Question

The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with mean 8 minutes.
(a) Find the probability that the time interval between two successive barges is less than 5 minutes.
(b) Find a time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t.

Answer 1

Answer:

a) 0.4647

b) 24.6 secs

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = 0.4647

Therefore the probability that the time interval between two successive barges is less than 5 minutes is 0.4647

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = 24.6 secs

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is 24.6 secs

Answer 2

a.) we can conclude that probability for the time interval between two successive barges is less than 5 minutes is 46.47%.

b.) the time interval 't' such that we can be 95% sure that the time interval between two successive barges will be greater than 't' is 24.6 secs.

Given to us:

exponential distribution with mean = E(T) = 8,

The mean of the exponential

E(T) = 1/λ

 1/λ = 8

λ = 1/8

∴ $t(t) =\frac{1}{8} \times e^\frac{-t}{8} }\ ;\ [ t > 0]$

(a) The probability that the time interval between two successive barges is less than 5 minutes.

Now the probability we need

$\begin{aligned} \ [T<5] &= \int\limits^5_0 {t(t)} \, dt\\&= \int\limits^5_0 { (\frac{1}{8} \times e^\frac{-t}{8} })dt\\&= \frac{1}{8} \int\limits^5_0 { (e^\frac{-t}{8} })dt\\&= \frac{1}{8} \int\limits^5_0 { (e^\frac{-t}{8} })dt\\\\&= [ \dfrac{(e^\frac{-t}{8})}{\frac{-1}{8}} ]_0^5\\&= - [ e^{-t/8}]_0^5\\&= - [ e^{-5/8} - 1 ]\\&= 1 - e^{-5/8}\\&= 0.4647\end{aligned}$

So, we can conclude that probability for the time interval between two successive barges is less than 5 minutes is 46.47%.

b)  Now we find t such that;

   

$\begin{aligned} \ &p[T>t]=0.95\\ & \int\limits^{10}_0 {t(t)} \, dt=0.95\\& \int\limits^{10}_0 { (\frac{1}{8} \times e^\frac{-t}{8} })dt=0.95\\& \frac{1}{8} \int\limits^{10}_0 { (e^\frac{-t}{8} })dt = 0.95\\& \frac{1}{8} \int\limits^{10}_0 { (e^\frac{-t}{8} })dt=0.95 \\\\& [ \dfrac{(e^\frac{-t}{8})}{\frac{-1}{8}} ]_0^{10}=0.95\\\\& - [ 0 - e^{-t/8} ] = 0.95&\ \ e^{-t/8} = 0.95\\\end{aligned}$

taking logs,

t = 24.6 secs

Hence,the time interval 't' such that we can be 95% sure that the time interval between two successive barges will be greater than 't' is 24.6 secs.

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