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4.63
I 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.
17 − 16 ⎞
⎛
P ( X > 17 ) = P ⎜ Z >
= P ( Z > 1) = 1− Φ(1) = 1− 0.8413 = 0.1587
⎟
⎝
1 ⎠
4.69
The GPA’s of the college students have a mean of 2.4 and
a standard deviation of 0.8.
Let X denote the GPA of a randomly chosen college student.
1.9 − 2.4 ⎞
⎛
P ( X < 1.9 ) = P ⎜ Z <
= P ( Z < −0.625 ) = 1− Φ(0.625)
⎟
⎝
0.8 ⎠
= 1− 0.7324 = 0.2660
We extrapolated to find the value for 0.625.
4.71
Wires 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?
0.14 − 0.3 ⎞
⎛ 0.12 − 0.13
P ( 0.12 < X < 0.14 ) = P ⎜
≤Z≤
= P ( −2 ≤ Z ≤ 2 )
⎟
⎝
.005
.005 ⎠
= 2Φ(2) − 1 = 2 × 0.9772 − 1 = 0.9544
(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?
0.9544 4 = 0.8297
4.73
The 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?
958 − 950 ⎞
⎛ 947 − 950
P ( 947 < X < 95 ) = P ⎜ <
<Z<
= P ( −0.3 < Z < 0.8 )
⎟
⎝
⎠
10
10
= Φ(0.3) + Φ(0.8) − 1 = 0.6179 + 0.7881− 1 = 0.4060
(b). What is the appropriate value for C such that a randomly chosen
bolt has a width less than C with probability .8531?
C − 950 ⎞
C − 950
⎛
0.8531 = P ( X < C ) = P ⎜ Z <
⇒
= 1.05
⎟
⎝
10 ⎠
10
So, C = 950 + 10.5 = 960.5
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 0.3 ounce, give the setting for µ so that 8 ounce cups
will overflow only 1% of the time.
Let X denote the number of ounces dispensed.
8− µ⎞
⎛
We want to find µ so that, P ( X > 8 ) = P ⎜ Z >
= 0.01
⎟
⎝
0.3 ⎠
P ( Z > t ) = 0.01 when P ( Z ≤ t ) = 0.99 ⇔ t = 2.33
8−µ
So, we want
= 2.33 ⇔ µ = 8 − 0.3 × 2.33 = 7.301
0.3
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
ν
α = , and β = 2.
2
2
χ
We say that X is a chi-square ( ) random variable.
E ( X ) = ν , and σ 2 = 2ν .
Values for a chi-square distribution can be readily obtained from tables.
Markov’s Theorem
For any random variable X ≥ 0 and a > 0,
E(X)
P ( X > a) ≤
a
Chebyshev’s Inequality
For any random variable Z with finite mean µ and variance σ 2
1
for any k > 0, P ( Z − µ > kσ ) ≤ 2
k
Proof: Let X = ( Z − µ ) and a = k 2σ 2 and apply Markov's Theorem.
2
Chebyshev’s Inequality
For any random variable Z with finite mean µ and variance σ 2
1
for any k > 0, P ( Z − µ > kσ ) ≤ 2
k
Proof: Let X = ( Z − µ ) and a = k 2σ 2 and apply Markov's Theorem.
2
Corollary
For any random variable Z with finite mean µ and variance σ 2
1
for any k > 0, P ( Z − µ ≤ kσ ) > 1− 2 .
k
Corollary
For any random variable Z with finite mean µ and variance σ 2
1
for any k > 0, P ( Z − µ ≤ kσ ) ≥ 1− 2 .
k
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 ) ?
1
P (1 ≤ X ≤ 5 ) = P ( −2 ≤ X − 3 ≤ 2 ) = P ( X − 3 ≤ 2 ) > 1− = 0.75
4
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.
Y
X
f (x, y)
H
T
1
1
12
1
12
1
12
1
12
1
12
1
12
1
12
1
12
1
12
1
12
1
12
1
12
2
3
4
5
6
Example
Suppose that we have a box containing
3 red marbles, 3 blue marbles, and 2 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)
0
0
0
1
3
2
9
3
3
70
70
70
1
1
9
9
2
35
35
35
70
2
3
9
3
70
70
70
0
1
2
21
P( X = Y ) =
70
P( X ≤ Y ) =
3
P ( X = 2) =
7
Example
Suppose that we have a box containing
3 red marbles, 3 blue marbles, and 2 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)
0
0
0
1
3
2
9
3
3
70
70
70
1
1
9
9
2
35
35
35
70
2
3
9
3
70
70
70
0
1
14
3
P ( X = 1) =
7
3
P ( X = 2) =
7
1
P ( X = 3) =
14
P ( X = 0) =
Example
Suppose that we have a box containing
3 red marbles, 3 blue marbles, and 2 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)
0
0
0
1
3
2
9
3
3
70
70
70
1
1
9
9
2
35
35
35
70
2
3
9
3
70
70
70
0
3 ⎞
⎛ 3⎞ ⎛ 2 ⎞ ⎛
⎜⎝ x ⎟⎠ ⎜⎝ y ⎟⎠ ⎜⎝ 4 − x − y ⎟⎠
f ( x, y ) =
70
x ≤ 3, y ≤ 2, and x + y = 4
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 ( x0 , y0 ) = ∫−∞ ∫−∞ f (x, y)dx dy
x0
y0
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
Example
Suppose that X and Y are continuous random
variables with joint pdf given by
3
f ( x, y ) = x, for 0 ≤ y ≤ x ≤ 2.
8
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
2
1⎞
⎛
Find the value of P ⎜ X ≤ , Y ≥ ⎟ .
⎝
3
3⎠
Continuous Multivariate Distributions
3
2
1⎞
⎛
f ( x, y ) = x, for 0 ≤ y ≤ x ≤ 2. Find the value of P ⎜ X ≤ , Y ≥ ⎟ .
⎝
3
3⎠
8
(2,2)
2
⌠ ⌠x 3
x dy dx =
⎮ ⎮
⌡1 3 ⌡1 3 8
2
3
1
3
(0,0)
1
3
2
3
3
Continuous Multivariate Distributions
3
2
1⎞
⎛
f ( x, y ) = x, for 0 ≤ y ≤ x ≤ 2. Find the value of P ⎜ X ≤ , Y ≥ ⎟ .
⎝
3
3⎠
8
(2,2)
2
⌠ ⌠x 3
x dy dx =
⎮ ⎮
⌡1 3 ⌡1 3 8
2
3
1
3
(0,0)
2
1
3
2
3
3
3 2 1
⌠
⎮ x − x dx =
⌡1 3 8
8
3
= 0.0116
2
1⌠ 3 2
⎮ 3x − x dx
8 ⌡1 3