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2. (30 points) The height of an adult male black bear is normally distributed with a mean of 30 inches and standard deviation of 3 inches.
a. If a male bear is selected at random, what is the probability that his height is above 34 inches?
b. If a sample of 36 male bears is selected at random, what is the probability that their average height is above 34 inches?
c. How many male bears must be selected so that there is only a 1% chance that the average will be above 34 inches?
A. P(34
The probability that the bear heght is above 34 inches is 9.18%
B.
n=36 P= 0.0918
36×0.0918= 3.3
The probability that their average height is 34 inches is 3.3
Somebody knows how to do C????
Professor there is not examples like that in the book
A. P (x>34)
34-30/3 =1.33
P (z>1.33)= .9082
1-.9082= .0918
B. Larger sample = higher P
.0918(36)= 3.30
z=3.30 =.9995
B. This is a question about the sample mean. The standard deviation for the sample mean is sigma/sqrt(n)=3/sqrt(36)-.5.
P(Xbar>34) is (via standardizing) the same as P(Z<(34-30)/.5=8) which is essentially 0.
C. z_.01=2.33 Now (34-30)/(sigma/sqrt(n))=z or 4*sqrt(n)/3=2.33. Solving for n, we get sqrt(n)=3*2.33/4 or n=(3*2.33/4)^2=3 (or technically 4,since it is slightly above 3).
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