What is the area of the following circle

One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst

grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let X = number of telephone calls.

The average number of calls per minute is, λ = 3.0.

The random variable X follows a Poisson distribution with parameter λ = 3.0.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...$

(a)

Compute the probability of X = 0 as follows:

$P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498$

Thus, the probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of X > 5 as follows:

P (X > 5) = 1 - P (X ≤ 5)

$=1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084$

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of X = 3 as follows:

$P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09$

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of X ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

$=\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578$

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0

3 years ago

Iliana was part of a group that was working on changing 0.4 repeated to a fraction. Each member of the group had a different ans

amm1812

$\bf 0.444444444\overline{4}\impliedby \textit{and keeps on going}\\\\
-------------------------------\\\\
\textit{let's say }\boxed{x=0.444444444\overline{4}}\quad \textit{ thus }10\cdot x=4.44444444\overline{4}
\\\\\\
\textit{wait a minute! }4.44444444\overline{4}\textit{ is really just }4+0.444444444\overline{4}$

$\bf \textit{but we know }x=0.444444444\overline{4} \textit{ so then }4+0.444444444\overline{4}=\boxed{4+x}
\\\\\\
\textit{wait a second! }10\cdot x\implies 10x=4.444444444\overline{4}=4+x
\\\\\\
thus\qquad 10x=4+x\implies 10x-x=4\implies 9x=4\implies \boxed{x=\cfrac{4}{9}}$

you can check in your calculator.

anyhow, to get the "recurring decimal to fraction", you start by setting to some variable, "x" in this case, then move the repeating part to the left of the point by multiplying it by some power of 10, and then do the equating.

3 0

3 years ago

Read 2 more answers

What is the value of y in the equation 4y – 2(1 – y) = –44?

LekaFEV [45]

Y=-7 use distributive property add like terms

8 0

3 years ago

Please help me idk this

Natali [406]

Answer:9000

Step-by-step explanation: