Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 285 days
and standard deviation sigma equals 28 days. What is the probability a random sample of size 20 will have a mean gestation period within 10 days of the​ mean?
The probability that a random sample of size 20 will have a mean gestation period within 10 days of the mean is _____ ​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that,

mean = = 285

standard deviation = = 28

n = 20

= = 285

= / n = 28/ 20 = 6.2610

The probability a random sample of size 20 will have a mean gestation period within 10 days of the​ mean.

P(275 < < 295) = P((275 - 285) / 6.2610<( - ) / < (295 - 285) / 6.2610))

= P(-1.60 < Z < 1.60)

= P(Z < 1.60) - P(Z < -1.60) Using standard normal table,  

= 0.9452 - 0.0548

= 0.8904

Probability = 0.8904  

The probability that a random sample of size 20 will have a mean gestation period within 10 days of the mean is 0.8904.