Question of probability?

[匿名]

2016-09-01 5:25 am

Can help me with this exercise?
The time in minutes that a woman talks on the phone is a random variable with the following probability density function:
f(x)=
{k. e^(-x/5) if 0 < x
{0 in another case

a)Find k such that f(x) is a density function.
b)Calculate f(x).
c)Find the probability that she talks more than 2 minutes.
d)Find the probability that she talks, at most, 3 minutes.
e)If she talks, at least, 2 minutes. Find the probability that she talks, at most, 3 minutes.
f)Find the average time of the lady, talking. Calculate the mean and variance.

Thanks!

回答 (1)

Lopez

2016-09-02 2:14 pm

✔ 最佳答案

(a)
∫ ke^(-x/5) dx , from x = 0 to x = ∞
= [ - 5ke^(-x/5) ] , from x = 0 to x = ∞
= 5k
= 1
Thus, k = 1/5 ..... Ans

(b)
Ans :
f(x) =
{ (1/5)e^(-x/5) , if x > 0
{ 0 , in another case

(c)
P( x > 2 )
= ∫ (1/5)e^(-x/5) dx , from x = 2 to x = ∞
= [ - e^(-x/5) ] , from x = 2 to x = ∞
= e^(-0.4) ..... Ans
( ≒ 0.670 )

(d)
P( x ≦ 3 )
= ∫ (1/5)e^(-x/5) dx , from x = 0 to x = 3
= [ - e^(-x/5) ] , from x = 0 to x = 3
= 1 - e^(-0.6) ..... Ans
( ≒ 0.451 )

(e)
P( 2 ≦ x ≦ 3 )
= ∫ (1/5)e^(-x/5) dx , from x = 2 to x = 3
= [ - e^(-x/5) ] , from x = 2 to x = 3
= e^(-0.4) - e^(-0.6)

P( x ≦ 3 ｜x ≧ 2 )
= P( x ≦ 3 and x ≧ 2 ) / P( x ≧ 2 )
= P( 2 ≦ x ≦ 3 ) / P( x ≧ 2 )
= [ e^(-0.4) - e^(-0.6) ] / e^(-0.4) ..... Ans
( ≒ 0.181 )

(f)
f(x) =
{ (1/β)e^(-x/β) , if x > 0
{ 0 , in another case
is called the exponential distribution with parameter β .
E(X) = β , V(Y) = β^2

For this case, β = 5
average time = mean = 5 ..... Ans
variance = 25 ..... Ans

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