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1 Math 39 Practice Questions for Final June. 8th 4 Name : 8. Continuous Probability Models You should know Continuous Random Variables Discrete Probability Distributions Expected Value of Discrete Random Variables and Binomial Random Variables Probability Density Functions Practice Questions. A delegation of 3 is selected from a city council made up of 5 women and 6 men. What is the expected number of women in the delegation? the expected number of men? Four possible outcomes: Men and 3 Women, Man and Women, Men and Woman, and 3 Men and Women. Find the probability associated with each outcomes: P ( Men and 3 Women) = (6 ) ( 5 3) P ( Men and Women) = (6 ) ( 5 ) P ( Men and Women) = (6 ) ( 5 ) P (3 Men and Women) = (6 3) ( 5 ) = /33.6 = /.36 = 5/.45 = 4/33. Thus E(# Women) =.35, and E(# Men) =.74.. Decide whether the following functions are probability density functions on the indicated intervals. If not, tell why. We have two conditions to check: f(x) for all x in the support f(x)dx =. support (a) f(x) = x ; [, ] This function is clearly non-negative. For the second condition:
2 x dx = 3 x3 So this is not a pdf. 3 (b) f(x) = 3 (x + ) 4/3 ; [, ) = Again, f is non-negative. The second condition is 3 3 (x + ) 4/3 dx = lim b 3 (x + ) /3 b = + =, so this is a valid pdf. 3. Find a value of k that will make f(x) a probability density function on the indicated intervals. Again, we need f(x)dx =, so we set this and solve for k. R (a) f(x) = kx 3 ; [4, 9] (b) f(x) = k( + x) 3/ ; [, ) 5 = 9 4 = 5 kx 5 kx 3 dx = k( = k(35 5 ) 43 3 = k. ) = k( + x) 3/ dx = k lim b ( + x) / b / = k.
3 4. Find the indicated probabilities with the probability density function f(x) = x e x ; [, ). To find probabilities with a continuous pdf, all we have to do is integrate over the interval. Here, tabular integration is most convenient: D I x e x x e x e x e x Then f(x)dx = x e x xe x e x. (a) P ( X ) (b) P (X ) P ( X ) = x e x = ( x e x xe x e x) = e e e ( e ) = 5 e 8%. x P (X ) = e x dx 3%. 8. Expected Value and Variance of Continuous Random Variables You should know Expected values for BOTH discrete and continuous random variables Variance and standard deviation for both discrete and continuous random variables A shortcut for variance Var(X) = E(X ) [E(X)] = b a x f(x)dx µ Median Practice Questions. Find the expected value, the variance, the standard deviation and median of the random variable whose probability density function is defined as the following.
4 (a) f(x) = (x ); [, ] E(X) = = = x3 x (x )dx x x dx 3 x 4 = 8 3 = 5 3. E(X ) = = = x4 x (x )dx x 3 x dx 4 x3 6 = = 8 3. Then V (X) = E(X ) E(X) = = 9. This is impossible, though!!! Variance cannot be negative. So what is happening here? There is an error in the problem, since the function f(x) < when x < /. (b) f(x) = x ; [, 4] E(X) = 4 x xdx = ( x 3 x3/) 4 = 7 6.
5 Then To find the median m, we set E(X ) = 4 x x 3/ dx = ( x x5/) 4 = V (X) = 43 5 (7 6 ). m = x / dx = (x x/ ) m = m m + = m 4 m + now set n = m and you get a quadratic equation. Solve for n and then you have m. 8.3 Special Probability Density Functions You should know Uniform distribution Exponential distribution Normal distribution Z-scores theorem How to use the standard normal distribution table for any normal distribution. Practice Questions. The time that it takes a driver to react to the brake lights on a decelerating vehicle is critical in helping to avoid rear-end collisions. Suppose the reaction time can be modeled by a normal distribution having mean value.5 sec and standard deviation of sec. Define a random variable T to be the reaction time. Then T N (.5, ). Then all of these problems are just Z transform problems.
6 (a) What is the probability that reaction time is less than. sec?..5 P (T <.) = P (Z < ) (b) What is the probability that reaction time is between. sec and.75 sec?.75.5 )..5 P (. < T <.75) = P ( < Z <.75.5 = P (Z < ) P (Z < (c) What is the probability that reaction time is exactly.5 sec? P (T =.5) =...5 ) (d) If the probability that reaction time is less than t sec is 86.65%, what is the value of t? This problem works in reverse. First find the z such that P (Z < z) = Then t = z +.5 by using the inverse Z transform.
Solution: First we need to find the mean of this distribution. The mean is. ) = e[( e 1 e 1 ) 0] = 2.
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