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If a random variable waiting time in min...

If a random variable waiting time in minutes for bus and probability density function of x is given by
`f(x)={{:(1/5","0lexle5),(0",""otherwise"):}`
Then probability of waiting time not more than 4 minutes is equal to

A

0.3

B

0.8

C

0.2

D

0.5

Text Solution

Verified by Experts

The correct Answer is:
B
Given `f(x)={{:(1/5","0lexle5),(0",""otherwise"):}`
`thereforeP(0lexle4)=overset4underset0intf(x)dx=overset4underset0int1/5dx=1/5[x]_0^4`
`=1/5(4-0)=4/5=0.8`
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Knowledge Check

  • IF r.v X : waiting time in minutes for bus and p.d.f of X is given by f(x) ={{:((1)/(5),0 le x le 5),(0,"otherwise ,"):} then probabaility of waiting time not more than 4 minutes is

    A
    `0.3`
    B
    `0.8`
    C
    `0.2`
    D
    `0.5`

  • Suppose r.v. X = waiting time in minutes for a bus and its p.d.f. is given by f(x)={{:((1)/(5)","0lexle5),(0", otherwise"):} , then probability that waiting time is more than 4 minutes is

    A
    `(2)/(5)`
    B
    `(3)/(5)`
    C
    `(1)/(5)`
    D
    `(4)/(5)`

  • Suppose r.v. X = waiting time in minutes for a bus and its p.d.f. is given by f(x)={{:((1)/(5)","0lexle5),(0", otherwise"):} , then probability that waiting time is between 1 and 3 minutes is

    A
    `(4)/(5)`
    B
    `(2)/(5)`
    C
    `(3)/(5)`
    D
    `(1)/(5)`