The amount of paint required to paint a surface with an area of 50 
is normally distributed with mean 6 L and standard deviation 0.3 L.
a. If 6.2 L of paint are available, what is the probability that the entire surface can be painted?
b. How much paint is needed so that the probability is 0.9 that the entire surface can be painted?
c. What must the standard deviation be so that the probability is 0.9 that 6.2 L of paint will be sufficient to paint the entire surface?
(a) The z-score of 6.2 is (6.2?6)/0.3 = 0.67. The area to the left of z= 0.67 is 0.7486. The probability is 0.7486.
(b) The amount of paint needed is the 90th percentile of the distribution. Let p be the 90th percentile. The z-score of the 90th percentile is approximately z= 1.28. Therefore 1.28 = (p? 6)?0.3. Solving for p yields p= 6.384 L.
(c) Let ? be the required standard deviation. The value of ? must be chosen so that the 90th percentile of the distribution is 6.2. The z-score of the 90th percentile is approximately z= 1.28. Therefore 1.28 = (6.2?6)/ ?.Solving for ?yields ?= 0.156 L.
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