Notice that [tex]f(x)=0.1[/tex] for [tex]1200\le x\le1210[/tex] implies that [tex]f(x)=0[/tex] elsewhere, since
[tex]\displaystyle\int_{1200}^{1210}f(x)\,\mathrm dx=1[/tex]
where [tex]X=x[/tex] is a random variable representing cable lengths according to the PDF [tex]f(x)[/tex].
a. By definition of expectation, the mean is
[tex]E[X]=\displaystyle\int_{-\infty}^\infty x\,f(x)\,\mathrm dx=0.1\int_{1200}^{1210}x\,\mathrm dx=1205[/tex]
The variance is
[tex]\operatorname{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]
where
[tex]E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f(x)\,\mathrm dx=0.1\int_{1200}^{1210}x^2\,\mathrm dx=\frac{4,356,100}3[/tex]
so that the variance is [tex]\frac{25}3[/tex], making the standard deviation [tex]\frac5{\sqrt3}[/tex].
b. The proportion of cables within specs is
[tex]P(1195<X<1205)=\displaystyle\int_{1195}^{1205}f(x)\,\mathrm dx=0.1\int_{1200}^{1205}\mathrm dx=0.5[/tex]
