If the c.d.f (cumulative distribution function) is given by `F(x)=(x-25)/(10)`, then `P(27 le x le 33)=` . . .

Options

A. `3/5`
B. `3/10`
C. `1/5`
D. `1/10`

Answer

A

Explanation

We have,
c.d.f as F(x)=`(x-25)/10`
`thereforeP(27lexle33)=F(33)-F(27)`
`=((33-25)/10)-((27-25)/10)`
`=8/10-2/10=6/10=3/5`
Second method,
Probability density function f(x)=`d/(dx)(F(x))`
`=d/(dx)((x-25)/10)=1/10`
`thereforeP(27lexle33)=int_(27)^(33)f(x)dx`
`=int_(27)^(33)(1/(10))dx=1/10intdx`
`=1/(10)(33-27)`
`=6/10=3/5`