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probability distribution

properties of binomial,poission and their appllication in bussiness. formulaes of binomial and poission distribution.

Research, Knowledge and Information :


Probability distribution - Wikipedia


In probability theory and statistics, a probability distribution is a mathematical function that, stated in simple terms, can be thought of as providing the ...
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Probability Distribution - Statistics and Probability


This lesson explains what a probability distribution is. Shows how to find probabilities of random variables. Includes problems with solutions.
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List of probability distributions - Wikipedia


Many probability distributions that are important in theory or applications have been given specific names.
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Probability Distribution in Statistics - ThoughtCo


A probability distribution is a function or rule that assigns probabilities to a random variable. This can be done in a list, a table or even a graph.
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1.3.6. Probability Distributions


Probability Distributions Probability distributions are a fundamental concept in statistics. They are used both on a theoretical level and a practical level.
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Probability Distribution - Investopedia


There are many different classifications of probability distributions. Some of them include the normal distribution, chi square distribution, binomial distribution ...
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Lesson 5: Probability Distributions | STAT 200


In this lesson we will begin to explore the concept of statistical inference. We will look at both discrete and continuous probability distributions. The concepts of ...
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Stats: Probability Distributions - Richland Community College


Probability Functions. A probability function is a function which assigns probabilities to the values of a random variable. All the probabilities must be between 0 ...
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Constructing a probability distribution for random variable ...


Sal breaks down how to create the probability distribution of the number of "heads" after 3 flips of a fair coin.
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Random variables | Statistics and probability | Math | Khan ...


Random variables can be any outcomes from some chance process, ... Constructing a probability distribution for random variable. Constructing probability distributions.
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Suggested Questions And Answer :


probability question

Out of 50 students, X=30, Y=15, making up 90% of the students. a) Tree diagram consists at the top of the 50 students, branching out to the second level 30, 15 and 5 for the three disciplines. The project team tree has three branches but, because students are randomly chosen, we cannot say what disciplines they represent, but we can represent the probabilities as another level of branches 1/10, 3/5 and 3/10 corresponding to 10%, 60% and 30% attached to each of the branches for the first level branches. (Strictly speaking, the selection process may change the probabilities, because the first student selected increases the probabilities for the remaining students by reduction of the pool of students to 49. If, for example, an X student was the first to be selected for the project team, then out of the 49 remaining, there would be 29 X students, changing the probabilities to 29/49, 15/49 and 5/49.) b) The joint probability distribution can be represented binomially as (p+(1-p))^3=p^3+3p^2(1-p)+3p(1-p)^2+(1-p)^3, where p=0.90 (for combined decision sciences and industrial statistics majors). So 1=0.729+0.243+0.027+0.001. The terms in this series separately show the make up of the project team as probabilities of all X or Y, 2 X or Y plus a business mathematics student, 2 of the latter and one X or Y, or all business maths students, c) If X=1 then 49 students remain, consisting of 29 X type, 15 Y type and 5 business maths majors. The binomial distribution of selecting Y is given by (p1-(1-p1))^2=p1^2+2p1(1-p1)+(1-p1)^2, where p1=15/49, the probability of selecting a Y student, and 1-p1=34/49 is the probability of selecting another X or a business maths student. So 1=0.0937+0.4248+0.4815. That means 9.37% probability of 2 Y students, 42.48% probability of 1Y and either of the other two types, and 48.15% probability of no Y type students. d) The mean in a binomial distribution is np, where n in this case is 3 (size of the project team) and p=3/5 or 0.6, so the mean is 1.8. This assumes that the mean applies before any member of the team has been selected, rather than when X=1, after a selection has already been made. e) For Y the mean is 3*15/49=0.92 approx., given that X=1. f) Before any member of the team has been selected, the probabilities are independent. The only dependence X and Y share is that once a team member has been selected, and is found to be X, for example, the probability changes because there is one fewer of that type left in the pool of students.
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Two dice are being thrown at once and getting three is considered a success. Obtain the probability distribution of the number of 3's.

The number of possible outcomes of throwing 2 dice together is 36, but only 2 of these outcomes add up to 3: 1+2 or 2+1. So the probability is 2/36=1/18. A probability distribution of all possibilities is given through the table below:  1+1=2 1+2=3 1+3=4 1+4=5 1+5=6 1+6=7 2+1=3 2+2=4 2+3=5 2+4=6 2+5=7 2+6=8 3+1=4 3+2=5 3+3=6 3+4=7 3+5=8 3+6=9 4+1=5 4+2=6 4+3=7 4+4=8 4+5=9 4+6=10 5+1=6 5+2=7 5+3=8 5+4=9 5+5=10 5+6=11 6+1=7 6+2=8 6+3=9 6+4=10 6+5=11 6+6=12 FREQUENCY DISTRIBUTION (OCCURRENCES): 2 1 3 2 4 3 5 4 6 5 7 6 8 5 9 4 10 3 11 2 12 1 The first column is the sum of the dice and the second is the number of ways that sum can be made. From the frequency distribution we get the triangular probability distribution: The dotted line just shows the general triangular shape of the distribution. The horizontal markings are the dice sums and the vertical markings are the number of occurrences. The fractions show the probabilities, which sum to 1 (100%) so the area of the triangle is 1 when measured in terms of the probabilities (NOT by counting squares!). From the table we can see that the probability of either die being 3 is as follows (probabilities in brackets: 1/6 = chance of a 3, 5/6 = chance of some other number): No 3's: 25 (25/36=5/6*5/6) One 3: 10 (10/36=5/18=5/6*1/6+1/6*5/6) Two 3s: 1 (1/36=1/6*1/6) TOTAL 36 (1)  
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Does the distribution represent probability distribution X 1 2 3 4 5 P (x) 1/10 3/10 1/10 2/10 3/10

P(x) for x=2 appears to be a glitch in an otherwise ascending series of values. If, for example the P values were in order: 1/10, 2/10, 3/10, 3/10, 1/10, they could be a skew distribution. The usual form for a probability distribution is a rise to a summit value and then a falling off; or a cumulative rise to a levelling off: 1/10, 1/10, 2/10, 3/10, 3/10. Although the values add up to 1, the implied shape of the supposed probability curve suggest that it isn't a probability curve at all.
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please solve this

f(t) gives the probability as a point on a distribution curve while F(t) is the area beneath the distribution curve to the left of the point defined by f(t), representing the sum of the probabilities up to the point f(t). A typical example is a normal distribution curve (bell curve). Along the t axis we may have the discrete (non-continuous) values, 0, 1, 2, 3, etc. The curve shows the probability of 0, 1, 2, 3, ... successes or hits. (In fact, the curve is really a series of blocks, one block for each value of t). So we have f(0), f(1), etc., for individual probabilities, t=T. F(t) gives the sum of the probabilities up to and including a particular value t=T, i.e., t0), as indicated in your question, the same rule applies to PDF (individual probability) and CDF (cumulative probability). The curve demonstrates or illustrates that t is continuous.
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What is the probability that a tree’s height exceeds the mean if normally distributed.

(a) 50% by definition of the mean as the central point of the distribution. (b) For Z=1 the table gives the value 0.8413 which is 0.3413 above 0.5000 (mean). So 34.13% is the probability of being in the range from the mean and 1 SD from the mean. 0.5000-0.3413=1-0.8413=0.1587 (15.87%) is the probability of the height being at least 1 SD from the mean. (c) 0.9987 is the probability for Z=3 SD from the mean, so 1-0.9987=0.0013 (0.13%) is the probability for the height in excess of 3 SD. (d) 0.9332 is the probability for Z=1.5, so 0.4332 (43.32%) is the probability of being within 1.5 SD of the mean.  
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how to solve it need help

a) The table below shows the probability distribution, where P is probability: RAINFALL P Profit (RM) Heavy 0.3 28000 Moderate 0.35 55000 Little 0.2 15000 None 0.15 7000 The table below shows cumulative probability from heavy to no rainfall: RAINFALL P Profit (RM) Heavy 0.3 8400 Heavy/moderate 0.65 27659 Heavy/moderate/little 0.85 30650 All 1 31700 Explanation of profit: e.g., Heavy/moderate/little 0.3*28000+0.35*55000+0.2*15000=8400+19250+3000=30650 Cumulative probability from no to heavy rainfall: RAINFALL P Profit (RM) None 0.15 1050 None/little 0.35 4050 None/little/moderate 0.7 23300 All 1 31700 b) Both tables give expected profit as RM31700. The figure is obtained by multiplying the probability of each possibility by the individual profit then adding all these products together. c) The average (mean) of the figures is 31700/4=7925. The dataset is 8400, 19250, 3000, 1050 To obtain the SD we work out the difference of each of these from the mean and square it: 225625, 128255625, 24255625, 47265625 then add these together: 200002500, divide by 4: 50000625 and take the square root: 7071.11 approx. d) The SD gives an effective margin of error on the expected profit: RM7925+7071 which puts the range of the profit between RM854 and RM14996. This does seem too much of a range to be true, so I guess I've made a wrong assumption. To make a better estimate, we need to look at all the rainfall possibilities in more detail. Consider how the meteorologist got his figures. Let's say he measured rainfall over a period of 365 days and discovered the following: No rainfall for 54.75 days Little rainfall for 73 days Moderate rainfall for 127.75 days Heavy rainfall for 109.5 days These periods of time would give the probabilities quoted. Based on these, and using a prorata value for the farmer's profit, we can calculate his profit for these periods: 54.75/365*7000+73/365*15000+127.75/365*55000+109.5/365*28000=31700, as expected from calculations earlier. However, the cumulative probabilities are wrong, because, apart from when the cumulative probability is 1, the time periods for other values needs to be adjusted prorata for the whole year. No rainfall for the whole year produces a profit of RM7000; but little or no rainfall requires adjustment. The time period is 54.75+73=127.75 which is 0.35 year. So the profit has to be extrapolated based on the factor 1/0.35. So we have (54.75/365*7000+73/365*15000)/0.35=RM11571.43. For rainfall ranging from none to moderate, the time period is 0.7 year. The adjustment or extrapolation factor is 1/0.7. This gives us 23300/0.7=RM33285.71. And, of course, all types of rainfall give us RM31700. Working from heavy to no rainfall: RAINFALL P Profit (RM) Heavy 0.3 28000 Heavy/moderate 0.65 42538.46 Heavy/moderate/little 0.85 36058.82 All 1 31700 And from no to heavy rainfall: RAINFALL P Profit (RM) None 0.15 7000 None/little 0.35 11571.43 None/little/moderate 0.7 33285.71 All 1 31700 In both the last two tables we have datasets for which we can calculate mean and SD for 7 rainfall patterns: No rainfall at all; little or no rain; all but heavy rain; a mixture of all types; heavy to moderate; some daily rain; only heavy rain. Using figures from the tables we arrive at a mean of RM27164.92 and a SD of RM12094.61, which gives a range for the profit of RM15070.31 to RM39259.53. (SD is square root of VARIANCE. Both mean and variance are calculated by dividing the sums of the relevant columns by 7, the number of rainfall types.) RAINFALL Profit Profit-mean (Profit-mean)^2 None 7000 -20164.92 406624000 Little or none 11571.43 -15593.49 243156930 No heavy rain 33285.71 6120.79 37464070  Mixed 31700 4535.08 20566951  Heavy or moderate 42538.46  15373.54  236345732 Some daily rain 36058.82  8893.9  79101457  Only heavy rain 28000 835.08  697359  MEAN: 27164.92  VARIANCE: 146279500  
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Poisson distribution question

The probability of no mail in a day when µ=10 and x=0 is P(10,0)=(e^-10)(10^0)/0!=e^-10=0.0000454 approx. This is 1/22026 approx., or an expectation of one day without mail in about 22,026 days. If we scale this up for 3000 days we get 3000*0.0000454=0.1362. So a new Poisson distribution can be set up where µ=0.1362. P(0.1362,x)=(e^-0.0000454)0.0000454^x/x! where x is 0, 1, 2, days, etc. Also, if x=0, (no days without mail) then the probability of at least one day in 3,000 days without mail is 1-P(0.1362,0)=1-e^-0.1362=1-0.8726=0.1273 or 12.73% approx.  
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may i know how to solve this question??

a) The table below shows the probability distribution, where P is probability: RAINFALL P Profit (RM) Heavy 0.3 28000 Moderate 0.35 55000 Little 0.2 15000 None 0.15 7000   Consider how the meteorologist got his figures. Let's say he measured rainfall over a period of 365 days and discovered the following: No rainfall for 54.75 days Little rainfall for 73 days Moderate rainfall for 127.75 days Heavy rainfall for 109.5 days These periods of time would give the probabilities quoted. Based on these, and using a prorata value for the farmer's profit, we can calculate his profit for these periods: 54.75/365*7000+73/365*15000+127.75/365*55000+109.5/365*28000=31700. So the answer to b) (your second a) is RM31700. No rainfall for the whole year produces a profit of RM7000; but little or no rainfall requires adjustment. The time period is 54.75+73=127.75 which is 0.35 year. So the profit has to be extrapolated based on the factor 1/0.35. So we have (54.75/365*7000+73/365*15000)/0.35=RM11571.43. For rainfall ranging from none to moderate, the time period is 0.7 year. The adjustment or extrapolation factor is 1/0.7. This gives us 23300/0.7=RM33285.71. And, of course, all types of rainfall give us RM31700. The tables below show cumulative probability. Working from heavy to no rainfall: RAINFALL P Profit (RM) Heavy 0.3 28000 Heavy/moderate 0.65 42538.46 Heavy/moderate/little 0.85 36058.82 All 1 31700 And from no to heavy rainfall: RAINFALL P Profit (RM) None 0.15 7000 None/little 0.35 11571.43 None/little/moderate 0.7 33285.71 All 1 31700 In both the last two tables we have datasets for which we can calculate mean and SD for 7 rainfall patterns: c) and d) No rainfall at all; little or no rain; all but heavy rain; a mixture of all types; heavy to moderate; some daily rain; only heavy rain. Using figures from the tables we arrive at a mean of RM27164.92 and a SD of RM12094.61, which gives a range for the profit of RM15070.31 to RM39259.53. (SD is square root of VARIANCE. Both mean and variance are calculated by dividing the sums of the relevant columns by 7, the number of rainfall types.) RAINFALL Profit Profit-mean (Profit-mean)^2 None 7000 -20164.92 406624000 Little or none 11571.43 -15593.49 243156930 No heavy rain 33285.71 6120.79 37464070  Mixed 31700 4535.08 20566951  Heavy or moderate 42538.46  15373.54  236345732 Some daily rain 36058.82  8893.9  79101457  Only heavy rain 28000 835.08  697359  MEAN: 27164.92  VARIANCE: 146279500
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statistics

(a)(1) The probability of a sale is 0.20 so the probability of no sale is 0.80. The probability of no sales with 8 firms is 0.80^8=16.78% approx. (2) We can use the binomial distribution with coefficients: 1 8 28 56 70 56 28 8 1. For no more than 2 sales we need the first 3 coefficients. (p+(p-1))^8=p^8+8p^7(1-p)+28p^6(1-p)^2+... where p=0.80. The first term is no sales at all, the second term just one sale, and the third term two sales. The total probability is 79.69%. (b)(1) Z score for X=10 is (10-7.5)/2.1=1.19 (approx) corresponding to 88.30%. So the percentage of students studying more than 10hrs per week is 11.70%. (2) If X=9, Z=0.7143, corresponding to 76.25%; and  if X=7, Z=-0.2380, corresponding to 1-0.5940=40.60%. So the proportion of students spending 7-9 hours is 76.25-40.60=35.65% approx. (3) If X=3 then Z=-2.143, corresponding to 1-0.9839=0.0161 or about 1.61% of students spend less than 3 hrs a week studying. (4) In the distribution table Z=1.645 corresponding to 95%. Therefore Z=-1.645 corresponds to 5% and -1.645=(X-7.5)/2.1, X=7.5-3.4545=4.04 or about 4 hours. So 5% of students spend less than 4 hours studying per week. (c) We don't know how many customers the clothing stores sees in one day. Should 2.7 be 2.7%?
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How to plot

I guess you want to use Excel to plot a binomial distribution where the probability is p=0.25. Excel will plot a graph for non-cumulative and cumulative distribution. You use the BINOMDIST statistical function, which has 4 parameters:   number of successes (x axis); total number of trials, n; the probability, p; whether cumulative or not. You set these up in a table for Excel to use to do the plots, with a fixed cell for the probability and columns for the number of successes and the corresponding values to contain the results of applying BINOMDIST.  The question asks for three graphs, so what could they be? The BINOMDIST function has 4 parameters, and the question supplies only one. So we can use Excel to plot graphs where the other parameters are different. We can change the number of trials; we can choose whether cumulative or non-cumulative; we can change the range of successes. So we set up three tables. The first table has fixed cells for n, p and cumulative=FALSE, and values in a column from 0 to n for the number of successes to be plotted; second table has the same but cumulative=TRUE; the third has a different n and successes column and cumulative can be either TRUE or FALSE. You would specify the continuous curve type of graph for Excel to plot the data as a curve and the result would show the typical bell-shaped binomial distribution curve for the non-cumulative distribution and the hill-shaped curve for the cumulative. To use the graphs, you read off for each x value (number of successes required) the percentage or fraction of the expected results. You would also title the graph, label its axes and show that p=0.25 is the active probability. Excel will allow you to customise how you want the graphs to look. [Non-cumulative means the exact number of successes expected in a given number of trials; cumulative means at least or at most the specified number of successes in a given number of trials.]
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