this is Poisson distribution with parameter λ=4.5 |
a)
P(X=0)=e-4.5*4.50/0! =0.0111
b)
P(X>3) =1-P(X<=3) =1-(P(X=0)+P(x=1)+P(X=2)+P(X=3))
=1-(e-4.5*4.50/0!+e-4.5*4.51/1!+e-4.5*4.52/2!+e-4.5*4.53/3!) =1-0.3423 =0.6577
c)
P(3<=X<=5) =P(X=3)+P(X=4)+P(X=5)
=e-4.5*4.53/3!+e-4.5*4.54/4!+e-4.5*4.55/5!
=0.1687+0.1898+0.1708 =0.5294
Problem(12)[10 pts. A Random variable X denotes numbers of child born with some geneti tain hospital...
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ive males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the X-linked genetic disorder. Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied. x P(x) 0 0.0290.029 1 0.1550.155 2 0.3160.316 3 0.3160.316 4 0.1550.155 5 0.0290.029 Does the table show...
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Poisson Distribution Question
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