Why do we need the standard deviation? 1- The standard deviation reflects dispersion of data values, so that the dispersion of different distributions.

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 x̄1 - 2s and x̄1 + 2s;

3- About 99.7% of the values will lie within 3 standard deviation of the mean, that is, between x̄1 - 3s and x̄1 + 3s;

BOX PLOT

A Box̄ Plot is a graphical summary of data that is based on five number summary.

To construct a box̄ plot use the following STEPS

1- A box̄ is drawn with the ends of the box̄ located at the first (Q1)

and third quartile ( Q3)

2- A vertical line is drawn in the box̄ at the location of the median.

3- By using IQR=Q3-Q1, limits (also called fences) are located.

The limits for the box̄ 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 box̄ 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! = 12! = 2*1 = 23! = 3*2*1 = 64! = 4*3*2*1 = 24.....n! = n*(n-1)* (n-2)*……….3*2*1

Suppose that we have a sample space:

S = {E1, E2, E3, E4, E5, E6, E7} where Ei = Sample points P(E1) = 0.05 P(E2) = 0.20 A = {E1, E4, E6} P(E3) = 0.20 B = {E2,E4, E7} P(E4) = 0.25 C = {E2, E3, E5, E7} P(E5) = 0.15 P(E6) = 0.10 P(E7) = 0.05

• 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 ProbabilitiesAssigning Probabilities

Basic Requirements for Assigning ProbabilitiesBasic Requirements for Assigning Probabilities

2. The sum of the probabilities for all experimental2. The sum of the probabilities for all experimental outcomes must equal 1.outcomes must equal 1. 2. The sum of the probabilities for all experimental2. The sum of the probabilities for all experimental outcomes must equal 1.outcomes must equal 1.

PP((EE11) + ) + PP((EE22) + . . . + ) + . . . + PP((EEnn) = 1) = 1

where:where:nn is the number of experimental outcomes is the number of experimental outcomes

Multiplication LawMultiplication Law

The The multiplication lawmultiplication law provides a way to compute the provides a way to compute the probability of the intersection of two events.probability of the intersection of two events. The The multiplication lawmultiplication law provides a way to compute the provides a way to compute the probability of the intersection of two events.probability of the intersection of two events.

The law is written as:The law is written as: The law is written as:The law is written as:

PP((AA BB) = ) = PP((BB))PP((AA||BB))PP((AA BB) = ) = PP((BB))PP((AA||BB))

Do not confuse the notion of mutually exclusiveDo not confuse the notion of mutually exclusive events with that of independent events.events with that of independent events. Do not confuse the notion of mutually exclusiveDo not confuse the notion of mutually exclusive events with that of independent events.events with that of independent events.

Two events with nonzero probabilities cannot beTwo events with nonzero probabilities cannot be both mutually exclusive and independent.both mutually exclusive and independent. Two events with nonzero probabilities cannot beTwo events with nonzero probabilities cannot be both mutually exclusive and independent.both mutually exclusive and independent.

If one mutually exclusive event is known to occur,If one mutually exclusive event is known to occur, the other cannot occur.; thus, the probability of thethe other cannot occur.; thus, the probability of the other event occurring is reduced to zero (and theyother event occurring is reduced to zero (and they are therefore dependent).are therefore dependent).

If one mutually exclusive event is known to occur,If one mutually exclusive event is known to occur, the other cannot occur.; thus, the probability of thethe other cannot occur.; thus, the probability of the other event occurring is reduced to zero (and theyother event occurring is reduced to zero (and they are therefore dependent).are therefore dependent).

Mutual Exclusiveness and IndependenceMutual Exclusiveness and Independence

Two events that are not mutually exclusive, mightTwo events that are not mutually exclusive, might or might not be independent.or might not be independent. Two events that are not mutually exclusive, mightTwo events that are not mutually exclusive, might or might not be independent.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