Question 1189726
You need the z-value
z=(x-mean)/sd where x is the specific value asked for.
for more than 2.5 minutes, you need z>(2.5-2.31)/0.6.
You can then take the z and use the table or the calculator 2ndVARS2 normal cdf(z value, 6) or any number greater than 6 standard deviations, everything to the right of the value.
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For a mean value, use the 2.31 as comparison and the sd is 0.36/sqrt(n). You will learn that while one value may be relatively extreme, the probability of the mean of a number of values being more extreme is a lot less.


z>(2.5-2.31)/0.36=0.53 and the probability of that is 0.2981
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this is z>(x bar-mean)/sigma/sqrt(25)
so z=0.19/0.072=2.64
that probability is 0.0041 (I rounded the sd; unrounded, the answer is 0.0042)
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z(0.97)=1.8807
1.8807=(x-2.31)/0.36
0.6771=x-2.31
x=2.987 minutes or 2 min 59 sec