SOLUTION: Youth volleyball practice times are normally distributed with a mean of 75 minutes and standard deviation of 10 minutes. Which of the following would be considered an unusual pra

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Question 847359: Youth volleyball practice times are normally distributed with a mean of 75 minutes and standard deviation of 10 minutes.
Which of the following would be considered an unusual practice?
50 minutes
65 minutes
90 minutes
all of the above
none of the above

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
mean of 75 minutes and standard deviation of 10 minutes.
50 minutes z = -2.5 an unusual practice
65 minutes z = -1
90 minutes z = 1.5
For the normal distribution:
one standard deviation from the mean accounts for about 68.2% of the set
two standard deviations from the mean account for about 95.4%
and three standard deviations from the mean account for about 99.7%.
Important to Understand z -values as they relate to the Standard Normal curve:
Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.
Note: z = 0 (x value the mean) 50% of the area under the curve is to the left and 50% to the right



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