(a) No doubt your textbook tells you that for a uniform distribution on the interval [a, b]:
.. mean = (a+b)/2 = (2.5 +30)/2 = 16.25
.. variance = (b -a)^2/12 = 27.5^2/12 = 63 1/48
(b) Likewise, from your text
.. p(x) = (x-a)/(b-a) = (x -2.5)/27.5
.. cdf(x) = p(x)^2 = (x -2.5)^2/756.25 . . . cumulative distribution function
(c) cdf(15) ≈ 0.2066 . . . . . . . . . . . probability of being 15 or less
.. cdf(20) -cdf(15) ≈ 0.1983 . . . . . probability of being between 15 and 20
(d) cdf(μ+σ) -cdf(μ-σ) ≈ 0.5774 . . . probability of being withing 1σ of the mean
.. cdf(μ+2σ) -cdf(μ-2σ) = 1 . . . . . . . the entire distribution is within 2σ of the mean
.. mean = (a+b)/2 = (2.5 +30)/2 = 16.25
.. variance = (b -a)^2/12 = 27.5^2/12 = 63 1/48
(b) Likewise, from your text
.. p(x) = (x-a)/(b-a) = (x -2.5)/27.5
.. cdf(x) = p(x)^2 = (x -2.5)^2/756.25 . . . cumulative distribution function
(c) cdf(15) ≈ 0.2066 . . . . . . . . . . . probability of being 15 or less
.. cdf(20) -cdf(15) ≈ 0.1983 . . . . . probability of being between 15 and 20
(d) cdf(μ+σ) -cdf(μ-σ) ≈ 0.5774 . . . probability of being withing 1σ of the mean
.. cdf(μ+2σ) -cdf(μ-2σ) = 1 . . . . . . . the entire distribution is within 2σ of the mean