## 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

=

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 |

= 15.5.

## Mean

**Mean**

The mean of the n numbers x_{1}, x_{2}……..x_{n }is

Example

Let x_{1} = 1179, x_{2} = 1404, x_{3 }= 1443,x_{4} = 1319 x_{5} = 1253 and x_{6 } = 1492.

Find mean ?.

= 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 x_{1},x_{2},x_{3},……..x_{n}.

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

P_{1}.P_{2},……….P_{n}. Then the expected value of the random variable is

E(x) = x_{1} p_{1 }+ x_{2}p_{2}+……..x_{n}p_{n.}

## 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

*p^{x}*(1 – p)^{n – x}.

## Combinations

**Combinations**

If denotes the number of combinations of n elements takes r at a time, where r ≤ n, then

=

## 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) =

## 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

m_{1} ways to make choice 1,

and for each of these ways,

m_{2} ways to make choice 2,

and so on, with

m_{n} ways to make choice n.

Then there are

m_{1} * m_{2} * …..m_{n}

Different ways to make the entire sequence of choices.

## Bayes Theorem (Special Case)

**Bayes Theorem (Special Case)**

P(F│E) = .

**Using Bayes’ Theorem**

**a. **Start a tree diagram with branches representing F_{1},F_{2},………….F_{n.} 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│F_{i}).

**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(F_{i}) and that the second branch occurs, P(E│F_{i})that is, the

product P(F_{i})*P(E│F_{i})which equals P(F_{i} ∩ E).

**d. P(F _{i}│E)**is found by dividing the probability of the branch for F

_{i}by the

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