A continuous random variable X has the p.d.f., f (x) = 3x^2; 0 ≤ x ≤ 1. The value of a constant λ that satisfies the relation Pr {X ≤ λ} = Pr {X > λ} is ? a) (1/3)^1/2 b) (1/2)^1/3 c) (2/3)^1/2 d) (2/3)^1/3

1 Answer
Dec 6, 2017

lamda = root(3)(1/2)

Or, Equivalently, lamda = (1/2)^(1/3) , making (b) the correct answer.

Explanation:

If f(x) is a probability density function then we require that:

int_(-oo)^(oo) \ f(x) \ dx = 1

We note that:

int_(0)^(1) \ 3x^2 \ dx = [x^3]_0^1 = 1

Hence, the complete probability density function is:

f(x) = { (3x^2, 0 le x le 1), (0, "otherwise") :}

So we seek the value of lamda such that:

P(X le lamda) = P(X gt lamda )

As X is a continuous random variable this is equivalent to:

P(X le lamda) = P(X ge lamda )
=> P(X le lamda) = 1 - P(X le lamda )
=> 2P(X le lamda) = 1
:. P(X le lamda) = 1/2

Thus we require that:

int_(0)^(lamda) \ 3x^2 \ dx = 1/2
=> [x^3]_(0)^(lamda) = 1/2
:. lamda^3 - 0 = 1/2
:. lamda^3 = 1/2

:. lamda = root(3)(1/2)

Or, Equivalently, lamda = (1/2)^(1/3) , making (b) the correct answer.