COMPETENCIAS DE LIDERAZGO EN LA COMUNIAD

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σ = S.D. = 1.6 = 1.265

3.2 DISCRETE PROBABILITY DISTRIBUTIONS

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If we let the random variable X equal the number of observed successes in n Bernoulli trials, the possible values of x are 0, 1, 2,..., n. If x successes occur, where x = 0, 1, 2,

…., n, then n–x failures occur. The pdf of x, say f(x) is

f(x) =nC_xP^x (1 – P)^n-x, x = 0, 1, 2, ….. n (5.5) where

nC_x = n!

x! (n – x)!

The number of ways of selecting x positions for the x successes in the n trials. Where (5.5) is the binomial distribution of the random variable X.

The binomial experiment must possess the following properties:

i. Each trial has two possible outcomes (success, failure) ii. There are n repeated independent trial

iii. The probability of success on each trial is a constant p; the probability of failure is q = 1 – p.

iv. The random variable X equals the number of successes in the n trials.

v. The random variable x defined above is said to be a Binomial distribution with parameters n and p denoted as X~ Bi (n, p)

Example

If X is a binomial random variable, calculate the probability of x for i. n = 3, x = 2, P = 0.3

ii. n = 4, x = 0, P = 0.4 Solution

P(X = x) = nC_xp^x (1 – p)^n-x

i. P(X = 2) =3C₂ (0.3)² (1-0.3)^3-2

= 3 x 0.09 x 0.7 = 0.189 ii. P(x = 0) = 4C₀ (0.4)⁰ (1 – 0.4)^4-0

= (0.6)⁴ = 0.1296 Example

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Suppose that in a certain malarious area past experience indicates that the probability of a person with a high fever will be positive for malaria is 0.7. Consider 3 randomly selected patients (with high fever) in that same area.

i. What is the probability that no patient will be positive for malaria?

ii. What is the probability that exactly one patient will be positive for malaria?

iii. What is the probability that exactly two of the patients will be positive for malaria?

iv. What is the probability that all patients will be positive for malaria?

v. Find the mean and the SD of the probability distribution given above.

Solution: (i) 0.027 (ii) 0.189 (iii) 0.441 (iv) 0.343 (v) μ = 2.1 and σ = 0.794

Poisson Distribution

Poisson distribution is another discrete probability distribution. Some experiments result in counting the number of times particular events occur in a given time. For example, we could count the number of road accidents that occur on the third mainland bridge in Lagos between 2 and 4pm, the number of phone calls arriving at an office between 5 and 6pm, or the number of defective antenna produced by a company in a day. All these three examples have certain characteristics in common.

First, in each case, we are looking at situations where there are relatively few successes during the indicated time period. Thus, the probability of success in any sufficiently small time interval will be quite small.

Second, the individual successes are independent of one another. That is the fact that one accident occurs a month ago should not influence any other accident from occurring today.

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Third, it is assumed that the event occurring is uniformly distributed over the entire time period under consideration. Thus, we assume that a phone call arriving into an office will arrive at any time during the day with equal likelihood.

Under the above conditions, the probabilities involved follow a Poisson probability distribution. The Poisson probability formula giving the probability that x success occur during a given time interval is:

P(x successes) =

! x e^^x

x = 0, 1, 2,… (5.6)

where  is the average number of successes occurring in the given time interval and the symbol ℮ represents the number

℮ ≈ 2.71828

When a random variable x has this distribution we write in shorthand X~ P₀(). For the Poisson distribution,

Mean (µ) = variance = 

Table II in the Appendix gives values of the distribution function of a Poisson random variable for selected values of .

Example

Suppose X is a Poisson random variable with parameter 4 [X~ P₀ (4)]. Compute the following probabilities.

i. P(X ≤ 4) ii. P(X  5) iii. P(X = 7) iv. P(X  6) v. P(X  5) vi. P(2 ≤ X ≤ 5) vii. P(2  X ≤ 5)

viii. P(2 ≤ X  5) Solution

We use the cumulative Poisson probability table to solve these problems. Here  = 4 and he probability P(x ≤ x) for selected values of  and x are given in statistical table.

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i. P(X≤ 4) = 0.629

ii. P(X  5) = P(X ≤ 4) = 0.629 iii. P(X = 7) = P(X ≤ 7) – P(X ≤ 6)

= 0.949 – 0.889

= 0.060 iv. P(X  6) = 1 – P(X ≤ 6)

= 1 – 0.889 = 0.111 v. P(X  5) = P(X  5) = 1 – P(X ≤ 4)

= 1 – 0.629 = 0.371

vi. P(2 ≤ X ≤ 5) = P(X ≤ 5) – P(X ≤ 1)

= 0.785 – 0.092

= 0.693

vii. P(2  X ≤ 5) = P(X ≤ 5) – P(X ≤ 2)

= 0.785 – 0.238 = 0.547 viii. P(2 ≤ X  5) = P(X ≤ 4) – P(X ≤ 1)

= 0.629 – 0.092 = 0.537 3.3 NORMAL DISTRIBUTION

The Normal Distribution, also called Gaussian distribution, is most important probability distribution in statistics. The distributions of many medical measurements in populations follow a normal distribution (e.g. Serum uric acid levels, cholesterol levels, blood pressure, height and weights e.t.c.).

For the normal distribution there are two parameters (μ and σ²) that define and describe it. The normal distribution is a theoretical, continuous probability distribution whose probability density function (PDF), f(x) is:

f(x) = _



 



 









)2

( 2 exp 1 2

1 x

 x  set of Real numbers

The area that represents the probability between two points c and d is defined by:

P (a ≤ x ≤ = b) =



a

f (x)dx

The important characteristics of the Normal Distribution are:

i. It is a continuous probability distribution with continuous variable.

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ii. The random variable extends from minus infinity(-∞) to plus infinity (+∞).

iii. It is unimodal, bell-shaped and symmetrical about x = u.

iv. It is determined by two quantities: its mean (μ) and SD (σ). Changing μ alone shifts the entire normal curve to the left or right. Changing σ alone changes the degree to which the distribution is spread out.

v. The height of the frequency curve, which is called the probability density, cannot be taken as the probability of a particular value. This is because for a continuous variable there are infinitely many possible values so that the probability of any specific value is zero.

vi. It is centered at the mean μ vii.







f(x) dx = 1, that is, the total area under the normal distribution curve is 1.