Find the value of λ such that the function f (x) is a valid probability density function. _______ \(f\left( x \right)\begin{array}{*{20}{c}} { = \lambda \left( {x - 1} \right)\left( {2 - x} \right)}&{for1 \le x \le 2}\\ { = 0}&{otherwise} \end{array}\)
Concept:
We know for probability density function
\(\mathop \smallint \nolimits_{ - \infty }^\infty f\left( x \right)dx = 1\)
Calculation:
Given:
f(x) = λ (x – 1) × (2 – x) 1 ≤ x ≤ 2, 0 otherwise
\(\mathop \smallint \nolimits_1^2 \lambda \left( { - {x^2} + 3x - 2} \right)\;dx = 1\)
\(\Rightarrow \lambda \left[ {\dfrac{{ - {x^3}}}{3} + 3 \times \dfrac{{{x^2}}}{2} - 2x} \right]_1^2 = 1\)
\(\Rightarrow \lambda \left[ {\dfrac{{ - 14 + 27 - 12}}{6}} \right] = 1\)
λ = 6Important Questions from Continuous Distributions
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs 500 is _______
The number of parameters in the univariate exponential and Gaussian distributions, respectively are
- The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000 mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is
