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If mean and variance of Y are equal, does y necessarily have Poisson diistribution?

Y =  aX + b  where   x has Poisson distribution

Research, Knowledge and Information :

Lecture 5 : The Poisson Distribution

Lecture 5 : The Poisson Distribution ... 4 Mean and Variance of the Poisson distribution In general, there is a formula for the mean of a Poisson distribution.
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Poisson distribution - Wikipedia

The positive real number λ is equal to the expected value ... distribution of Y follows a Poisson distribution ... distribution with mean λ and variance λ ...
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Variance - Wikipedia

... as is the case for the Cauchy distribution, it does not have a variance ... Poisson distribution ... its mean. A square with sides equal to the ...
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Why is the mean of normal distribution equal to zero? - Quora

Why is the mean of normal distribution equal ... What is the meaning of mean is equal to zero and variance ... Normal distributions do not necessarily have the same ...
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Mean and Variance of Random Variables

Mean and Variance of Random Variables Mean ... which gives each observation equal weight, the mean of a random ... The variance for this distribution, with mean ...
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poisson - Can variance be equal to mean? - Cross Validated

Can variance be equal to mean? ... you have a distribution whose variance is equal to ... illustrating that even though for a Poisson mean and and variance are ...
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Lesson 12: The Poisson Distribution - Statistics

Lesson 12: The Poisson Distribution Introduction. In this lesson, we learn about another specially named discrete probability distribution, namely the Poisson ...
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Stats Flashcards | Quizlet

The mean of the t distribution is equal to that of ... A Poisson random variable can ... The pooled sample variance will not necessarily equal the F statistic ...
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13. The Poisson Probability Distribution -

Mean and Variance of Poisson Distribution. If ... then the mean and the variance of the Poisson distribution are both equal to ...
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Poisson Distribution - Statistics and Probability

Poisson Distribution. A Poisson distribution is the probability distribution that results from a Poisson experiment. Attributes of a Poisson ...
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Suggested Questions And Answer :

If mean and variance of Y are equal, does y necessarily have Poisson diistribution?

If mean and variance are equal, Y can have an exponential distribution rather than a Poisson distribution, because the mean and SD are the same. Therefore, if the mean is 1 the SD=1 and variance=SD^2=1. The correlation is a linear relationship and affects the mean for Y: E(Y)=aE(X)+b=µ (or lambda) for Po(µ,Y).  
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A real estate agent claims that there is a difference between the mean household

For sample 1: n1=12, m1=32750, s1=1900, variance, v1=s1^2=3610000 For sample 2: n2=10, m2=31200, s2=1825, v2=s2^2=3330625 Variance of source population=(v1+v2)/((n1-1)+(n2-1))=347031.25, because we're told that it is equal for both samples. We can now estimate the standard deviation S for the sampling distribution = √(V(1/n1+1/n2))=252.2348 approx. The t-score=(m1-m2)/S=(32750-31200)/252.2348=6.145 approx. This tells us that the difference in means is more than 6 standard deviations. The degrees of freedom are n1-1+n2-1=20. The value of over 6 well exceeds the confidence level of 99.9% (critical value for t=1.725 - 2-tail). So the claim is false: there is no significant difference between the income levels.  
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mean # of strokes/hole is 3.3 variance and standard deviation using Poisson distri?

Variance=mean=3.3, SD=sqrt(3.3)=1.82 approx. Poisson probability P(X)=(e^-µ)µ^X/X!, where µ=mean. P(72/18)=P(4)=e^-3.3*3.3^4/4!=0.182 is the probability of averaging 72 in 18 holes. Exclude P(0) in this case because it has no meaning to average zero strokes per hole. Probability of more than 4 strokes per hole is (1-probability of 1, 2, 3 or 4 strokes per hole). Probability of more than 4 strokes per hole is 1-(P(1)+P(2)+P(3)+P(4))= 1-e^-3.3(3.3+3.3^2/2+3.3^3/6+3.3^4/24)= 1-3.3e^-3.3(1+3.3/2+3.3^2/6+3.3^3/24)=0.2743 or 27.43%.
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probabity of 3 chances of a event is 2/3 times the probabity of 4 chances. Find mean & SD

Po(m;x)=e^-m * m^x/x!  symbolises the Poisson probability of x events for a mean m events. When x=3 Po(m;3)=e^-m * m^3/6 and when x=4 Po(m;4)=e^m * m^4/24. Po(m;3)=(2/3)Po(m;4), so m^3/6=(2/3)m^4/24, 1/6=m/36 and m, the mean=36/6=6. In a Poisson distribution the mean and variance are the same, and SD=√variance, so SD=√6=2.4495 approx.
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solve: P(X=3) = (2/3) P(X=4)

In a Poisson distribution, the probability P of x successes is e^-m*m^x/x!, where m is the mean. So if x=3, P=e^-m*m^3/3! and if x=4, P=e^-m*m^4/4!. If P(3)=(2/3)P(4), then m^3/6=(2/3)*m^4/24, so m=3*24/(2*6)=6. The standard deviation=sqrt(variance), and in a Poisson distribution, variance=mean, so SD=sqrt(6)=2.45.
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The random variable X follows a Poisson distribution with standard deviation 2. Find P(X ≤ 3).

In a Poisson distribution, variance=mean, and SD^2=variance, so mean=4. P(X)=e^-m*m^X/X!. P(X<3)=P(0)+P(1)+P(2)+P(3)=e^-4(1+4+16/2+64/6)=71(e^-4)/3=0.4335.
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let the random variable X have a mean of m and variance n^2. define Y=a+bx where a and b are constants satisfying -∞<a<∞ and b>0. select a and b so that Y has a mean of zero and variance equal to one.

Suppose the set of data for X is { Xsub1 Xsub2 Xsub3 ... XsubN }, then m=S(XsubJ)/N where S represents the sum of XsubJ for J=1 to N. But X=(Y-a)/b and XsubJ=(YsubJ-a)/b. Therefore m=S(XsubJ)/N=S((YsubJ-a)/b)/N=S(YsubJ)/N-Na/bN=m'-a/b where m' is the mean of Y. We know that m'=0. m'=m+a/b=0, m=-a/b, and a=-bm. S((XsubJ-m)^2)/N is the variance=n^2. So n^2=S(XsubJ-m)^2/N=S(((YsubJ-a)/b-m)^2)/N= S((YsubJ/b-a/b+a/b)^2/N=S((YsubJ)^2)/Nb^2. Variance of Y=1=S((YsubJ)^2)/N. So n^2=1/b^2 and b^2=1/n^2, b=1/n>0. Therefore, since a=-bm, then a=-m/n. ANSWER: a=-m/n and b=1/n.  
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Pls answer this question for me

I've only just approved your question. But I can begin to answer it. The PDF has an area of 1 because the sum total of all probabilities must be equal to 1. This enables us to find k. ∫f(x)dx=1 between the limits 1 and 4, where f(x)=k(x-1)(4-x)=k(5x-x^2-4). So integrating between the limits we have [5kx^2/2-kx^3/3-4kx]{1,4}= ((5/2)(16-1)-(64-1)/3-4(4-1))k=1 (37.5-21-12)k=1 4.5k=1 so k=1/4.5=2/9 (answer 18C) and f(x)=(2/9)(x-1)(4-x). The mean is given by ∫xf(x)dx{1,4}=∫k(5x^2-x^3-4x)dx between the limits= (2/9)[5x^3/3-x^4/4-2x^2]{1,4}=(2/9)(105-63.75-30)= (2/9)(45/4)=2.5, which is to be expected since the PDF is symmetrical about x=2.5. Mean=2.5 (answer 19C) We need the mean to calculate the variance, the square root of which is the standard deviation. VAR=∫(x-2.5)^2(f(x))dx{1,4} The answer choices for 20 are the same as for 19. This seems suspicious. So the SD may not be listed. I calculate it to be 0.6368. The computation for the variance can be simplified: VAR=∫x^2f(x)dx-5∫xf(x)dx+6.25∫f(x)dx between the limits=∫x^2f(x)dx-5*2.5+6.25=∫x^2f(x)dx-6.25, because we have already computed the integral for the mean and we know ∫f(x)dx=1.  VAR=(2/9)∫(5x^3-x^4-4x^2)dx{1,4}-6.25=(2/9)[5x^4/4-x^5/5-4x^3/3]{1,4}-6.25= (2/9)(599/20)-6.25=599/90-6.25=73/180=0.405556 approx, making SD=√0.405556=0.6368 approx. This is not a listed answer choice for 20.
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find the moment generating function of S to independent random variable?

If you want re a l time help, go to and your question will be answered there.
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C = 7.00 + 0.0742N. where N denotes the number of units, N is modelled as a random variable with mean 600 and variance 250. fine the mean and variance of the charges C in their respective units

N=(C-7.00)/0.0742. Mean of N=S(Nsubj)/n=600, where S is the sum of elements Nsubj for j between 1 and n, where n is the number of elements of N={ N1 N2 N3 ...}. So S(Csubj-7.00)/0.0742n=600=(mean of C)/0.0742-7.00n/0.0742n. (Mean of C)/0.0742=600+7.00/0.0742=694.33962; mean of C=51.52. This is the same as C=7.00+0.0742*600=51.52, just substituting the mean of N for N in the equation. I just showed the logic behind it. The logic for the variance is similar. Var of N = 250=S((Nsubj-600)^2)/n=S(((Csubj-7.00)/0.0742 - 600)^2)/n. 250=S((Csubj-51.52)^2)/0.0742^2n=(var of C)/0.0742^2, so var of C=250*0.0742^2=1.38 approx.
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