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

 

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