0,   elsewhere 


a.P(x >0)


c.P(X<0.5or >0.5)

d.probability that X exceed 1.5


asked Aug 12 in Other Math Topics by McCquabena Bannor Level 4 User (6,220 points)

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

1 Answer

The area under the curve = 1 the sum total of all probabilities.

For continuous random variable, summation is replaced by integration.

(a) P(x>0)=0.09375∫(4-x^2)dx{0,2}=0.09375[4x-x^3/3]{0,2}=0.09375(8-8/3)=0.5 (confirmed by picture)

(b) P(-1<x<1)=0.09375[4x-x^3/3]{-1,1}=0.09375(4-1/3-(-4+1/3)=0.09375(8-2/3)=0.6875. Note that because of symmetry this is the same as 2*0.09375[4x-x^3/3]{0,1}.

(c) 1, because this covers the whole of x between -2 and 2. The area to the left of x=0.5 and the area to the right include all x. So x has to be in one or other of the areas.

(d) P(0<x<1.5)=0.09375[4x-x^3/3]{0,1.5}=0.09375(6-1.125)=0.45703125.

Therefore P(x>1.5)=0.5-0.45703125=0.04296875, because 0.5 of the area is on the right (>0) so we merely subtract P(0<x<1.5). The picture makes this clearer.



answered Aug 13 by Rod Top Rated User (493,580 points)
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
80,062 questions
83,884 answers
66,810 users