SOLUTION: The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8.6 cm.
a. Find the probability that an individual dis
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-> SOLUTION: The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8.6 cm.
a. Find the probability that an individual dis
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Question 1128414: The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8.6 cm.
a. Find the probability that an individual distance is greater than
210.00cm.
b. Find the probability that the mean for 20 randomly selected distances is greater than 196.00 cm.
A. 0.0735 I got this one.
B. Need help with this and please show work so I can understand. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
z>(210-197.5)/8.6 or z>1.45
That probability is 0.0735
For 20, the denominator is sigma/sqrt(n) That is 8.6/sqrt(20)
that would be 1.92, the z value is >-1.5/1.92 or >-0.78
The -1.5 is for 196-197.5 and that is divided by the 1.92
The probability is 0.7823
