The area covered by 1 L of a certain stain is normally distributed with mean 10 
a. What is the probability that 1 L of stain will be enough to cover 10.3 
?
b. What is the probability that 2 L of stain will be enough to cover 19.9 ?
(a) The z-score of 10.3 is (10.3 ? 10)/0.2 = 1.5. The area to the right of z = 1.5 is 1 ? 0.9332 = 0.0668. The probability is 0.0668.
(b) The mean area covered by 2 L is 2(10) = 20, and the standard deviation is 
. The z score of 19.9 is (19.9 ? 20)/0.28284 = ?0.35. The area to the right of z = ?0.35 is 1 ? 0.3632 = 0.6368. The probability is 0.6368.
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