I've only just approved your question. But I can begin to answer it.
The PDF has an area of 1 because the sum total of all probabilities must be equal to 1. This enables us to find k. ∫f(x)dx=1 between the limits 1 and 4, where f(x)=k(x-1)(4-x)=k(5x-x^2-4).
So integrating between the limits we have [5kx^2/2-kx^3/3-4kx]{1,4}=
((5/2)(16-1)-(64-1)/3-4(4-1))k=1
(37.5-21-12)k=1
4.5k=1 so k=1/4.5=2/9 (answer 18C) and f(x)=(2/9)(x-1)(4-x).
The mean is given by ∫xf(x)dx{1,4}=∫k(5x^2-x^3-4x)dx between the limits=
(2/9)[5x^3/3-x^4/4-2x^2]{1,4}=(2/9)(105-63.75-30)=
(2/9)(45/4)=2.5, which is to be expected since the PDF is symmetrical about x=2.5. Mean=2.5 (answer 19C)
We need the mean to calculate the variance, the square root of which is the standard deviation.
VAR=∫(x-2.5)^2(f(x))dx{1,4}
The answer choices for 20 are the same as for 19. This seems suspicious. So the SD may not be listed. I calculate it to be 0.6368. The computation for the variance can be simplified:
VAR=∫x^2f(x)dx-5∫xf(x)dx+6.25∫f(x)dx between the limits=∫x^2f(x)dx-5*2.5+6.25=∫x^2f(x)dx-6.25, because we have already computed the integral for the mean and we know ∫f(x)dx=1.
VAR=(2/9)∫(5x^3-x^4-4x^2)dx{1,4}-6.25=(2/9)[5x^4/4-x^5/5-4x^3/3]{1,4}-6.25=
(2/9)(599/20)-6.25=599/90-6.25=73/180=0.405556 approx, making SD=√0.405556=0.6368 approx. This is not a listed answer choice for 20.