Mathematics: A. X is a random variable denotes number of customers visiting a local coffee shop, which follows a Poisson distribution. The mean number of customers per 10 minutes is 6. a. What is the probability that there are 8 or less customers in the next 20 minutes? b. What is the probability that there are more than 4 customers in the next 10 minutes? B. X is a random variable denotes number of customers visiting a local coffee shop, which follows a Poisson distribution. The mean number of customers per

A. X is a random variable denotes number of customers

Ответ:

benallyadam2067

11.10.2021

Mathematics

(a) You're looking for

$P(X\le 8) = \displaystyle \sum_{x=0}^8 P(X=x)$

where

$P(X=x) = \begin{cases}\dfrac{\lambda^x e^{-\lambda}}{x!}&\text{if }x\in\{0,1,2,\ldots\}\\0&\text{otherwise}\end{cases}$

Customers arrive at a mean rate of 6 customers per 10 minutes, or equivalently 12 customers per 20 minutes, so

$\lambda = \dfrac{12\,\rm customers}{20\,\rm min}\times(20\,\mathrm{min}) = 12\,\mathrm{customers}$

Then

$\displaystyle P(X\le 8) = \sum_{x=0}^8 \frac{12^x e^{-12}}{x!} \approx \boxed{0.155}$

(b) Now you want

$P(X\ge4) = 1 - P(X$

This time, we have

$\lambda = \dfrac{6\,\rm customers}{10\,\rm min}\times(10\,\mathrm{min}) = 6\,\mathrm{customers}$

so that

$P(X\ge4) = 1 - \displaystyle \sum_{x=0}^3 \frac{6^x e^{-6}}{x!} \approx \boxed{0.849}$

(a) In other words, you're asked to find the probability that more than 1 customer shows up in the same minute, or

$P(X 1) = 1 - P(X \le 1) = 1 - P(X=0) - P(X=1)$

with

$\lambda = \dfrac{6\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 1\,\mathrm{customer}$

So we have

$P(X 1) = 1 - \dfrac{1^0 e^{-1}}{0!} - \dfrac{1^1 e^{-1}}{1!} \approx \boxed{0.264}$

(a) Similar to B, you're looking for

$P(X \le 1) = P(X=0) + P(X=1)$

with

$\lambda = \dfrac{12\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 2\,\mathrm{customers}$

so that

$P(X\le1) = \dfrac{2^0e^{-2}}{0!} + \dfrac{2^1e^{-2}}{1!} \approx \boxed{0.406}$

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Ответ:

anisagreen10

09.02.2020

Mathematics

145°

Step-by-step explanation:

As per the given information: quadrilateral ABCD is a parallelogram.

$m\angle B + m\angle DCB =180\degree \\$

(adjacent angles of a parallelogram)

$110\degree + m\angle DCB =180\degree \\ m\angle DCB =180\degree - 110\degree \\ m\angle DCB =70\degree\\m\angle DCG=\frac{1}{2} m\angle DCB \\(\because GC \: bisects \: \angle DCB) \\m\angle DCF=\frac{1}{2}\times 70\degree... (\because C-G-F) \\m\angle DCF=35\degree\\m\angle BFC = m\angle DCF = 35\degree\\ (Alternate \: \angle s) \\\\m\angle BFC + m\angle CFA = 180\degree\\ (straight \: line \: \angle s) \\\\35\degree + m\angle CFA= 180\degree\\m\angle CFA = 180\degree-35\degree \\\\m\angle CFA= 145\degree \\\\\huge\purple {\therefore m\angle GFA = 145\degree} \\(\because C-G-F) \\$

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