Question 973394: Suppose the lengths of the pregnanices of a certain animal are approximately normally distributed with a mena of u=207 and a standard deviation of 17 days.
A)what is the probability that a randomly selected pregancy lasts less than 201 days?
B)What is the probability that a random sample of 16 pregnancies has a mean gestation period of 201 days or less?
C)What is the probability that a random sample of 37 pregnancies has a mean gestation period of 201 days or less?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! mean is 207 and SD is 17
z=(x-mean)/SD
=(201-207)/17
want the value of z <(-6/17) or z< -0.353
This is 0.3620. Intuitively, that makes sense. About 16% of the probability is with a z-score <-1, so this should be more.
z=(x-mean(/sd/sqrt(n)) The probability should be much less, because the mean of 16 has to be less than 201, not just one.
=(201-207)/(17/4)= (-6/4.25)
Now, the z< -1.411
Probability is 0.079, much less as expected
I can write
z= (-6)*sqrt (37)/17 I have inverted the denominator here, and the sqrt (n) comes to the numerator.
z=-2.147
Probability is 0.0159.
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