Continuous Probability Distributions - Essay Example

Name Professor Title Date [] Continuous probability distributions A continuous random variable x can take any value in a particular or collection of intervals in a real number line (Peter 2011). The probability that the variable will be of a particular value is hard to tell. The lack of specificity calls for assumptions when establishing such occurrence. The probability of an assumption that the given random variable X will be of a certain value P within a given interval (A, B )is given by the area under the graph between the points A and B. There are three categories of continuous probability distribution. The categories are uniform probability distribution, normal probability distribution and exponential probability distribution (Patrick 24). Uniform probability distribution A valid claim that a random variable is uniformly distributed must show that the probability of the variable under explanation taking a particular value is proportional to the length of the interval between A and B. The distribution has a probability density function such that (Koehn 49) F (x) =; for A ≤x ≤B f(x) A B The function f(x) equals the reciprocal of the difference between B and A where the variable x lies only between the interval of B and A. The point B is where the greatest value that x exists whereas point A is where the smallest value of X exists. The mean or expected value of X is obtained by finding the average of the greatest value of X and the smallest value of X. The variance of the variable x is obtained by dividing the square of the difference of A and B and a by 12 (Koehn 52) An illustration about uniform probability distribution There is an assumption of a trader who can sell a minimum of 5 units of a commodity and a maximum of 20. Taking the commodity to be a random variable x, we can claim that the quantity of X cannot exist below 5 and above 20 hence the probability of taking value p is the reciprocal of 15. the expected/mean value of X is 12.5 and the variance is 225/12=18.75 when plotting the graph of the consumption function, there is consideration of all values between five and 20 on the X intercept while the y intercept contains only the function of X with the crest of the graph having a maximum of 1/15. the probability that a customer can by buy five to ten units is given by the reciprocal multiplied by the range given. That is 10-5=5. hence the probability becomes the product of 1/15 and 5=0.33 Normal probability distribution It enhances description of a continuous random variable such as heights of people, scientific measurements, and students test scores. Amounts of rainfall Applied in statistical inference. The variable X is distributed such that X is a function of meanµ, standard deviation (ɗ), pie (π) and exponential (е) (Tamás 87) Where µ is the mean and the square of ɗ is the variance. The constants of the function are Π=3.142 and е=2.71828 and they have fixed values. The following diagram shows normal distribution curves. F(x) x The distribution is symmetric and shaped like a bell. All normal distributions must encompass the mean and standard deviation. The symbol µ represents the mean and the symbol δ represents the standard deviation. When plotted, the highest point of the curve is at the mean. The mean is equal to the median and mode. The mean can be a positive, negative or zero unempirical value. The standard deviation defines the width of the curve. When values are large, the curves become wider and flat. A curve with standard deviation of 20 has a higher crest and a shorter range of values than a curve with a standard deviation of 30 (Tamás 91). The distribution probability of the values is determined by the areas under the curve. Either sides of the area from the mean have an area of 0.5. About 69% of the values of the random variable occur within the mean ±1 of the standard deviation 95 % of the values lie within the mean ±and 99% of the variables lie within the mean ±3 of the standard deviation. A random variable with standard normal probability distribution has a mean of zero and a standard deviation of 1 and is represented by Z to convert a variable to standard normal distribution, we use the formula; Where µ is the mean and ɗ is the standard deviation. The Microsoft excel program is used for normal distributions computation. An illustration of normal distribution computation There is an assumption of a company that supplies a particular good. At one point there is a drop in stock to a particular level say 20 units. The company makes an order for replenishment of the stock. Lack of punctuality in the replenishing worries the store manager due to the increased probability of discrepancy between demand and supply of the commodity that can inconvenience customers. The stock distribution during this time is normally distributed at a mean of 20 units and a standard deviation of 4 units. The manager wants to determine the probability of running out of stock. He specifies that running of stock occurs in the case whereby the company has stock amounting to twenty units. Then there is need to determine of the probability of the stock moving below 20 unit’s p (x≤20). In order to solve the problem, there is conversion of the stock distribution in to normal. The conversion is by dividing the difference of the existing stock and mean stock with the standard deviation. That is (23-20)/4=0.75 .Z=0.75. Next There is determination of there are existing between the mean and Z=0.75. From the Z distribution table, we move 0.7 unit’s vertical and 0.05 units horizontal to the values is 0.2734. The genera stock out probability under the curve is double the difference between the mean and the value obtained. For a manager to establish a maximum probability for stock shortage, say 0.5, there is establishment of the stock whose probability of stock-out is 0.5. The best way is by working it out with 99% confidence level, First there is subtraction of 0.1 from 1 an remain with 0.9 of the area under the total curve. The calculation involves only one side of the curve about the mean. The circumstance calls for division of the 0.9 by 2 to get is 0.45. We work backwards from the Z distribution table and find the z value that cuts of an area of 0.45 in the right tail of the standard normal distribution. The value is 1.645. we multiply the value with the mean to obtain the wanted value of stock. The reorder-point stock obtained is the one that can place the probability of the stock-out at the expected rate. From the above example, the probability of stock-out reduces from 0.2 to 0.05 customers satisfaction is high Microsoft Excel application is more accurate when calculating normal probabilities than using the manual method. Exponential probability distribution The distribution is used for computing time that a task takes to be completed. The time computed includes, the duration between arrivals of vehicles at the toll booth, time that a questionnaire take, a time that a commercial vehicle rakes to have a defect or default. The function of distribution encompasses the exponent and the mean of the variable. The value of exponent is 2.71828 (Holland et al. 46). The exponential functions apply the concept of lack of memory such that the probability of occurrence of an event in the past does not affect the probability of occurrence of the event in the current period. F(x) x The probability density function of an exponential distribution is f(x) =1/β [е-(x-µ)/β] Where µ is the location of the parameter in question must be .β is the scale parameter. A standard normal distribution has µ=0 and β=1. Standard normal distribution has a function of the form f(x) = е-x An illustration of the application of the distribution There is consideration of a firm that sells gas. Vehicles arrive at the pump at a particular interval. The mean arrival time is 3 minutes the manager want s to determine that the probability that it will take two minutes or less for a successive vehicle to arrive. Mathematically, the expression of the situation is p(x Read More