Pant 1 di 3
Suppose the lengths of the pregnancies of a certain animal are approuimately normaly distibuted with mean $\mu=282$ days and standard deviation $\sigma=15$ days.
Click here to view the standard normal distribution table ipage 11.
Click here to view the standard nocmal cistribution table ipage 2 .
(a) What is the probability that a randomly selected pregnancy lasts less than 276 days?

The probability that a randomly selected pregnancy laets less than 276 drys is approximately $\square$ (Round to four decimal places as needed)

Step 1 ：The problem is asking for the probability that a pregnancy lasts less than 276 days. This is a problem of normal distribution. We know that the mean is 282 days and the standard deviation is 15 days.

Step 2 ：We can calculate the z-score for 276 days using the formula \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we're interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：Substituting the given values into the formula, we get \(z = \frac{276 - 282}{15} = -0.4\).

Step 4 ：We then use the standard normal distribution table to find the probability corresponding to this z-score, which is approximately 0.3446.

Step 5 ：Thus, the probability that a randomly selected pregnancy lasts less than 276 days is approximately \(\boxed{0.3446}\) (rounded to four decimal places as needed).