Week 1: Descriptive Statistics

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Transcript Week 1: Descriptive Statistics

Quick review of
some key ideas
CEE 11 Spring 2002
Dr. Amelia Regan
These notes draw liberally from the class text, Probability and Statistics for
Engineering and the Sciences by Jay L. Devore, Duxbury 1995 (4th edition)
mean

The mean of a sample or data set is simply
the arithmetic average of the values in the
set, obtained by summing the values and
dividing by the number of values.
x1  x 2  ... x n
1 n
x 
  xi
n
n i 1
The mean of the sample of weights is
144.63 pounds
mean of a frequency distribution

When we summarize a data set in a frequency
distribution, we are approximating the data set by
"rounding" each value in a given class to the class
mark.
n
1 n
x   fi xi   pi xi
n i 1
i 1
where fi  the frequency of the ith observation and
pi = the proportion associated with the ith observation
The mean of the weight data obtained in this way is
146.67
shortcut formula for the
variance

Its sometimes more convenient to use
the following formula for the variance


x
 i 
n
2
 i 1 
x


i
n
 i 1
n 1
n
n
s2 
 x  x 
i 1
i
n 1
2
2
properties of S2



Let x1, x2, x,...,xn be a sample and c be any nonzero
constant.
If y1 = x1 + c, y2 = x2 + c,...,yn = xn + c, then S2y = S2x
If y1 = cx1, y2 = cx2,...,yn = cxn, then S2y = c2S2x, Sy
= |c|S2x
In other words -- if we add a constant to a
sample we do not increase the variance -- if we
multiply by a constant we increase the variance
by the square of the constant
related properties of the
sample mean

Let x1, x2, x,...,xn be a sample and c be any nonzero
constant.

If y1 = x1 + c, y2 = x2 + c,...,yn = xn + c then y  x  c

If y1 = cx1, y2 = cx2,...,yn = cxn, then
y  cx
In other words if we add or multiply the sample
by a constant we add or multiply the mean by
the same constant
Class exercise (new)
Without using your calculators, calculate the mean
and variance of the following data

Xi | 33
27
18
36
21
---------------------------------------------fi | 4
3
2
5
1

Hint, shift the observations “to the left” by
subtracting a constant and then divide by another
constant
Class exercise (new)
Without using your calculators, calculate the mean
and variance of the following data

Step 1 yi | 11
9
6
12
---------------------------------------------Step 2 yi | 2
0
-3
3
yi | 2
0
-3
3
-2
---------------------------------------------fi | 4
3
2
5
1

Divide by 3, then subtract 9
7
-2
xi
yi   9
3
yi |
2
0
-3
3
-2
-------------------------------------------------------(yi)2 |
4
0
9
9
4
-------------------------------------------------------fi |
4
3
2
5
1
 f y  8  0  6  15  2  15
f y 15

y
  1.0
i
i
i

n
i
15
fi y 2i  16  0  18  45  4  83
xi
yi   9
3
2


  fi yi 
2
2
i 1
15


fi yi 
83 
n
15  4.847

n 1
14
n
n

i 1
xi  3( yi  9)
x  3( y  9)  30.0
s  9 s  43.714
2
x
2
y
conditional
probability

For any two events A and B with P(B) > 0, the
conditional probability of A given that B has occurred is
defined by
P( A  B)
P( A | B) 
P( B)

The multiplication rule for P(A intersection B) follows
directly
P( A  B)  P( A | B) P( B)
independence
Remember that in general

P( A  B)  P( A | B) P( B)

However, if A and B are independent then
P( A  B)  P( A) P( B)

In fact, A and B are independent if and
only if the above is true
Counting Techniques

When the various outcomes of an experiment are
equally likely then the task of computing probabilities
reduces to counting. In particular, if N is the number
of outcomes in the sample space and N(A) is the
number of outcomes contained in an event A, then
N ( A)
P ( A) 
N
Permutations


Any ordered sequence of k objects taken from a set of n
distinct objects is called a permutation of size k of the
objects. The number of permutations of size k that can be
constructed from the n objects is denoted by Pk,n
The number of permutations of size k that can be
constructed from n objects is equal to n(n-1)(n-2)…(nk+1)
Pk , n  n(n  1),...(n  k  1)
n(n  1),..., (n  k  1)(n  k )(n  k  1)
(n  k )(n  k  1),...(2)(1)
n!

(n  k )!

combinations

Given a set of n distinct objects,
any unordered subset of size k of
the objects is called a combination.
The number of combinations of
size k that can be formed from n
distinct objects is denoted by  nk  or
sometimes by Ck,n
 n  Pk ,n
n!


 
k!
k !(n  k )!
k