Suppose XX is a random variable with a distribution that may be known or unknown (it
can be any distribution). Using a subscript that matches the random variable, suppose:
- a. μXμX = the mean of XX
- b. σXσX = the standard deviation of XX
If you draw random samples of size
nn, then as
nn increases, the random variable
X¯
X
which consists of sample means, tends to be
normally distributed and
X¯
X
~
N
(
μ
X
,
σ
X
n
)
N(
μ
X
,
σ
X
n
)
The Central Limit Theorem for Sample Means (Averages) says that if you keep drawing
larger and larger samples (like rolling 1, 2, 5, and, finally, 10 dice) and calculating their means
the sample means (averages) form their own normal distribution. This distribution has the same
mean as the original distribution and a variance that equals the original variance divided by nn, the
sample size. nn is the number of values that are averaged together not the number of times the
experiment is done.
The random variable
X¯
X
has a different z-score associated with it.
x¯
x
is one sample
z
=
x¯
-
μ
X
(
σ
X
n
)
z=
x
-
μ
X
(
σ
X
n
)
(1)
μXμX is both the average
of XX and of
X¯
X
.
σ
X¯
=
σ
X
n
=
σ
X
=
σ
X
n
=
standard deviation of
X¯
X
and is called the standard error of the mean.
An unknown distribution has a mean of 90 and a standard deviation of 15.
Samples of size nn = 25 are drawn randomly from the population.
Find the probability that the sample mean is between 85 and 92.
Let XX = one value from the original unknown population.
The probability question asks you to find a probability for the sample mean (or average).
Let
X¯
=
X
=
the mean or average of a sample of size 25.
Since
μ
X
=
90
μ
X
=90,
σ
X
=
15
σ
X
=15,
and
n
=
25
n
=25;
then
X¯
X
~
N
(
90
,
15
25
)
N(90,
15
25
)
Find
P
(
85
<
X¯
<
92
)
P
(
85
X
92
)
Draw a graph.
P
(
85
<
X¯
<
92
)
=
0.6997
P
(
85
X
92
)
=0.6997
The probability that the sample mean is between 85 and 92 is 0.6997.

TI-83: normalcdf(lower value, upper value, mean for averages, stdev for averages)
stdev = standard deviation
The parameter list is abbreviated (lower, upper,
μμ,
σ
n
σ
n
)
normalcdf(85,92,90,
15
25
) = 0.6997
(85,92,90,
15
25
) = 0.6997
Find the average value that is 2 standard deviations above the the mean of the averages.
To find the average value that is 2 standard deviations above the mean of the
averages, use the formula
value =
μ
X
+
(#ofSTDEVs)
(
σ
X
n
)
μ
X
+(#ofSTDEVs)(
σ
X
n
)
value =
90
+
2
⋅
15
25
=
96
90+2⋅
15
25
=96
So, the average value that is 2 standard deviations above the mean of the averages is
96.
The length of time, in hours, it takes an "over 40" group of people to play
one soccer match is normally distributed with a mean of 2 hours and a standard
deviation of 0.5 hours. A sample of size nn = 50 is drawn randomly from the population.
Find the probability that the sample mean is between 1.8 hours and 2.3 hours.
Let XX = the time, in hours, it takes to play one soccer match.
The probability question asks you to find a probability for the sample mean or average
time, in hours, it takes to play one soccer match.
Let
X¯
X
= the average time, in hours, it takes to play one soccer match.
If
μ
X
=
μ
X
=
_________,
σ
X
=
σ
X
=
__________,
and n=n= ___________,
then
X¯
~
N
(______, ______)
X
~N(______, ______)
by the Central Limit Theorem for Averages of Sample Means.
μ
X
=
μ
X
=
2,
σ
X
=
σ
X
=
0.5,
n=n= 50,
and
X
~
N
(
2
,
0.5
50
)
X~N(2 ,
0.5
50
)
Find
P
(
1.8
<
X¯
<
2.3
)
P
(
1.8
X
2.3
)
.
Draw a graph.
P
(
1.8
<
X¯
<
2.3
)
=
0.9977
P
(
1.8
X
2.3
)
=0.9977
normalcdf(1.8,2.3,2,
.5
50
) = 0.9977
(1.8,2.3,2,
.5
50
) = 0.9977
The probability that the sample mean is between 1.8 hours and 2.3 hours is ______.
- Average:
A number that describes the central tendency of the data. There are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean.
- Central Limit Theorem:
Given a random variable (RV) with known mean μμ and known variance σσ
22 size 12{ {} rSup { size 8{2} } } {}, we are sampling with size n and we are interested in two new RV - sample mean,
XˉXˉ size 12{ { bar {X}}} {},and sample sum,ΣΣ
XX size 12{X} {}. If the size n of the sample is sufficiently large, then
XˉXˉ size 12{ { bar {X}}} {}∼
N
nμ
σ
2
n
N
nμ
σ
2
n
and
ΣXΣX size 12{X} {} ∼
N
nμ
n
σ
2
N
nμ
n
σ
2
. In words, 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. And even more, the mean of the sampling distribution will equal the population mean and mean of sampling sums will equal n times the population mean. The standard deviation of the distribution of the sample means,
σ
n
σ
n
, is called standard error of the mean.
- Normal Distribution:
A continuous random variable (RV) with
pdf=1σ2πe−(x−μ)2/2σ2pdf=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
2
N
μ
σ
2
. If μ=0μ=0 and σ=1σ=1, the RV is called standard normal distribution, or z-score.
- Standard Error of the Mean:
The standard deviation of the distribution of the sample means,
σ
n
σ
n
.
"This is the course textbook for Biology 502 at CSU Dominguez Hills"