Answer
$$f\left( x \right){\text{ }}is{\text{ }}not{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2}}}{{16}};\,\,\,\,\,\,\,\left[ { - 2,2} \right] \cr
& {\text{The function f is a probability density function of a random variable X in the interval }} \cr
& \left[ {a,b} \right]{\text{ if}} \cr
& 1{\text{ condition}}:f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right].{\text{ then}} \cr
& \frac{{{x^2}}}{{16}}{\text{ is positive for all real number }}x,{\text{ then }}f\left( x \right) \geqslant 0{\text{ for the given interval}} \cr
& \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then}} \cr
& \int_{ - 2}^2 {\left( {\frac{{{x^2}}}{{16}}} \right)} dx \cr
& {\text{integrating}} \cr
& = \left( {\frac{{{x^3}}}{{16\left( 3 \right)}}} \right)_{ - 2}^2 \cr
& = \frac{1}{{48}}\left( {{{\left( 2 \right)}^3} - {{\left( { - 2} \right)}^3}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{{48}}\left( {8 + 8} \right) \cr
& = \frac{{16}}{{48}} \cr
& = \frac{1}{3} \cr
& \int_a^b {f\left( x \right)} dx \ne 1 \cr
& \cr
& {\text{
Condition 2 is not satisfied}}{\text{, so }} \cr
& f\left( x \right){\text{ }}is{\text{ }}not{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function \cr} $$
