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 flock
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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 flock
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Question 520336: 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 be expected to be smaller than b.)?
I can't tell if I need to use the Z-Score Table in my Statistics book for a.) or if I need to use a formula?
Thanks in advance!! Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! You don't need a table to find the z-score for a). You could use the formula but you don't have to either. A z-score of -1.5 means 1.5 standard deviations below the mean, or 1.5(4) = 6 below the mean. The size of the flock corresponding to z = -1.5 would be 38-6 = 32.
For part b), use a calculator and the normal cdf function.
