Question

Babies’ birth weights are approximately normally distributed with mean 3375 grams and standard deviation of 493 grams. An obstetrician wants to conduct a study of babies that are unusually heavy or unusually light, to try to understand what genetic and nutritional factors might contribute to this. He decides that babies with weights in the top 2.5% will be considered unusually heavy, and babies with weights in the bottom 2.5% will be considered unusually light.
Babies that weigh more than ___ grams or less than ___ grams will be included in the study.

Answer 1

Answer:

Babies that weigh more than 2409 grams or less than 4341 grams will be included in the study.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 3375, \sigma = 493$

Top 2.5%

More than X when Z has a pvalue of 1-0.025 = 0.975. So more than X when Z = 1.96.

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

$1.96 = \frac{X - 3375}{493}$

$X - 3375 = 493*1.96$

$X = 4341$

More than 4341 grams is the top 2.5%.

Bottom 2.5%.

Less than X when Z has a pvalue of 0.025. So less than X when Z = -1.96.

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

$-1.96 = \frac{X - 3375}{493}$

$X - 3375 = 493*(-1.96)$

$X = 2409$

Less than 2409 grams is in the bottom 2.5%.

Babies that weigh more than 2409 grams or less than 4341 grams will be included in the study.