Suppose XX is a random variable with a distribution that may be known or unknown (it
can be any distribution) and suppose:
- a.
μ
X
=
μ
X
=
the mean of
XX
- b.
σ
X
=
σ
X
=
the standard deviation of
XX
If you draw random samples of size
nn, then as
nn increases, the random variable
ΣXΣX which consists of sums tends to be
normally distributed and
Σ
XΣX ~
N
(
n
⋅
μ
X
,
n
⋅
σ
X
)
N(n⋅
μ
X
,
n
⋅
σ
X
)
The Central Limit Theorem for Sums says that if you keep drawing larger and larger samples
and taking their sums, the sums form their own normal distribution. The distribution has a
mean equal to the original mean multiplied by the sample size and a standard deviation
equal to the original standard deviation multiplied by the square root of the sample size.
The random variable Σ
XΣX has the following z-score associated with it:
- a. ΣxΣx
is one sum.
- b.
z
=
Σ
x
-
n
⋅
μ
X
n
⋅
σ
X
z=
Σ
x
-
n
⋅
μ
X
n
⋅
σ
X
- a.
n
⋅
μ
X
=
n⋅
μ
X
=
the mean of ΣXΣX
- b.
n
⋅
σ
X
=
n
⋅
σ
X
=
standard deviation of ΣXΣX
An unknown distribution has a mean of 90 and a standard deviation of 15. A
sample of size 80 is drawn randomly from the population.
- a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more
than 7500.
- b. Find the sum that is 1.5 standard deviations below the mean of the sums.
Let XX = one value from the original unknown population.
The probability question asks you to find a probability for the sum (or total of) 80 values.
ΣXΣX
= the sum or total of 80 values. Since
μ
X
=
90
μ
X
=90,
σ
X
=
15
σ
X
=15,
and
σ
X
=
80
σ
X
=80,
then
Σ
XΣX ~
N
(
80
⋅
90
,
80
⋅
15
)
N(80⋅90,
80
⋅15)
- a. mean of the sums =
n
⋅
μ
X
=
(
80
)
(
90
)
=
7200
n⋅
μ
X
=(80)(90)=7200
- b. standard deviation of the sums =
n
⋅
σ
X
=
80
⋅
15
n
⋅
σ
X
=
80
⋅15
- c. sum of 80 values =
Σx
=
7500
Σx=7500
Find
P
(
ΣX
>
7500
)
P
(
ΣX
7500
)
Draw a graph.
P
(
ΣX
>
7500
)
=
0.0127
P
(
ΣX
7500
)=0.0127

normalcdf(lower value, upper value, mean of sums, stdev of sums)
The parameter list is abbreviated (lower, upper,
n
⋅
μ
X
,
n
⋅
σ
X
n⋅
μ
X
,
n
⋅
σ
X
)
normalcdf(7500,1E99,
80
⋅
90
,
80
⋅
15
)
=
0.0127
80⋅90,
80
⋅15)=0.0127
Reminder:
1E99
=
10
99
1E99=
10
99
. Press the EE key for E.
- Central Limit Theorem:
Given a random variable (RV) with known mean μμ and known standard deviation σσ. We are sampling with size n and we are interested in two new RVs - the sample mean,
X¯X¯,
and the sample sum, ΣXΣX.
If the size nn of the sample is sufficiently large, then
X¯X¯ size 12{ { bar {X}}} {}∼
N
μ
σ
n
N
μ
σ
n
and
ΣXΣX size 12{X} {} ∼
N
(
nμ
,
n
σ
)N(nμ,nσ).
If the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means,
σ
n
σ
n
, is called the standard error of the mean.
- Normal Distribution:
A continuous random variable (RV) with pdf
f(x)=1σ2πe−(x−μ)2/2σ2f(x)=1σ2πe−(x−μ)2/2σ2 size 12{ ital "pdf"= { {1} over {σ sqrt {2π} } } e rSup { size 8{ - \( x - μ \) rSup { size 6{2} } /2σ rSup { size 6{2} } } } } {}, where μμ is the mean of the distribution and σσ is its standard deviation. Notation: XX ~ N
μ
σ
N
μ
σ
. If μ=0μ=0 and σ=1σ=1, the RV is called the standard normal distribution.
"Collaborative Statistics was written by two faculty members at De Anza College in Cupertino, California. This book is intended for introductory statistics courses being taken by students at two- […]"