( ) = P Z > = P( Z > 1) = 1 Φ(1) = = P X > 17

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1 4.6 I company that manufactures and bottles of apple juice uses a machine that automatically fills 6 ounce bottles. There is some variation, however, in the amounts of liquid dispensed into the bottles that are filled. The amount dispensed has been observed to be approximately normally distributed with mean 6 ounces and standard deviation ounce. Use tables to determine the proportion of bottles that will have more than 7 ounces dispensed into them. Let X denote the amount of juice dispensed. ( ) = P Z > P X > = P( Z > ) = Φ() = 0.84 = 0.587

2 4.69 The GPA s of the college students have a mean of.4 and a standard deviation of 0.8. Let X denote the GPA of a randomly chosen college student. ( ) = P Z < P X <.9 = 0.74 = = P( Z < 0.65) = Φ(0.65) We extrapolated to find the value for 0.65.

3 4.7 Wires manufactured for use in a computer system are specified to have resistances between. and.4 ohms. The actual measured resistances of the wires produced by company A have a normal probability distribution with mean. and standard deviation.005. (a). What is the probability that a randomly selected wire from Company A s production will meet the specifications? P( 0. < X < 0.4) = P = Φ() = = Z = P Z ( ) (b). If four of these wires are used in each computer system and all are selected from Company A, what is the probability that all four in a randomly selected system will meet the specifications? = 0.897

4 4.7 The width of bolts of fabric is normally distributed with mean 950 mm and standard deviation 0 mm. (a). What is the probability that a randomly chosen bolt has a width of between 947 and 958 mm? ( ) = P < P 947 < X < < Z < = Φ(0.) + Φ(0.8) = = = P( 0. < Z < 0.8) (b). What is the appropriate value for C such that a randomly chosen bolt has a width less than C with probability.85? 0.85 = P X < C ( ) = P Z < C 950 So, C = = C =.05

5 4.75 A soft drink machine can be regulated so that it discharges an average of µ ounces per cup. If the ounces of fill are normally distributed with standard deviation 0. ounce, give the setting for µ so that 8 ounce cups will overflow only % of the time. Let X denote the number of ounces dispensed. We want to find µ so that, P X > 8 ( ) = P Z > 8 µ P( Z > t) = 0.0 when P( Z t) = 0.99 t =. So, we want 8 µ = 0.0 =. µ = = 7.0

6 Let v be a positive integer. A random variable X is said to have a chi-square distribution with v degrees of freedom if and only if X is gamma-distributed with α = ν, and β =. We say that X is a chi-square ( χ ) random variable. E( X) = ν, and σ = ν. Values for a chi-square distribution can be readily obtained from tables.

7 Markov s Theorem For any random variable X 0 and a > 0, ( ) E ( X ) P X > a Chebyshev s Inequality a For any random variable Z with finite mean µ and variance σ for any k > 0, P( Z µ > kσ ) k Proof: Let X = ( Z µ ) and a = k σ and apply Markov's Theorem.

8 Chebyshev s Inequality For any random variable Z with finite mean µ and variance σ for any k > 0, P( Z µ > kσ ) k Proof: Let X = ( Z µ ) and a = k σ and apply Markov's Theorem. Corollary For any random variable Z with finite mean µ and variance σ for any k > 0, P( Z µ kσ ) > k.

9 Corollary For any random variable Z with finite mean µ and variance σ for any k > 0, P( Z µ kσ ) k. Suppose that X is a random variable with unknown distribution, but we know that its mean is and standard deviation is. What can we say about the value of P( X 5)? P( X 5) = P( X ) = P( X ) > 4 = 0.75

10 Multivariate Distributions If X and Y are discrete random variables, then the joint probability function for X and Y is f (a,b) = P( X = a, Y = b) for a,b in the support of X,Y respectively. Example We roll a die and let X denote the number that appears and flip a fair coin and let Y indicate Heads or Tails. X f (x, y) Y H T

11 Example Suppose that we have a box containing red marbles, blue marbles, and green marbles. We choose 4 marbles from the box. Let X denote the number of red marbles, and let Y denote the number of green marbles. Y X f (x, y) P( X Y ) = P( X = Y ) = 70 P( X = ) = 7

12 Example Suppose that we have a box containing red marbles, blue marbles, and green marbles. We choose 4 marbles from the box. Let X denote the number of red marbles, and let Y denote the number of green marbles. Y f (x, y) 0 X P( X = 0) = 4 P( X = ) = P( X = ) = P X = ( ) = 4

13 Example Suppose that we have a box containing red marbles, blue marbles, and green marbles. We choose 4 marbles from the box. Let X denote the number of red marbles, and let Y denote the number of green marbles. X Y f (x, y) f ( x, y) = x y 4 x y 70 x, y, and x + y =

14 Continuous Multivariate Distributions If X and Y are any two random variables, then their joint distribution function F(x, y) is defined by: F( x, y) = P( X x, Y y), for < x <, < y < Suppose that X and Y are continuous having joint distribution function F(x, y). If there exists a function f (x, y) 0 such that F x 0, y 0 y 0 ( ) = f (x, y)dx dy then X and Y are said to be jointly continuous random variables. In this case, f (x, y) 0 is said to be their joint probability density function. Naturally,we must have f (x, y)dxd =. x 0

15 Continuous Multivariate Distributions Example Suppose that X and Y are continuous random variables with joint pdf given by f ( x, y) = 8 x, for 0 y x. It is a consequence of the definition of the distribution function that for any subset A is the plane, the probability that ( X,Y ) A is f (x, y)da. A Find the value of P X, Y.

16 Continuous Multivariate Distributions ( ) = 8 x, for 0 y x. Find the value of P X, Y f x, y (,). x 8 x dydx = (0,0)

17 Continuous Multivariate Distributions ( ) = 8 x, for 0 y x. Find the value of P X, Y f x, y (,). (0,0) x 8 x dydx = x 8 8 x dx = 8 x x dx = 0.06

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