## Question 401:

1## Answer:

No answer provided yet.We'd calculate the z-score for this event, then find the probability for this normal score. To get a z-score, subtract the data-point from the mean and divide that result by the standard deviation.

- So (308-268)/15 = a z-score of 2.667.
- Now we lookup the area under the normal curve up to 2.667 standard units using the z-score to percentile calculator. Select 1-sided area and you should get 99.617%.

In other words, this pregnancy is longer than more than 99% of all pregnancies, so I'd say that's unusual.