EduFarming
Your list · 0हिं

Type at least two letters. Press Esc to close.

↑↓ navigate↵ openTAB cycle filter

Binomial & Poisson Distributions · Question 47 of 68

Question 47

EasyOriginal question

A shop gets an average of 3 complaints a day, following a Poisson distribution. What is the probability of no complaint on a given day? (Take e−3=0.0498e^{-3} = 0.0498.)

Options for question 47

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. P(X=r)=e−λλrr!P(X = r) = \frac{e^{-\lambda}\lambda^{r}}{r!} with λ=3\lambda = 3, r=0r = 0.

  2. =0.0498×300!=0.0498= \frac{0.0498 \times 3^{0}}{0!} = 0.0498.

Worth noting

0.9502 is the probability of at least one complaint.

Concept
The probability of no event
Typical time
60 s
Your time
—
The rule this testsThe Poisson distribution
Practise this chapter →

Ten questions from Binomial & Poisson Distributions, at your own pace.

Practise this chapter →

Ten questions from Binomial & Poisson Distributions, at your own pace.

Practise this chapterAll 68 questions in Binomial & Poisson DistributionsRevise the notes for Binomial & Poisson Distributions