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Let the diameter be x and its probability density be P(X)= 1/(b-a), (a≤X≤b).

The volume of the sphere v = 4π(x/2)3/3 =π×3/6.

ev=∫[a,b]πx^3/6(b-a)dx=[π(a+b)(a^2+b^2)]/24

Solution: probability density function of x: f(x)= 1/(b-a) x:[a, b]

F(x)=0 Other x

E(v)=∫(b,a) πx? /6/(b-a) dx

= π/[6(b-a)] ∫(b,a) x? Advanced (short for deluxe)

= π/[24(b-a)] x^4 |(b,a)

= π/[24(b-a)] (b^4 - a^4)

= π(a+b)(a? +b? )/24 ( 1)

That is, the mathematical expectation of sphere volume: E(v) = π(a+b)(a? +b? )/24

Extended data:

In mathematics, the probability density function of a continuous random variable is a function that describes the possibility that the output value of this random variable is near a certain value point.

This refers to one-dimensional continuous random variables, and multi-dimensional continuous variables are similar.

Probability density function of random data: it represents the probability that the instantaneous amplitude falls within the specified range, so it is a function of amplitude. It varies with the range taken.

Baidu Encyclopedia-Density Function