Question

The length of human pregnancies from conception to birth varies according to a distribution that can be modeled by a normal random variable with mean 265 days and standard deviation 14 days. Question 1. What percent of pregnancies last less than 240 days? Note that the answer is requested as a percent. Use 2 decimal places in your answer. % Question 2. What percent of pregnancies last between 240 and 270 days? Note that the answer is requested as a percent. Use 2 decimal places in your answer. % Question 3. The longest 20% of pregnancies last at least how many days? (round to the nearest whole day)

Answer 1

Answer:

1) 3.67%

2) 60.39%

3) The longest 20% of pregnancies last at least 277 days.

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:

$\mu = 265, \sigma = 14$

Question 1. What percent of pregnancies last less than 240 days?

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

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

$Z = \frac{240 - 265}{14}$

$Z = -1.79$

$Z = -1.79$ has a pvalue of 0.0367

3.67% of pregnancies last less than 240 days

Question 2. What percent of pregnancies last between 240 and 270 days?

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

X = 270

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

$Z = \frac{270 - 265}{14}$

$Z = 0.36$

$Z = 0.36$ has a pvalue of 0.6406

X = 240

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

$Z = \frac{240 - 265}{14}$

$Z = -1.79$

$Z = -1.79$ has a pvalue of 0.0367

0.6406 - 0.0367 = 0.6039

60.39% of pregnancies last between 240 and 270 days

3. The longest 20% of pregnancies last at least how many days?

They last at least X days, in which X is found when Z has a pvalue of 1-0.2 = 0.8. So it is X when Z = 0.84.

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

$0.84 = \frac{X - 265}{14}$

$X - 265 = 0.84*14$

$X = 277$

The longest 20% of pregnancies last at least 277 days.