Question

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with a mean of 266 days and standard deviation of 16 days.a. What percent of pregnancies last less than 240 days?b. What percent of pregnancies last between 240 days and 270 days?c. How long do the longest 20% of pregnancies last?

Answer 1

Answer:

a) 5.16% of pregnancies last less than 240 days.

b) 54.71% of pregnancies last between 240 days and 270 days.

c) The longest 20% of pregnancies last at least 279.44 days.

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:

$\mu = 266, \sigma = 16$

a. What percent of pregnancies last less than 240 days?

This is the pvalue of Z when X = 240. So:

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

$Z = \frac{240 - 266}{16}$

$Z = -1.63$

$Z = -1.63$ has a pvalue of 0.0516.

So 5.16% of pregnancies last less than 240 days.

b. What percent of pregnancies last between 240 days and 270 days?

This is the pvalue of Z when X = 270 subtracted by the pvalue of Z when X = 240.

X = 270

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

$Z = \frac{270 - 266}{16}$

$Z = 0.25$

$Z = 0.25$ has a pvalue of 0.5987.

X = 240

From a., has a pvalue of 0.0516.

So 0.5987 - 0.0516 = 0.5471 = 54.71% of pregnancies last between 240 days and 270 days.

c. How long do the longest 20% of pregnancies last?

This is the value of Z when X has a pvalue of 1-0.2 = 0.8.

So it is X when Z = 0.84.

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

$0.84 = \frac{X - 266}{16}$

$X - 266 = 0.84*16$

$X = 279.44$

The longest 20% of pregnancies last at least 279.44 days.