of the weights are between what two values?

Answer 1

Answer:

About 34% of the weights are between 3280 grams and 3720 grams.

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.

Mean of 3500 grams, standard deviation of 500 grams.

This means that $\mu = 3500, \sigma = 500$

About 34% of the weights are between what two values?

Between the 50 - (34/2) = 33rd percentile and the 50 + (34/2) = 67th percentile.

33rd percentile:

X when Z has a p-value of 0.33, so X when Z = -0.44.

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

$-0.44 = \frac{X - 3500}{500}$

$X - 3500 = -0.44*500$

$X = 3280$

67th percentile:

X when Z has a p-value of 0.67, so X when Z = 0.44.

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

$0.44 = \frac{X - 3500}{500}$

$X - 3500 = 0.44*500$

$X = 3720$

About 34% of the weights are between 3280 grams and 3720 grams.