A discrete probability distribution is defined over a set value (such as a value of 1 or 2 or 3, etc). A continuous probability distribution is defined over an infinite number of points (such as all values between 1 and 3, inclusive).
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(1) That the probabilities lie between 0 and 1. (2) The sum of all probabilities of the distribution sum up to 1.
The probability is determined by the binomial distribution. We consider p = probability of defect, q = probability of not defect, n = sample size, and x= number of defects in sample, in this case x=2. We calculate the probability as P(X = x) = n!/[(n-x)! x!] pxqn-x If sample size = 10 and p = 0.1 then: P(x= 2) = 10!/(8!x2!)(0.1)2(0.9)8 = 0.1937 You can find more about the binomial distribution under Wikipedia. It is important also to note the assumptions when using this distribution. It must be a random sample and the probability of defects is known.
It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.
A discrete probability distribution is defined over a set value (such as a value of 1 or 2 or 3, etc). A continuous probability distribution is defined over an infinite number of points (such as all values between 1 and 3, inclusive).
Given U_i~χ_(ν_i)^2, (U_1/ν_1)/(U_2/ν_2 ) follows which distribution? F_(ν_1,ν_2 ) F Probability Distribution with ν degree of freedom Given T=Z/√(U/ν), Z~N(0,1) and U~χ_ν^2, T^2 follows an F-Distribution F_(1,ν) F Probability Distribution with one degree of freedom in the numerator and ν in the denominator
The answer depends on what the distribution is!
calulate each of the foollowing each of the following distribution x p(X)2 0 .2 1 .8
Given Z~N(0,1), Z^2 follows χ_1^2 Chi-square Probability Distribution with one degree of freedom Given Z_i~N(0,1), ∑_(i=1)^ν▒Z_i^2 follows χ_ν^2 Chi-square Probability Distribution with ν degree of freedom Given E_ij=n×p_ij=(r_i×c_j)/n, U=∑_(∀i,j)▒(O_ij-E_ij )^2/E_ij follows χ_((r-1)(c-1))^2 Chi-square Probability Distribution with ν=(r-1)(c-1) degree of freedom Given E_i=n×p_i, U=∑_(i=1)^m▒(O_i-E_j )^2/E_i follows χ_(m-1)^2 Chi-square Probability Distribution with ν=m-1 degree of freedom
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That would depend on the conditions of the experiment.
(1) That the probabilities lie between 0 and 1. (2) The sum of all probabilities of the distribution sum up to 1.
This is a binomial probability distribution; n=12, r=2 & P=.05. Read directly from the table probability of 2 is .099 (plugging this data into my calculator gives 0.09879).
If the question is to do with a probability distribution curve, the answer is ONE - whatever the values of mu and sigma. The area under the curve of any probability distribution curve is 1.
The probability is determined by the binomial distribution. We consider p = probability of defect, q = probability of not defect, n = sample size, and x= number of defects in sample, in this case x=2. We calculate the probability as P(X = x) = n!/[(n-x)! x!] pxqn-x If sample size = 10 and p = 0.1 then: P(x= 2) = 10!/(8!x2!)(0.1)2(0.9)8 = 0.1937 You can find more about the binomial distribution under Wikipedia. It is important also to note the assumptions when using this distribution. It must be a random sample and the probability of defects is known.
States that to determine a probability, we multiply the probability of one event by the probability of the other event. Ex: Probability that two coins will land face heads up is 1/2 x 1/2 = 1/4 .