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If the c.d.f (cumulative distribution function) is given by `F(x)=(x-25)/(10)`, then `P(27 le x le 33)=` . . .A. `3/5`B. `3/10`C. `1/5`D. `1/10` |
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Answer» Correct Answer - A 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` |
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