Question

The lengths of pregnancies

VARIANCE of 225 days.
A) 0.2375
of humans are normally distributed with a mean of 268 days and a
Find the probability of a pregnancy lasting more than 300 days.
B) 0.0166
C) 0.9834
D) 0.3189

Answer 1

The probability of a pregnancy exceeding 300 days is found by calculating the z-score with the value of 300, the mean of 268, and the standard deviation of 15. The z-score is 2.13, and using the z-table we find the probability associated with this z-score, which in this case is 0.0166.

Step-by-step explanation:

To find the probability of a pregnancy lasting more than 300 days, given that the lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation (which is the square root of the variance) of 15 days (since the variance is 225), we need to calculate the z-score for 300 days.

First, calculate the z-score using the formula:

Z = (X - μ) − σ, where X is the value we are interested in (300 days), μ is the mean (268 days), and σ is the standard deviation (15 days).

Z = (300 - 268) − 15 = 32 − 15 = 2.13

After calculating the z-score, we use the z-table to find the probability that Z is greater than 2.13. This gives us the probability of a pregnancy lasting more than 300 days. Let's assume this probability is found to be 0.0166 (this value would need to be looked up in a z-table or found using a statistical software).

The correct answer is B) 0.0166, which is the probability of a pregnancy lasting more than 300 days.