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Unformatted text preview: Math 218 Final Examination: Fall 2005 Problem 1 (20 pts) . Matilda throws a big party to celebrate the A she got in Math 218. Among many other things she prepared a very special fruit punch with fresh fruits from her garden. The punch is dispensed by a dispensing machine that she received as a gift when she successfully passed Math 118. The amount of fruit punch dispensed varies from cup to cup, and it may be looked upon as a random variable having a normal distribution with a mean of 7 . 7 oz and a standard deviation of . 25 oz. (a) Find the probability that the machine will dispense between 7 . 33 oz and 7 . 51 oz of fruit punch. (b) Find the probability that the machine will dispense more than 8 oz. (c) If 20 cups are filled with fruit punch, what is the expected number of cups that are filled with more than 8 oz? (d) The machine has a knob which allows the mean to be reset, without changing the standard deviation. Matilda wants to reset the machine to make sure that 98 . 5% of the time it will not dispense more than 8 oz. What should the new mean be set to? Solution. (a) The answer is . 1542 . This is computed by standarding X by subtracting its mean and dividing by its standard deviation: P (7 . 33 ≤ X ≤ 7 . 51) = P (7 . 33 7 . 7 ≤ X 7 . 7 ≤ 7 . 51 7 . 7) = P . 37 . 25 ≤ X 7 . 7 . 25 ≤ . 19 . 25 = P ( 1 . 48 ≤ Z ≤  . 76) = P (0 ≤ Z ≤ 1 . 48) P (0 ≤ Z ≤ . 76) = 0 . 4306 . 2764 = 0 . 1542 The plot is shown in Figure 1. (b) The answer is . 1151 . We have P ( X ≥ 8) = P ( X 7 . 7 ≥ . 3) = P X 7 . 7 . 25 ≥ . 3 . 25 = P ( Z ≥ 1 . 2) = 0 . 5 . 3849 = 0 . 1151 . This is illustrated in Figure 2.321 1 2 3 0.1 0.2 0.3 0.4 Figure 1. The normal curve between 1 . 48 and . 76 has area 0 . 1542.321 1 2 3 0.1 0.2 0.3 0.4 Figure 2. The normal curve between 1 . 2 and ∞ has area 0 . 1151. (c) The answer is 2 . 302 . This is a binomial probability prob lem: there are 20 independent experiments, pouring a cup, with a probability of overfilling (more than 8 oz) of 0 . 1151; the expected number of cups which will be overfilled is therefore 20 × . 1151 = 2 . 302. (d) The desired mean is 7 . 4575 . To see this, let μ be the desired mean. We want P ( X ≥ 8) = 0 . 015 . Therefore . 015 = P ( X μ ≥ 8 μ ) = P X μ . 25 ≥ 8 μ . 25 = P ( Z ≥ 32 4 μ ) . But z . 015 = 2 . 17 (you have to get this by doing an inverse lookup in the Z table, since there is no 0 . 015 cutoff in any of 1 2 the tables), and therefore we want 32 4 μ = 2 . 17 , an equation which is easily solved to get 7 . 4575. Problem 2 (20 pts) . The graph of a probability density function f ( x ) is shown below. The function has the properties that it is zero outside the interval [0 , 1]; that f (1 / 2) = 0 and f (0) = f (1) = c , a constant to be determined; that it is linear on the interval [0 , 1 / 2]; that it is linear on [1 / 2 , 1] with equation y = 4 x 2; and that it is continuous on [0 , 1]....
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This homework help was uploaded on 02/05/2008 for the course MATH 218 taught by Professor Haskell during the Fall '06 term at USC.
 Fall '06
 Haskell
 Math, Probability

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