Transcript Annual Incomes of 10 Families

Why do we need the standard deviation?
1- The standard deviation reflects dispersion of data values,
so that the dispersion of different distributions may be
compared by using standard deviations.
2- The standard deviation permits the precise interpretation of
data values within a distributions.
3- The standard deviation, like the mean, is a member od a
mathematical system which permits its use in more advanced
statistical considerations.
EMPIRICAL RULES
1- About 68% of the values will lie within 1 standard deviation of the mean, that
is, between x̄ - s and x̄ + s;
2- About 95% of the values will lie within 2 standard deviation of the mean, that
is, between x̄ - 2s and x̄ + 2s;
3- About 99.7% of the values will lie within 3 standard deviation of the mean,
that is, between x̄ - 3s and x̄ + 3s;
BOX PLOT
A Box Plot is a graphical summary of data that is based on five number summary.
To construct a box plot use the following STEPS
1- A box is drawn with the ends of the box located at the first (Q1)
and third quartile ( Q3)
2- A vertical line is drawn in the box at the location of the median.
3- By using IQR=Q3-Q1, limits (also called fences) are located.
The limits for the box plot are Q1 - 1.5*IQR and Q3 + 1.5*IQR
DATA OUTSIDE THESE LIMITS ARE CONSIDERED OUTLIERS
4- The dashed lines (also called whiskers) are drawn from the ends of the box to the
smallest (L) and highest (H) values inside the limits computed in STEP 3
5- Finally, the location of each outlier is shown with a symbol
*
Experiment
Experimental Outcome (Possibilities)
Toss a Coin
Head, Tail
Select a part for inspection
Defective, Non-Defective
Conduct a Sales
Purchase, No Purchase
Roll a die
1, 2, 3, 4, 5, 6
Play a football game
Win, Lose, Tie
FACTORIAL NOTATION
0! = 1(By definition)
1! = 1
2! = 2*1 = 2
3! = 3*2*1 = 6
4! = 4*3*2*1 = 24
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n! = n*(n-1)* (n-2)*……….3*2*1
Suppose that we have a sample space:
S = {E1, E2, E3, E4, E5, E6, E7}
P(E1) = 0.05
P(E2) = 0.20
P(E3) = 0.20
P(E4) = 0.25
P(E5) = 0.15
P(E6) = 0.10
P(E7) = 0.05
where Ei = Sample points
A = {E1, E4, E6}
B = {E2,E4, E7}
C = {E2, E3, E5, E7}
• Example:
Consider a recent study conducted by the personnel manager of a
major computer software company.
The study showed that 30% of employees who left the firm within
two years did so primarily because they were dissatisfied with
their salary, 20% left because they were dissatisfied with their
work assignments, 12% of the former employees indicated
dissatisfaction with both their salary and their work assignments.
• Question:
What is the probability that an employee who leaves within two
years does so because of dissatisfaction with salary, dissatisfaction
with work assignment or both?
Assigning Probabilities
 Basic Requirements for Assigning Probabilities
2. The sum of the probabilities for all experimental
outcomes must equal 1.
P(E1) + P(E2) + . . . + P(En) = 1
where:
n is the number of experimental outcomes
Multiplication Law
The multiplication law provides a way to compute the
probability of the intersection of two events.
The law is written as:
P(A B) = P(B)P(A|B)
Mutual Exclusiveness and Independence
Do not confuse the notion of mutually exclusive
events with that of independent events.
Two events with nonzero probabilities cannot be
both mutually exclusive and independent.
If one mutually exclusive event is known to occur,
the other cannot occur.; thus, the probability of the
other event occurring is reduced to zero (and they
are therefore dependent).
Two events that are not mutually exclusive, might
or might not be independent.
The Sales of Automobiles for 300 days
0
automobile sold 54 days
1
automobile sold 117 days
2 automobile sold 72 days
3 automobile sold 42 days
4 automobile sold 12 days
5 automobile sold 3 days
Total: 300 days