A manager at a company that manufactures cell phones has noticed that the number of faulty cell phones in a production run of cell phones is usually small and that the quality of one day's run seems to have no bearing on the next day.
a) What model might you use to model the number of faulty cell phones produced in one day?
b) If the mean number of faulty cell phones is 1.8 per day, what is the probability that no faulty cell phones will be produced tomorrow?
c) If the mean number of faulty cell phones is 1.8 per day, what is the probability that 3 or more faulty cell phones were produced in today's run?

a) Poisson probability distribution

b) The probability of no faulty cell phones will be produced tomorrow is 0.1653

c) The probability of 3 or more faulty cell phones were produced in today's run is 0.2694

Step-by-step explanation:

Poisson distribution is used for independent events which occur at a constant rate within a given interval of time

The Poisson probability distribution formula is P(x; μ) = (e^-μ) (μ^x) / x! where

x is the actual number of successes that result from the experiment

e is approximately equal to 2.71828

μ  is the mean of the distribution

The number of faulty cell phones in a production run of cell phones is usually small and that the quality of one day's run seems to have no bearing on the next day

That means the probability of finding faulty phones in first day not depend on finding another days

Then the model might you use to model the number of faulty cell phones produced in one day is Poisson probability distribution

a) Poisson probability distribution

∵ The mean number of faulty cell phones is 1.8 per day

∴ μ = 1.8

∵ There is no faulty cell phones will be produced tomorrow

∴ x = 0

- Use the formula above to find the probability

∵ P(0 ; 1.8) = (e^-1.8) (1.8^0) / 0!

- Remember 1.8^0 = 1 and 0! = 1

∴ P(0 ; 1.8) = (e^-1.8)(1)/(1)

∴ P(0 ; 1.8) = 0.1653

b) The probability of no faulty cell phones will be produced tomorrow is 0.1653

∵ The mean number of faulty cell phones is 1.8 per day

∴ μ = 1.8

∵ There is 3 or more faulty cell phones were produced in today's run

∴ x ≥ 3

∵ P(x ≥ 3) = 1 - P(x = 0) - P(x = 1) - P(x = 2)

- Let us find P(1 ; 1.8) and P(2 ; 1.8)

∵ P(1 ; 1.8) = (e^-1.8) (1.8^1) / 1!

∵ 1.8^1 = 1.8

∵ 1! = 1

∴ P(1 ; 1.8) = (e^-1.8)(1.8)/(1)

∴ P(1 ; 1.8) = 0.2975

∵ P(2 ; 1.8) = (e^-1.8) (1.8^2) / 2!

∵ 1.8^2 = 3.24

∵ 2! = 2

∴ P(2 ; 1.8) = (e^-1.8)(3.24)/(2)

∴ P(2 ; 1.8) = 0.2678

Substitute them in the rule above

∵ P(x ≥ 3 ; 1.8) = 1 - 0.1653 - 0.2975 - 0.2678

∴ P(x ≥ 3 ; 1.8) = 0.2694

c) The probability of 3 or more faulty cell phones were produced in today's run is 0.2694

a. the residual plot for model 1 has a random pattern and is a good fit for the data.

the diameter is 16 cm. are you a teacher?

step-by-step explanation: