Archive for August, 2014

Mean of a Grouped Distribution

Mean of a Grouped Distribution

 

The mean of a distribution, where x represents the midpoints, f the frequencies, and n = ∑f , is

\overline{x} = \frac{\sum xf}{n}.

 

Example

Listed below is the grouped frequency distribution for the 30 business executives

Find the mean from the grouped frequency distribution.

 

Interval Midpoint, x Frequency, f Product, xf
0-4 2 3 6
5-9 7 4 28
10-14 12 6 72
15-19 17 8 136
20-24 22 5 110
25-29 27 3 81
30-34 32 1 32
    Total 30 Total 465

 

\overline{x} = \frac{465}{30} = 15.5.

 

Mean

Mean

The mean of the n numbers x1, x2……..xn is

\overline{x}= \frac{\sum x}{n}.

Example

Let x1 = 1179, x2 = 1404, x3 = 1443,x4 = 1319 x5 = 1253 and x6  = 1492.

Find mean ?.

\overline{x}= \frac{1179+1404+1443+1319+1253+1492}{6}

\overline{x}=\frac{8090}{6} = 1348

 

Expected Value For Binomial Probability

Expected Value For Binomial Probability

 

For binomial probability, E(x) = np.

In other words, the expected number of successes is the number of trials times the probability of success in each trial.

 

 

Random Variable & Expected Value

Random Variable

A random variable is a function that assigns a real number to each outcome of an experiment.

 

Expected Value

 

Suppose  the random variable x can take on the n values x1,x2,x3,……..xn.

Also, suppose the probabilities that values occur are, respectively,

P1.P2,……….Pn. Then the expected value of the random variable is

E(x) = x1 p1 + x2p2+……..xnpn.

Binomial Experiment & Binomial Probability

Binomial Experiment

a. The same experiment is repeated several times.

 

b. There are only two possible outcomes, success and failure.

 

c. The repeated trials are independent, so that the probability of success remains the same for each trial.

 

 

Binomial Probability

If p is the probability of success in a single trial of a binomial experiment, the

probability of x successes and n – x failures in n independent repeated trials of

the experiment is

\begin{pmatrix}  n\\  x  \end{pmatrix}*px*(1 – p)n – x.

 

Combinations

Combinations

If \begin{pmatrix}  n\\  r  \end{pmatrix} denotes the number of combinations of n elements takes r at a time, where r ≤ n, then

\begin{pmatrix}  n\\  r  \end{pmatrix} = \frac{n!}{(n-r)!r!}.

Permutations

Permutations

If P(n,r) (where r ≤ n) is the number of permutations of n elements takes r at a time, them

P(n,r) = \frac{n!}{(n-r)!}.

 

Factorial Notation

Factorial  Notation

For any natural number n,

n! = n(n – 1)(n – 2)…………(3)(2)(1).

Also,

0! = 1.

Multiplication Principle

 

Multiplication Principle

Suppose n choices must be made, with

m1 ways to make choice 1,

and for each of these ways,

m2 ways to make choice 2,

and so on, with

mn ways to make choice n.

Then there are

m1 * m2 * …..mn

Different ways to make the entire sequence of choices.

 

Bayes Theorem (Special Case)

Bayes Theorem (Special Case)

P(F│E) = \frac{P(F)\cdot P(E\mid F)}{P(F)\cdot P(E\mid F)+P({F}')\cdot P(E\mid {F}')}.

 

Using Bayes’ Theorem

a.  Start a tree diagram with branches representing  F1,F2,………….Fn.   Label each

branch with its corresponding probability.

b.  From the end of each of these branches, draw a branch for event E. Label

this branch with the probability of getting to it, P(E│Fi).

c. You now have n different paths that result in event E. Next to each path,

put its probability—the product of the probabilities that the first branch

occurs,P(Fi) and that the second branch occurs, P(E│Fi)that is, the

product P(Fi)*P(E│Fi)which equals P(Fi ∩ E).

d.  P(Fi│E)is found by dividing the probability of the branch for Fi by the

sum of the probabilities of all the branches producing event E.