A continuous random variable X has a probability density function

$f (x) = e^{- 2 x}, 0 < x < \infty .$

Then P[X > 1] is

0.270

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.067

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.034

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.135

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 0.067
$P [X > 1] = \int_{1}^{\infty} f (x) d x = \int_{1}^{\infty} e^{- 2 x} d x = {\begin{matrix} \frac{- e^{- 2 x}}{2} \end{matrix} ∣ ∣ ∣}_{1}^{\infty}$
$= - (\frac{e^{- 2 \infty}}{2} - \frac{e^{- 2}}{2}) = \frac{e^{- 2}}{2} = 0.067$

Suggest Corrections

Similar questions

Q. A continuous random variable X has a probability density function

f (x) = e^{- x}, 0 < x < \infty .

Then P{X > 1} is

Q. If X is a continuous random variable whose probability density function is given by

f (x) = (\begin{matrix} k (5 x - 2 x^{2}), & 0 \leq x \leq 2 0, & otherwise \end{matrix}

Then P(x > 1) is

Q. A random variable

X

has probability density function as below:

f_{X} (x) = ⎧ ⎨ ⎩ \begin{matrix} 8 x; & 0 \leq x \leq \frac{1}{2} 0; & o t h e r w i s e \end{matrix}

If random variable

Y = 2 + 4 X^{2}

then the probability density function

f_{Y} (y)

y = 3

A random variable X has probability density function f(x) as give below:
$f (x) = (\begin{matrix} a + bx for 0 < x < 1 0 otherwise \end{matrix}$

If the expected value E[X] = 2/3, then Pr[X < 0.5] is .

Q. A normal random variable X has the following probability density function

f_{x} (x) = \frac{1}{\sqrt{8 π}} e^{- ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \frac{(x - 1)^{2}}{8} ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭}, - \infty < x < \infty

Then

\int_{1}^{\infty} f_{x} (x) d x ?

Join BYJU'S Learning Program

A continuous random variable X has a probability density function f(x)=e−2x,0<x<∞. Then P[X > 1] is

The correct option is B 0.067 P[X>1]=∫∞1f(x)dx=∫∞1e−2xdx=−e−2x2∣∣∣∞1 =−(e−2∞2−e−22)=e−22=0.067

A continuous random variable X has a probability density function

$f (x) = e^{- 2 x}, 0 < x < \infty .$

Then P[X > 1] is

The correct option is B 0.067
$P [X > 1] = \int_{1}^{\infty} f (x) d x = \int_{1}^{\infty} e^{- 2 x} d x = {\begin{matrix} \frac{- e^{- 2 x}}{2} \end{matrix} ∣ ∣ ∣}_{1}^{\infty}$
$= - (\frac{e^{- 2 \infty}}{2} - \frac{e^{- 2}}{2}) = \frac{e^{- 2}}{2} = 0.067$