SOLUTION: The average number of turkeys per flock in Kennebec County is 38 with a standard deviation of 4.
A. What size flock would have a z score of --1.5?
B. What percent of flocks would
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-> SOLUTION: The average number of turkeys per flock in Kennebec County is 38 with a standard deviation of 4.
A. What size flock would have a z score of --1.5?
B. What percent of flocks would
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Question 586713: The average number of turkeys per flock in Kennebec County is 38 with a standard deviation of 4.
A. What size flock would have a z score of --1.5?
B. What percent of flocks would have expected to be smaller than A)? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The average number of turkeys per flock in Kennebec County is 38 with a standard deviation of 4.
A. What size flock would have a z score of --1.5?
Use x = z*s + u
x = -1.5*4 + 38
x = 32
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B. What percent of flocks would have expected to be smaller than A)?
Solution: P(z < -1.5) = normalcdf(-100,-1.5) = 0.0668 = 6.68%
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Cheers,
Stan H.
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