Afactory dumps an average of 2.43 tons of pollutants into a river every week. if the standard deviation is 0.88 tons, what is the probability that in a week more than 3 tons are dumped? (assume normal distribution)

25.8%  
First, determine how many standard deviations from the norm that 3 tons are. So:
 (3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273  
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%  
So the probability that more than 3 tons will be dumped in a week is 25.8%

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