P X >17( ) = P Z > 17 −161

⎛⎝⎜

⎞⎠⎟ = P Z >1( ) = 1−Φ(1) = 1− 0.8413= 0.1587

4.63I company that manufactures and bottles of apple juice uses a machine that automatically fills 16 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 16 ounces and standard deviation 1 ounce.

Use tables to determine the proportion of bottles that will have more than 17 ounces dispensed into them.

Let X denote the amount of juice dispensed.

P X <1.9( ) = P Z < 1.9 − 2.40.8

⎛⎝⎜

⎞⎠⎟ = P Z < −0.625( ) = 1−Φ(0.625)

= 1− 0.7324 = 0.2660

4.69

Let X denote the GPA of a randomly chosen college student.

We extrapolated to find the value for 0.625.

The GPA’s of the college students have a mean of 2.4 anda standard deviation of 0.8.

4.71Wires manufactured for use in a computer system are specified to have resistances between .12 and .14 ohms. The actual measured resistances of the wires produced by company A have a normal probability distribution with mean .13 and standard deviation .005.

(a). What is the probability that a randomly selected wire from Company A’s production will meet the specifications?

(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?

P 0.12 < X < 0.14( ) = P 0.12 − 0.13.005

≤ Z ≤ 0.14 − 0.3.005

⎛⎝⎜

⎞⎠⎟ = P −2 ≤ Z ≤ 2( )

= 2Φ(2)−1= 2 × 0.9772 −1= 0.9544

0.95444 = 0.8297

P 947 < X < 95( ) = P < 947 − 95010

< Z < 958 − 95010

⎛⎝⎜

⎞⎠⎟ = P −0.3< Z < 0.8( )

= Φ(0.3)+Φ(0.8)−1= 0.6179 + 0.7881−1= 0.4060

4.73The width of bolts of fabric is normally distributed with mean 950 mm and standard deviation 10 mm.

(a). What is the probability that a randomly chosen bolt has a width of between 947 and 958 mm?

(b). What is the appropriate value for C such that a randomly chosen bolt has a width less than C with probability .8531?

0.8531= P X <C( ) = P Z < C − 95010

⎛⎝⎜

⎞⎠⎟ ⇒

C − 95010

= 1.05

So, C = 950 +10.5 = 960.5

4.75A soft drink machine can be regulated so that it discharges an average of ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.3 ounce, give the setting for so that 8 ounce cups will overflow only 1% of the time.

µµ

Let X denote the number of ounces dispensed.

We want to find µ so that, P X > 8( ) = P Z > 8 − µ0.3

⎛⎝⎜

⎞⎠⎟ = 0.01

P Z > t( ) = 0.01 when P Z ≤ t( ) = 0.99 ⇔ t = 2.33

So, we want 8 − µ0.3

= 2.33⇔ µ = 8 − 0.3× 2.33= 7.301

Let v be a positive integer.

α = ν2

, and β = 2.

We say that X is a chi-square random variable.χ 2( )E X( ) = ν , and σ 2 = 2ν .

Values for a chi-square distribution can be readily obtained from tables.

A random variable X is said to have a chi-square distribution withv degrees of freedom if and only if X is gamma-distributed with

Chebyshev’s Inequality

Markov’s TheoremFor any random variable X ≥ 0 and a > 0,

P X > a( ) ≤ E X( )a

For any random variable Z with finite mean µ and variance σ 2

for any k > 0, P Z − µ > kσ( ) ≤ 1k2

Proof: Let X = Z − µ( )2 and a = k2σ 2 and apply Markov's Theorem.

Chebyshev’s Inequality For any random variable Z with finite mean µ and variance σ 2

for any k > 0, P Z − µ > kσ( ) ≤ 1k2

Proof: Let X = Z − µ( )2 and a = k2σ 2 and apply Markov's Theorem.

For any random variable Z with finite mean µ and variance σ 2

for any k > 0, P Z − µ ≤ kσ( ) >1− 1k2 .

Corollary

For any random variable Z with finite mean µ and variance σ 2

for any k > 0, P Z − µ ≤ kσ( ) ≥1− 1k2 .

Corollary

Suppose that X is a random variable with unknown distribution,but we know that its mean is 3 and standard deviation is 1.

What can we say about the value of P 1≤ X ≤ 5( )?

P 1≤ X ≤ 5( ) = P −2 ≤ X − 3≤ 2( ) = P X − 3 ≤ 2( ) >1− 14 = 0.75

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.

ExampleWe roll a die and let X denote the number that appears and flip a fair coin and let Y indicate Heads or Tails.

H T123456

X

Yf (x, y)

112

112

112

112

112

112

112

112

112

112

112

112

ExampleSuppose that we have a box containing 3 red marbles, 3 blue marbles, and 2 green marbles.

Let X denote the number of red marbles, andlet Y denote the number of green marbles.

We choose 4 marbles from the box.

X

Y

f (x, y) 0 1 2

0 0

1

2

3 0

135

370

370 9

35970

970 9

35370

370 2

70

P X ≤Y( ) = 12

P X = Y( ) = 2170

P X = 2( ) = 37

ExampleSuppose that we have a box containing 3 red marbles, 3 blue marbles, and 2 green marbles.

Let X denote the number of red marbles, andlet Y denote the number of green marbles.

We choose 4 marbles from the box.

X

Y

f (x, y) 0 1 2

0 0

1

2

3 0

135

370

370 9

35970

970 9

35370

370 2

70

P X = 2( ) = 37

P X = 0( ) = 114

P X = 1( ) = 37

P X = 3( ) = 114

ExampleSuppose that we have a box containing 3 red marbles, 3 blue marbles, and 2 green marbles.

Let X denote the number of red marbles, andlet Y denote the number of green marbles.

We choose 4 marbles from the box.

X

Y

f (x, y) 0 1 2

0 0

1

2

3 0

135

370

370 9

35970

970 9

35370

370 2

70

f x, y( ) =

3x

⎛⎝⎜

⎞⎠⎟2y

⎛⎝⎜

⎞⎠⎟

34 − x − y

⎛⎝⎜

⎞⎠⎟

70x ≤ 3, y ≤ 2, and x + y = 4

Continuous Multivariate DistributionsIf 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 x0, y0( ) = f (x, y)dxdy−∞

y0∫−∞

x0∫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 = 1.−∞

∞

∫−∞

∞

∫

Continuous Multivariate Distributions

ExampleSuppose that X and Y are continuous random variables with joint pdf given by

f x, y( ) = 38x, for 0 ≤ y ≤ x ≤ 2.

Find the value of P X ≤ 23

,Y ≥ 13

⎛⎝⎜

⎞⎠⎟ .

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)dAA∫∫ .

23

13

Continuous Multivariate Distributions

f x, y( ) = 38x, for 0 ≤ y ≤ x ≤ 2. Find the value of P X ≤ 2

3,Y ≥ 1

3⎛⎝⎜

⎞⎠⎟ .

(2,2)

(0,0)

38xdydx

13

x⌠⌡⎮1

3

23⌠

⌡⎮=

13

23

23

13

Continuous Multivariate Distributions

f x, y( ) = 38x, for 0 ≤ y ≤ x ≤ 2. Find the value of P X ≤ 2

3,Y ≥ 1

3⎛⎝⎜

⎞⎠⎟ .

(2,2)

(0,0)

38xdydx

13

x⌠⌡⎮1

3

23⌠

⌡⎮=

13

23

381

3

23⌠

⌡⎮x2 − 1

8xdx = 1

83x2 − xdx

13

23⌠

⌡⎮

= 0.0116