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Let the probability density of \(X\) be given by $$ f(x)=\left\\{\begin{array}{ll} c\left(4 x-2 x^{2}\right), & 0

Short Answer

Expert verified
(a) The value of c is \(\frac{3}{8}\). (b) \(P\left(\frac{1}{2} < X < \frac{3}{2}\right) = \frac{9}{32}\).

Step by step solution

01

Find the value of c that makes f(x) a valid pdf

To find the value of c, we need to integrate f(x) over its entire domain (0 to 2) and set the integral equal to 1. This means: \[ \int_{0}^{2} c(4x - 2x^{2}) dx = 1 \] Now, solve for c: 1. Find the antiderivative of the above function with respect to x: \[ \int (4x - 2x^{2}) dx = 2x^2 - \frac{2}{3}x^3 + C \] 2. Now evaluate the integral with the limits 0 and 2: \[ c \left[ \left( 2(2)^2 - \frac{2}{3}(2)^3 \right) - \left( 2(0)^2 - \frac{2}{3}(0)^3 \right) \right] = 1 \] 3. Simplify and solve for c: \[ c(8 - \frac{16}{3}) = 1 \] \[ c = \frac{1}{\frac{8}{3}} = \frac{3}{8} \] The value of c that makes f(x) a valid pdf is 3/8.
02

Find the probability P(1/2 < X < 3/2)

Now that we have found the value of c, we can find the probability of X lying within the given interval by integrating the pdf over the interval (1/2 to 3/2): \[ P\left(\frac{1}{2} < X < \frac{3}{2}\right) = \int_{\frac{1}{2}}^{\frac{3}{2}} f(x) dx \] Plug in the value of c into f(x): \[ f(x) = \frac{3}{8} (4x - 2x^2) \] Now, compute the integral: \[ P\left(\frac{1}{2} < X < \frac{3}{2}\right) = \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{3}{8} (4x - 2x^2) dx \] 1. Find the antiderivative of the above function with respect to x: \[ \int \frac{3}{8} (4x - 2x^2) dx = \frac{3}{8} \left( 2x^2 - \frac{2}{3}x^3 \right) + C \] 2. Now evaluate the integral with the limits 1/2 and 3/2: \[ P\left(\frac{1}{2} < X < \frac{3}{2}\right) = \frac{3}{8} \left[ \left( 2\left(\frac{3}{2}\right)^2 - \frac{2}{3}\left(\frac{3}{2}\right)^3 \right) - \left( 2\left(\frac{1}{2}\right)^2 - \frac{2}{3}\left(\frac{1}{2}\right)^3 \right) \right] \] 3. Simplify the expression: \[ P\left(\frac{1}{2} < X < \frac{3}{2}\right) = \frac{3}{8} \left[ \left( \frac{9}{2} - \frac{27}{4} \right) - \left( \frac{1}{2} - \frac{1}{4} \right) \right] = \frac{3}{8} \left( -\frac{3}{4} \right) = \frac{9}{32} \] So, the probability P(1/2 < X < 3/2) is 9/32.

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Most popular questions from this chapter

If \(X\) is a nonnegative random variable, and \(g\) is a differential function with \(g(0)=0\), then $$ E[g(X)]=\int_{0}^{\infty} P(X>t) g^{\prime}(t) d t $$ Prove the preceding when \(X\) is a continuous random variable.

Suppose that two teams are playing a series of games, each of which is independently won by team \(A\) with probability \(p\) and by team \(B\) with probability \(1-p .\) The winner of the series is the first team to win four games. Find the expected number of games that are played, and evaluate this quantity when \(p=1 / 2\).

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A television store owner figures that 50 percent of the customers entering his store will purchase an ordinary television set, 20 percent will purchase a color television set, and 30 percent will just be browsing. If five customers enter his store on a certain day, what is the probability that two customers purchase color sets, one customer purchases an ordinary set, and two customers purchase nothing?

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