Question 624015: I am having trouble understanding "at most" and "at least"
If, for example, the problem was " ... Their times are normally distributed with a mean of 184 minutes and a standard deviation of 55 minutes. For a randomly selected person from the control group, find the probability that the time is at least 210 minutes."
I used the following function on my TI-84:
2ND, VARS, normalcdf(209.5,1E99,184,55)
=0.321454
=32.14%
is this correct?
Thanks for your time.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
for at least 210, using 209.5 is the correct spproach
z = (209.5 - 184)/25 = .4636 which results in .6786 or 67.86% to the left
Note: we know as 210 >184 that this P has to be greater than 50% for
Also as to Your P(x≥210) =  Note (1-.3214) = .6786
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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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