The probability density function of evaporation E on any day during a year in a watershed is given by
$f\left(E\right)=\left\{\begin{array}{ll}\frac{1}{5}& 0\le E\le 5mm/day\\ 0& otherwise\end{array}\right.$
The probability that E lies in between 2 and 4 mm/day in a day in the watershed is (in decimal) _____________
If {x} is a continuous, real valued random variable defined over the interval (− ∞,+ ∞) and its occurrence is defined by the density function given as: $f\left(x\right)=\frac{1}{\sqrt{2\mathrm{\pi}}*b}{e}^{\frac{1}{2}{\left(\frac{x-a}{b}\right)}^{2}}$where 'a' and 'b' are the statistical attributes of the random variable {x}. The value of the integral ${\int}_{-\infty}^{a}\frac{1}{\sqrt{2\mathrm{\pi}}*b}{e}^{\frac{1}{2}{\left(\frac{x-a}{b}\right)}^{2}}dx$ is
Find the value of λ such that the function f(x) is a valid probability density function. __________
$f\left(x\right)=\lambda (x-1)(2-x)for1\le x\le 2\phantom{\rule{0ex}{0ex}}$
$=0otherwise$
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
The standard normal probability function can be approximated as
$\mathrm{F}\left({x}_{\mathrm{N}}\right)=\frac{1}{1+\mathrm{exp}\left(-1.7255x{}_{\mathrm{N}}{\left|{x}_{\mathrm{N}}\right|}^{0.12}\right)}$
Where x_{N} = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102 cm is
If probability density function of a random variable X is
f(x) = x^{2} for -1 ≤ x ≤ 1, and = 0 for any other value of x
Then, the percentage probability $P\left(-\frac{1}{3}\le x\le \frac{1}{3}\right)$ is