Question

A continuous random variable x that can assume value between x= 2 and x=5 has a
probability density function
p(x) = k( 1+x) . Determine p(x<4)

​

Answer 1

RONIT

...

..

..

.

.

.

.Answer

Answer 2

$\displaystyle \sf P(x < 4) = \frac{16}{27}$

Given :

A continuous random variable x that can assume value between x = 2 and x = 5 has a probability density function p(x) = k( 1+x)

To find :

The value of P(x < 4)

Solution :

Step 1 of 2 :

Find the value of k

Here it is given that the continuous random variable x that can assume value between x = 2 and x = 5 has a probability density function p(x) = k( 1+x)

By the given condition

$\displaystyle \sf \int\limits_{ - \infty}^{\infty} p(x) \, dx = 1$

$\displaystyle \sf \implies \int\limits_{2}^{5} k(1 + x) \, dx = 1$

$\displaystyle \sf \implies k \bigg[x + \frac{ {x}^{2} }{2} \bigg]_{2}^{5} = 1$

$\displaystyle \sf \implies k \bigg[5 + \frac{ {5}^{2} }{2} \bigg] - k \bigg[2 + \frac{ {2}^{2} }{2} \bigg] = 1$

$\displaystyle \sf \implies k \bigg[3 + \frac{ 21}{2} \bigg] = 1$

$\displaystyle \sf \implies k \bigg[ \frac{6 + 21}{2} \bigg] = 1$

$\displaystyle \sf \implies k \times \frac{27}{2} = 1$

$\displaystyle \sf \implies k = \frac{2}{27}$

Step 2 of 2 :

Find value of P(x < 4)

P(x < 4)

$\displaystyle \sf = \int\limits_{ - \infty}^{4} p(x) \, dx$

$\displaystyle \sf = \int\limits_{ 2}^{4} p(x) \, dx$

$\displaystyle \sf = \int\limits_{ 2}^{4} k(1 + x) \, dx$

$\displaystyle \sf = k \bigg[x + \frac{ {x}^{2} }{2} \bigg]_{2}^{ 4}$

$\displaystyle \sf = k \bigg[4 + \frac{ {4}^{2} }{2} \bigg] - k \bigg[2 + \frac{ {2}^{2} }{2} \bigg]$

$\displaystyle \sf = k \bigg[4 + 8 \bigg] - k \bigg[2 + 2 \bigg]$

$\displaystyle \sf = 12 k -4 k$

$\displaystyle \sf =8k$

$\displaystyle \sf =8 \times \frac{2}{27}$

$\displaystyle \sf = \frac{16}{27}$