Odds

Odds

1. The odds in favor of an event E occurring is the ratio:

p(E) / p(EC) ; provided p(EC) in not 0

Notations:The odds is, often, expressed in the

form a:b, (where a and b are natural numbers) and read a to b.

2. Given the nonzero odds a:b in favor of an event E occurring, we can find p(E), by substituting the relevant values in the definition formula:

a/b = p(E) / [ 1 – p(E) ]

Or use the following derived formula:

P(E) = a / (a + b)

Assignment: Prove this formula

Example (1)

Omaima believes that the probability E of getting an A in this course is 0.8.

What’s the odds that Omaima will get an A?

SolutionThe odds in favor of that occuring= p(E) / p(EC) = 0.8 / ( 1 – 0.8)= 0.8 / 0.2 = 4The odds : 4 : 1

Example (2)

If the odd of Omaima’s getting an A in this course is 4:1

What’s the probability of that happening?Solution

Let the event of getting an A in the exam be E

Method (1) : Using the derived formulap(E) = 4 / (4 + 1) = 4/5 = 8/10 = 0.8

Method (2): From the definition of the odds

4 / 1 = p(E) / p(EC) = p(E) / [ 1 – p(E) ]→ p(E) = 4 [1 - p(E) ]→ p(E) + 4 p(E) = 4→ 5p(E) = 4→ p(E) = 4/5 = 0.8

Median,& Mode

In addition to the mean, there are two other measures of central tendency of a group of numerical data:

The median of a group of numbers&The mode of a group of numbers

Median, Mode

1. The median of a group of n numbers arranged in increasing or decreasing order is:

The middle number if n is odd &The mean (average) of the two middle

numbers if n is even

2. The mode of a group of n numbers is the most frequently occurring number of these numbers.

Examples (3)

Find the median of the following set of numbers:a. 46, 42, 49, 40, 52, 48, 45, 43, 50b. 37, 36, 39, 37, 34, 38, 41, 40Solution:a. Arranging the numbers in increasing order, we get:40, 42, 43, 45, 46, 48, 49, 50, 52We have 9 numbers (odd)Thus the median is the number in the middle, which

is 46

b. Arranging the numbers in increasing order, we get:34, 36, 37, 37, 38, 39, 40, 41We have 8 numbers (even)Thus the median = (37+38)/2 = 37.5

Examples (4)

Find the mode of the following set of numbers:a. 1, 2, 3, 4, 6b. 2, 3, 3, 4, 6, 8c. 2, 3, 3, 3, 4, 4 , 4, 8Solution:a. No mode; since no number occurs more

frequently than the others.b. The mode is 3; since it occurs more frequently

than the others.c. The modes are 3 and 4; each number occurs 3

times.

Variance & Standard Deviation

The variance is a measure of the degree of dispersion or spread of probability distribution associated with a random variable about its mean (expected value).

A small variance means small spread about the mean, while larger one means larger spread.

Thus, the variance gives more accurate picture of te probability distribution associated with the random variable.

Variance

Let X be a random variable with the probability distribution

p(X=xk) = pk; k = 1, 2, 3,……,n

& expected value E(X) = μ

Then the variance of the random variable X is:

Var(X) =p1(x1- μ)2 + p2(x2- μ)2 + p3(x3- μ)2 + .… + pn(xn- μ)2

Let X be a random variable with the probability distribution

p(X=xk) = pk; k = 1, 2, 3,……,n

& expected value E(X) = μ

Then the variance of the random variable X is:

Var(X) =p1(x1- μ)2 + p2(x2- μ)2 + p3(x3- μ)2 + .… + pn(xn- μ)2

Because the formula for the variance of the random variable X involves squaring the deviation of the values of the random variable (from the mean), then the unit measurement of Var(X) is the square of the measurement of these values, rather than that of the values themselves. Thus, it makes sense to consider the square root of the variance. This root will be called the standard deviation.

Standard Deviation

Let X be a random variable. The standard deviation of the random variable X is:

σ = √Var(X)

Example (5)

The shown table gives the probability distribution of a random variable X. Find the variance Var(X) and the standard deviation

σ of the random variable X

Solution:

The expected value E(X) = μ of the random variable X:

μ = 1(0.5) + 2(0.75) + 3(0.2) + 4(0.375) + 5(0.15) + 6(0.1) + 7(0.05) = 4

Var(X) = 0.5(1-4)2 + 0.75(2-4)2 +

0.2(3-4)2 + 0.375(4-4)2 + 0.15(5-4)2 + 0.1(6-4)2 + 0.05(7-4)2

= 1.95

σ = √ Var(X) = √ 1.95 ≈ 1.3964 ≈ 1.4

xp(X=x)

10.500

20.750

30.200

40.375

50.150

60.100

70.050

Remember that the expected value E(X) , denoted by μ of the random variable X is:

μ = E(X) = x1p1 + x2p2 + x3p3 + x4p4 + x5p5 + x6p6 + x7p7 = 1(0.5) + 2(0.75) + 3(0.2) + 4(0.375) + 5(0.15) + 6(0.1) + 7(0.05) = 4

& that he variance Var(X) is:Var(X) = p1(x1- μ)2 + p2(x2- μ)2 + p3(x3- μ)2 + p4(x4- μ)2 +

p5(x5- μ)2 + p6(x6- μ)2 + p7(x7- μ)2

= 0.5(1-4)2 + 0.75(2-4)2 +0.2(3-4)2 + 0.375(4-4)2 + 0.15(5-4)2 + 0.1(6-4)2 + 0.05(7-4)2

= 1.95

& that the Standard deviation σ is:σ = √ Var(X)

= √ 1.95 ≈ 1.3964 ≈ 1.4

Example (5)

y

Relative frequency

of occurrence

p(Y=y)

15.720.2

15.810.1

15.910.1

16.010.1

16.120.2

16.220.2

16.310.1

Let X & Y be the random variables defined to be as the weight of the packages of brands A & B of a product respectively. Find the mean and the standard deviation of X & Y and interpret the results.

x

Relative frequency

of occurrence

p(X=x)

15.810.1

15.920.2

16.040.4

16.120.2

16.210.1

Solution:Let μx and μy be the means or averages (the expected values) of the

random variables X & Y respectively, then:

μx = (15.8) (0.1) + (15.9) (0.2) + (16.0) (0.4) + (16.1) (0.2) + (16.2) (0.1)

= 16

→

Var(X) = (0.1) (15.8 -16)2 + (0.2) (15.6 -16)2 + (0.4) (16.0 -16)2 + (0.2) (16.1 -16)2 + (0.1) (16.2 -16)2 ≈ 0.012

→

σx = √(0.012) ≈ 0.11

μy = (15.7) (0.2) + (15.8) (0.1) + (15.9) (0.1) + (16.0) (0.1) + (16.1) (0.2) + (16.2) (0.2) + (16.3) (0.1)

Find μy , Var(Y) and σy !

Answers: μy = 16, Var(Y) = 0.042 and σy = 0.20

Interpretation:

The mean of X & Y are 16

→ The average weight of the packages of both brands is 16.

The standard deviation of Y is greater than that of X

→ The weight of a package of brand B is more widely dispersed about the common mean than the weight of brand A.