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Unformatted text preview: 8.7.4 Since the function takes on the value of 4, it cannot be a cdf (whose maximum value is 1). Also, the function decreases, which means that it is not a cdf because a cdf is increasing. Thus, this function is a pdf. Since the area under a pdf is 1, we have 4 c = 1 c = 1 / 4 The pdf is p ( x ) = 4 for ≤ x ≤ 1 / 4 . Using the fact that the cdf P of p is an antiderivative of p , i.e. P = p , the cdf is a straight line with slope 1/4 on the interval [0 , 1 / 4] . So the graph of the cdf P is given in the following figure. P(x) 1 1/4 x 8.7.8 This function increases and levels off to 3 c . The area under the curve is not finite, so it is not 1. Thus, the function cannot be a pdf and must be a cdf. Since the limit of a cdf as x → ∞ is 1, we have 3 c = 1 c = 1 / 3 The pdf p is the derivative, or slope, of the cdf. So p ( x ) = for x < (1 / 3 0) / (2 0) = 1 / 6 for ≤ x ≤ 2 (1 1 / 3) / (4 2) = 1 / 3 for 2 < x ≤ 4 for x > 4 See the following figure....
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This homework help was uploaded on 04/11/2008 for the course MATH 10C taught by Professor Hohnhold during the Fall '07 term at UCSD.
 Fall '07
 Hohnhold
 Calculus

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