1.3k views
4 votes
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

User Zmirc
by
8.2k points

2 Answers

3 votes

Answer:

The probability that a randomly selected pregnancy lasts less than 191 days is approximately 0.36317.

Explanation:

Given : Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 198 days and standard deviation is 20 days.

To find : What is the probability that a randomly selected pregnancy lasts less than 191 ​days?

Solution :

Mean is
\mu=198

Standard Deviation
\sigma=20

The formula to find z-score is


z=(x-\mu)/(\sigma)

We have to find the probability that the pregnancy lasts less than 191 days.

So, x=191

Substitute the value in the formula,


z<(191-198)/(20)


z<(-7)/(20)


z<-0.35

Referring to z-table we find the value of z less that -0.35

Value of z less than -0.35 is 0.36317

Therefore, The probability that a randomly selected pregnancy lasts less than 191 days is approximately 0.36317.

User Eunyoung
by
8.1k points
5 votes
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.


z= (x-u)/(s)= (191-198)/(20)=-0.35

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.
User DuyguK
by
7.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.