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 a standard deviation of 16 days.
(1) Using the 68-95-99.7% rule, between what two lengths do the most typical 68% of all pregnancies fall 95%, 99.7%?
(2) What percent of all pregnancies last less than 250 days?
(a) What percentage of pregnancies last between 241 and 286 days?
(b) What percentage of pregnancies last more than 286 days?
(c) What percentage of pregnancies last more than 333 days?
(3) What length cuts off the shortest 2.5% of pregnancies?
(4) Find the quartiles for pregnancy length.
(5) Between what two lengths are the most typical 72% of all pregnancies?

Answer 1

Answer:

1) The most typical 68% of pregnancies last between 250 and 282 days, the most typical 95% between 234 and 298 days, and the most typical 99.7% between 218 and 314 days.

2) 15.87% of all pregnancies last less than 250 days

2a) 83.5% of pregnancies last between 241 and 286 days

2b) 10.56% of pregnancies last more than 286 days.

2c) 0% of pregnancies last more than 333 days

3) A pregnancy length of 234.6 days cuts off the shortest 2.5% of pregnancies.

4) The first quartile of pregnancy lengths is of 255.2, and the third quartile is of 276.8 days.

5) The most typical 72% of all pregnancies last between 248.72 and 283.28 days.

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

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 266 days and a standard deviation of 16 days.

This means that $\mu = 266, \sigma = 16$

(1) Using the 68-95-99.7% rule, between what two lengths do the most typical 68% of all pregnancies fall 95%, 99.7%?

68%: within 1 standard deviation of the mean, so 266 - 16 = 250 days to 266 + 16 = 282 days.

95%: within 2 standard deviations of the mean, so 266 - 32 = 234 days to 266 + 32 = 298 days.

99.7%: within 3 standard deviations of the mean, so 266 - 48 = 218 days to 266 + 48 = 314 days.

The most typical 68% of pregnancies last between 250 and 282 days, the most typical 95% between 234 and 298 days, and the most typical 99.7% between 218 and 314 days.

(2) What percent of all pregnancies last less than 250 days?

The proportion is the p-value of Z when X = 250. So

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

$Z = \frac{250 - 266}{16}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.1587*100% = 15.87%.

15.87% of all pregnancies last less than 250 days.

(a) What percentage of pregnancies last between 241 and 286 days?

The proportion is the p-value of Z when X = 286 subtracted by the p-value of Z when X = 241. So

X = 286

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

$Z = \frac{286 - 266}{16}$

$Z = 1.25$

$Z = 1.25$ has a p-value of 0.8944.

X = 241

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

$Z = \frac{241 - 266}{16}$

$Z = -1.56$

$Z = -1.56$ has a p-value of 0.0594.

0.8944 - 0.0594 = 0.835*100% = 83.5%

83.5% of pregnancies last between 241 and 286 days.

(b) What percentage of pregnancies last more than 286 days?

1 - 0.8944 = 0.1056*100% = 10.56%.

10.56% of pregnancies last more than 286 days.

(c) What percentage of pregnancies last more than 333 days?

The proportion is 1 subtracted by the p-value of Z = 333. So

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

$Z = \frac{333 - 266}{16}$

$Z = 4.19$

$Z = 4.19$ has a p-value of 1

1 - 1 = 0% of pregnancies last more than 333 days.

(3) What length cuts off the shortest 2.5% of pregnancies?

This is the 2.5th percentile, which is X when Z = -1.96.

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

$-1.96 = \frac{X - 266}{16}$

$X - 266 = -1.96*16$

$X = 234.6$

A pregnancy length of 234.6 days cuts off the shortest 2.5% of pregnancies.

(4) Find the quartiles for pregnancy length.

First quartile the 25th percentile, which is X when Z = -0.675.

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

$-0.675 = \frac{X - 266}{16}$

$X - 266 = -0.675*16$

$X = 255.2$

Third quartile is the 75th percentile, so X when Z = 0.675.

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

$0.675 = \frac{X - 266}{16}$

$X - 266 = 0.675*16$

$X = 276.8$

The first quartile of pregnancy lengths is of 255.2, and the third quartile is of 276.8 days.

(5) Between what two lengths are the most typical 72% of all pregnancies?

Between the 50 - (72/2) = 14th percentile and the 50 + (72/2) = 86th percentile.

14th percentile:

X when Z = -1.08.

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

$-1.08 = \frac{X - 266}{16}$

$X - 266 = -1.08*16$

$X = 248.72$

86th percentile:

X when Z = 1.08.

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

$1.08 = \frac{X - 266}{16}$

$X - 266 = 1.08*16$

$X = 283.28$

The most typical 72% of all pregnancies last between 248.72 and 283.28 days.