find the probability for 158 min?

label Mathematics
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The average time a person spends in each visit to an online social networking service is

60 minutes. 
The standard deviation is 10 minutes. If a visitor is selected at random, find the probability that he or she will
spend at least 158 minutes?
The 158 seems to be too much, I can't get this one figured. Thanks for your help!
May 30th, 2015

60 minutes. s.d. =10 

so do 158-60= 98   98/10= 9.8 so the S.D. is 9.8 

now whenever you get a number greater than 3.4 or is bigger than the table in your textbook you assume 0.99999 now we are looking for someone who spends at least 158 minutes so we are looking for a person who spends 158 minutes or more which means were looking for the area to the right of the curve 1-0.99999=0.00001 then X100 for a percentage which is 0.001% which basically means noone spends that amount of time on social media.

May 30th, 2015

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May 30th, 2015
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May 30th, 2015
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