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

Question

49369 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-06T19:27:02

# lamda = root(3)(1/2) #

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

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.