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2 years ago

13

Network breakdowns are unexpected rare events that occur every 3 weeks, on the average. Compute the probability of more than 4 b

reakdowns during a 21-week period

1 answer:

allsm [11]2 years ago

5 0

Answer:

0.827

Step-by-step explanation:

Data provided in the question:

Probability of breakdown, p = once in 3 weeks i.e $\frac{1}{3}$

number of weeks n = 21

now,

mean, λ = np

= $\frac{1}{3}\times21$

= 7

P(X > 4) = 1 - ( P(X ≤ 4))

using Poisson distribution

P(X = x) = $\frac{e^{-\lambda}\lambda^x}{x!}$

Thus,

P(X = 0) = $\frac{e^{-7}7^0}{0!}$

= 0.00091

P(X = 1) = $\frac{e^{-7}7^1}{1!}$

= 0.00638

P(X = 2) = $\frac{e^{-7}7^2}{2!}$

= 0.02234

P(X = 3) = $\frac{e^{-7}7^3}{3!}$

= 0.05213

P(X = 4) = $\frac{e^{-7}7^4}{4!}$

= 0.09123

Thus,

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.00091 + 0.00638 + 0.02234 + 0.05213 + 0.09123

= 0.17299

Therefore,

P(X > 4) = 1 - 0.17299

= 0.827

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2 years ago

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2 years ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

(c) Parameterize $C$ by

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

8 months ago