Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 198 daysμ=198 days and standard deviation sigma equals 20 daysσ=20 days. complete parts​ (a) through​ (f) below. ​(a) what is the probability that a randomly selected pregnancy lasts less than 191191 ​days? the probability that a randomly selected pregnancy lasts less than 191191 days is approximately

We are given:
Population mean = u = 198
Population Standard Deviation = s = 20

Since population standard deviation is known, we will use z-distribution to solve this problem.

To find the probability that the pregnancy lasts less than 191 days, we have to first convert it into z score.

[tex]z= \frac{x-u}{s}= \frac{191-198}{20}=-0.35 [/tex]

So z score will be = - 0.35

From the z table, the probability of z score to be less than – 0.35 is 0.3632

Therefore, we can conclude that probability that a randomly selected pregnancy lasts less than 191 days is approximately 0.3632.