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Question 1182544: The mean height of an adult giraffe is 18 feet.
Suppose that the distribution is normally distributed with standard deviation 1 feet.
Let X be the height of a randomly selected adult giraffe.
Round to 4 decimal places
a. What is the distribution of X? X - N ( 18 1 )
b. What is the median giraffe height? 18 ft.
c. What is the Z-score for a giraffe that is 20 foot tall? 2
d. What is the probability that a randomly selected giraffe will be shorter than 17.3 feet tall? 0.242
e. What is the probability that a randomly selected giraffe will be between 18.8 and 19.5 feet tall?
___________
f. The 90th percentile for the height of giraffes is 19.282 ft.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The variance is 1 ft^2 as well, and that is what the 1 stands for in N~(18, 1)
Median is 18 ft.
z-score is 2
shorter than 17.3 is a z-score of -0.7/1 or -0.7. That probability is 0.2420.
e.is z between +0.8 and +1.5, with a probability of 0.1450. Can also do with 2nd VARS2normalcdf(18.8,19.5,18,1)ENTER
f. The 90th percentile is where z=1.282, and the height would therefore be 19.282 ft., as written
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