Answer:
(a) What is the mean waiting time? = 0.875
(b) What is the median waiting time? = 0.594
(c) Find P(T >2). = 0.102
(d) Find P(T< 0.1). = 0.108
(e) Find P(0.3 < T < 1.5). = 0.53
Step-by-step explanation:
a)
Let the particles are emitted at a mean rate of 3.5 per second, So, the mean waiting time is 0.875
b)
From the formula, the mean of the exponential distribution is
μ[tex]_{r}[/tex] = 1/λ = 0.875
λ = 1.143
Let m denote the median
Then P(T≤m) = 0.875
Using the formula for the cumulative distribution function of the exponential distribution
P(T≤m) = 0.875
1 - [tex]e^{-3.5m}[/tex] = 0.875
[tex]e^{-3.5m}[/tex] = 0.125
- 3.5m = Ln (0.125)
m = -2.079 / -3.5
m = 0.594
Solving for m, we obtain that the median is 0.594
c)
for P(T >2)
P(T >2) = 1 - P(T≤2)
= 1 - (1 - e^(-λ2))
= e^(-λ2)
= e^(-1.143x2)
= e^(-2.286)
= 0.102
d)
P(T<0.1) = 1 - e^(-λ*0.1)
= 1 - e^(-1.143*0.1)
= 1 - 0.892
= 0.108
e)
P(0.3 < T < 1.5)
P(T< 1.5) = 1 - e^(-λ(1.5))
= 1 - e^(-1.143(1.5))
= 1 - 0.18
= 0.82
P(T < 0.3) = 1 - e^(-λ(0.3))
= 1 - e^(-1.143(0.3))
= 1 - 0.71
= 0.29
Therefore, P(0.3 < T < 1.5) = P(T< 1.5) - P(T < 0.3)
= 0.82 - 0.29
= 0.53
