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Section 07


x has equal probability over entire interval
Parameters:
 a – beginning of interval
 b – end of interval

𝑓 𝑥

𝐸 𝑋

𝑉𝑎𝑟
1
=
𝑏−𝑎
𝑎+𝑏
=
2
(𝑏−𝑎)2
𝑋 =
12

Parameters:
 μ – mean
 σ2 – variance
 n – sample size
1
𝜎 2𝜋
∗𝑒
−(𝑥−𝜇)2
2𝜎2

𝑓 𝑥 =


𝐸 𝑋 =𝜇
𝑉𝑎𝑟 𝑋 = 𝜎 2

Standard normal distribution is 𝑍~(0,1)

Standardizing
 Given a normal random variable 𝑋~𝑁(𝜇, 𝜎 2 ), find
𝑃(𝑟 < 𝑋 < 𝑠)
 Define 𝑍 =


𝑋−𝜇
𝜎
𝑟−𝜇
𝑥−𝜇
𝑠−𝜇
Then 𝑃 𝑟 < 𝑋 < 𝑠 = 𝑃(
<
<
)
𝜎
𝜎
𝜎
𝑟−𝜇
𝑠−𝜇
𝑠−𝜇
𝑟−𝜇
=𝑃
<𝑍<
=φ
− 𝜑( )
𝜎
𝜎
𝜎
𝜎

Approximating another distribution
 A random variable X with mean μ and variance σ2
is sometimes approximated by assuming the
distribution of X is approximately 𝑁(𝜇, 𝜎 2 )
 Often the question will ask for an approximate
probability of some interval – assume normal
distribution if not specifically mentioned
 This is also often used for sums of random
variables

Integer correction for the normal approximation
 When estimating a discrete distribution using the
normal distribution
 If 𝑛 and 𝑚 are integers, to find discrete probability
𝑃(𝑛 ≤ 𝑋 ≤ 𝑚)
 We assign normal r. v. 𝑌 the same mean and
1
1
variance as X and find 𝑃(𝑛 − ≤ 𝑌 ≤ 𝑚 + )
2
2
 We do this because endpoints matter in discrete
distributions, but not continuous


x is often time between events
Parameters:
 λ
𝑓 𝑥 = λ𝑒 −λ𝑥
 𝐹 𝑋 = 1 − 𝑒 −λ𝑥
1
 𝐸 𝑋 =

λ


𝑉𝑎𝑟 𝑋 =
1
λ2
Exponential and Poisson are often
used in tandem; eg, exponential
represents time until an event
occurs, Poisson represents number
of events occurring in a period of
time. This is because they both use
the variable λ.
Can also be described with 𝜃 =
1
λ
so that 𝑓 𝑥 =
1 −𝑥
𝑒𝜃
𝜃
 Questions can be ambiguous – look for description of
mean

The minimum of a collection of independent
exponential random variables
 Suppose independent r.v. 𝑌1 , 𝑌2 , … 𝑌𝑛 each have
1 1
1
exponential distributions with means , , … ,
λ1 λ2
λ𝑛
 If 𝑌 = min{𝑌1 , 𝑌2 , … 𝑌𝑛 }
 then 𝑌 has an exponential distribution with mean
1
λ1 +λ2 +⋯+λ𝑛
 These types of word problems are pretty common
with other distributions as well

The minimum of a collection of independent
exponential random variables - alternative
▪ Suppose independent r.v. 𝑌1 , 𝑌2 , … 𝑌𝑛 each have
1 1
1
exponential distributions with means , , … , and
λ1 λ2
λ𝑛
𝑌 = min{𝑌1 , 𝑌2 , … 𝑌𝑛 }
 Convert 𝑌1 , 𝑌2 , … , 𝑌𝑛 to Poisson variables with
means λ1 , λ2 , … , λ𝑛
 Then add the means to get the mean of 𝑌
▪ 𝑌 now has a Poisson distribution with mean λ1 + λ2 +
⋯ + λ𝑛
▪ This is the same as an exponential distribution with
1
mean
λ1 +λ2 +⋯+λ𝑛

Parameters:
 α and β

𝑓 𝑥 =
βα ∗𝑥 α−1 ∗𝑒 −β𝑥
Γ(α)
for x>0 only
 Γ 𝑛 = 𝑛 − 1 ! for positive integers n


𝐸 𝑋 =
α
β
𝑉𝑎𝑟 𝑋 =
α
β2
𝑀𝑋 𝑡 =
𝛽 𝛼
( )
𝛽−𝑡
for t<β

DEFINITELY know
 Uniform
 Normal
 Exponential

Try to know
 Gamma
A student received a grade of 80 on a math final
where the mean grade was 72 and the standard
deviation was s. In the statistics final, he
received a 90, where the mean grade was 80
and the standard deviation was 15. If the
standardized scores (i.e., the scores adjusted to
a mean of 0 and standard deviation of 1) were
the same in each case, find s.
The lifetime of a printer costing 200 is
exponentially distributed with mean 2 years.
The manufacturer agrees to pay a full refund to
a buyer if the printer fails during the first year
following its purchase, and a one-half refund if
it fails during the second year.
If the manufacturer sells 100 printers, how
much should it expect to pay in refunds?
An actuary determines that the claim size for a
certain class of accidents is a random variable,
X, with a moment generating function
M(t) = 1/(1-2500t)^4
Calculate the standard deviation of the claim
size for this class of accidents.
The waiting time for the first claim from a good
driver and the waiting time for the first claim from
a bad driver are independent and follow
exponential distributions with means 6 years and 3
years, respectively.
What is the probability that the first claim from a
good driver will be filed within 3 years and the first
claim from a bad driver will be filed within 2 years?
Write as an expression, not a value
In an analysis of healthcare data, ages have been
rounded to the nearest multiple of 5 years. The
difference between the true age and the rounded
age is assumed to be uniformly distributed on the
interval from -2.5 years to 2.5 years. The
healthcare data are based on a random sample of
48 people.
Calculate the approximate probability that the
mean of the rounded ages is within 0.25 years of
the mean of the true ages
The minimum force required to break a
particular type of cable is normally distributed
with mean 12,432 and standard deviation 25. A
random sample of 400 cables of this type is
selected.
Calculate the probability that at least 349 of the
selected cables will not break under a force of
12,400.
A machine has two components and fails when
both components fail. The number of years
from now until the first component fails, X, and
the number of years from now until the
machine fails, Y, are random variables with joint
density function f(x,y) = 1/18 * exp[-(x+y)/6], for
0<x<y, and 0 otherwise.
Calculate Var(Y|X=2).