There are n machines in a factory, each of them has defective rate of 0.01. Some maintainers are hired to help machines working. Using Poisson approximations to answer the following questions and explain why we can use Poisson approximation. (a) If one maintainer is responsible for 20 machines, what’s the probability that machines go defective and cannot be repaired on time?
(b) If three maintainers are responsible for 80 machines, what’s the probability that machines go defective and cannot be repaired on time?
(c) Assume n=300, then at least how many maintainers are required such that the probability machines cannot be fixed on time is less than 0.01?

a) 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!

b) 1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!

c) ∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³

Step-by-step explanation:  

Given that;

if n ⇒ ∞

p ⇒ 0

⇒ np = Constant = λ,  we can apply poisson approximation

⇒ Here 'p' is small ( p=0.01)

⇒ if (n=large) we can approximate it as prior distribution

⇒ let the number of defective items be d

so p(d) = ((e^-λ) × λ) / d!

NOW

a)

Let there be x number of repairs, So they will repair 20x machines on time. So if the number of defective machine is greater than 20x they can not repair it on time.

λ[n0.01]

p[ d > 20x ] = 1 - [ d ≤ 20x ]

= 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!

b)

Similarly in this case if number of machines d > 80x/3;

Then it can not be repaired in time

p[ d > 80x/3 ]

1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!

c)

n = 300, lets do it for first case i.e;

p [ d > 20x } ≤ 0.01

1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.01

⇒ ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.99

⇒ ∑²⁰ˣ_k=0 (λ^k)/k! = 0.99e^λ

∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³

-1, -2, -3, -4, -6

step-by-step explanation:

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: