Question

HELP !!!!!!!! probability question

Measurements of the position x (in units of cm) of a particle are governed by the
probability density

P(x) = A(1 - x^2)

negative 1less than or equal to  x less than or equal to  1

(a) Determine A.
(b) Find the expectation values < x > and < x2 >.
(c) What is the probability that a measurement of the particle's position yields a
result between 0 and 0.5?
(d) If your classmates and you each performed one measurement of the position of
the particle, what do you expect the spread in your measurement results to be?

Answer 1

Of course the integral of P(x) from x=0 to 1 will equal 1. This is how to solve for A.

Answer 2

The expected value of x will be the integral of xP(x) over the same interval.
(c) integrate from x=0 to 0.5
Expected value of |x1-x2| is trickier. Try to nail down the first three first.

Answer 3

why use the bound from x= 0 to 1 ?

Answer 4

Oh, sorry. Integral of P(x) from x=-1 to 1 will be 1.

Answer 5

thanks

Answer 6

\[\int_{x=-1}^{1}A(1-x^2)dx=1\ \ \text{ (solve for A)}\]

Answer 7

Expected Spread of two observations would be:\[\int_{x=-1}^{1}\int_{y=-1}^{1}\sqrt{(x-y)^2}A^2(1-x^2)(1-y^2)dxdy\]

Answer 8

why is there two observations?

Answer 9

Oh, I misread that too. I was thinking you and one classmate and defining spread as |x-y| but they are probably just asking for variance or standard error which is a little easier, just integrate x^2 over the domain. (since the expected mean will be zero).